# Notes on Commutative Algebra 3.

This is a “short” study note of Atiyah-MacDonald.

In this post, we start focusing on studying the topological space $\text{Spec}(A)$. In particular, we try to see what it means for an ideal to be “radical.”

Again, recall that we exclude the possibility of the zero ring (mainly because $\text{Spec}(A)$ will be empty if we allow $A$ to be zero).

We say an ideal $\mathfrak{a} \subseteq A$ is a radical ideal if $\mathfrak{a = \sqrt{a}}$. We also say $\mathfrak{a}$ is a radical to mean this for the following reason. This is not in Atiyah-MacDonald, so we need to be careful for its proof.

A radical ideal is exactly an ideal that is radical of some ideal. Let $\mathfrak{a} \subseteq A$ be an ideal. Then

$\mathfrak{a}$ is a radical ideal if and only if $\mathfrak{a = \sqrt{b}}$ for some ideal $\mathfrak{b} \subseteq A$.

Proof. Suppose $\mathfrak{a = \sqrt{b}}$ for some ideal $\mathfrak{b} \subseteq A$. Then $\mathfrak{a = \sqrt{b} = \sqrt{\sqrt{b}} = \sqrt{a}}$. Thus $\mathfrak{a}$ is a radical ideal.

The converse is trivial. Q.E.D.

An ideal is radical exactly when it is an intersection of prime ideals (Ex 1.9 in AM). Let $\mathfrak{a} \subseteq A$ be an ideal. Then

$\mathfrak{a}$ is radical if and only if $\mathfrak{a}$ is an intersection of prime ideals.

Proof. Suppose that $\mathfrak{a} = \cap_{\alpha \in I}\mathfrak{p}_{\alpha}$ for a collection of prime ideals $\{\mathfrak{p}_{\alpha}\}_{\alpha \in I}$ in $A$. If $I = \empty$, our definition of intersection implies $\mathfrak{a} = (1)$, which is radical. Suppose that $I$ is nonempty.

Then for any $\alpha \in I$, we have $\mathfrak{p}_{\alpha} \supseteq \mathfrak{a}$, which means $\mathfrak{p}_{\alpha} \in V(\mathfrak{a})$. Thus

$\mathfrak{a} = \cap_{\alpha \in I} \mathfrak{p}_{\alpha} \supseteq \cap_{\mathfrak{p} \in V(\mathfrak{p})}\mathfrak{p} = \mathfrak{\sqrt{a}}$,

which implies $\mathfrak{a = \sqrt{a}}$.

The converse is true since $\mathfrak{\sqrt{a}} = \cap_{\mathfrak{p} \in V(\mathfrak{a})}\mathfrak{p}$. Q.E.D.

For any ideals $\mathfrak{a, b} \subseteq A$, we know that

$\mathfrak{a} \supseteq \mathfrak{b}$ implies $V(\mathfrak{a}) \subseteq V(\mathfrak{b})$.

When is the converse of above true? Let’s make up some name for the converse so that we can study it whenever we see it. For any sets $X, Y$ and a function $f$ from the class of sets to itself, we say $X$ can devour $Y$ via $f$ to mean:

$f(X) \subseteq f(Y)$ implies $X \supseteq Y$.

The two following lemmas do not appear in Atiyah-MacDonald, so we must be careful despite their simplicity.

A radical ideal can devour any ideals via Zariski-closed set operator. Let $\mathfrak{a} \subseteq A$ be radical. Then

$V(\mathfrak{a}) \subseteq V(\mathfrak{b})$ implies $\mathfrak{a \supseteq b}$,

for any ideal $\mathfrak{b} \subseteq A$.

Proof. We have $\mathfrak{a = \sqrt{a}} = \cap_{\mathfrak{p} \in V(\mathfrak{a})}\mathfrak{p} \supseteq \cap_{\mathfrak{p}\in V(\mathfrak{b})}\mathfrak{p} = \mathfrak{\sqrt{b} \supseteq b}$. Q.E.D.

A trivial corollary: for any ideals $\mathfrak{a, b} \subseteq A$,

$\mathfrak{\sqrt{a} \supseteq b}$ if and only if $V(\sqrt{\mathfrak{a}}) \subseteq V(\mathfrak{b})$.

Another corollary is as follows.

A radical operator is increasing with inclusion. For two ideals $\mathfrak{a, b} \subseteq A$, if $\mathfrak{a \subseteq b}$, then $\mathfrak{\sqrt{a} \subseteq \sqrt{b}}$.

Proof. If $\mathfrak{a \subseteq b}$, then $V(\mathfrak{\sqrt{a}}) = V(\mathfrak{a}) \supseteq V(\mathfrak{b}) = V(\mathfrak{\sqrt{b}})$. Thus we have $\mathfrak{\sqrt{a} \subseteq \sqrt{b}}$, by devouring property. Q.E.D.

This “devouring property via Zariski-closed set operator” actually characterizes the definition of a radical ideal.

A radical ideal is precisely an ideal with “devouring property” via Zariski-closed set operator. For any ideal $\mathfrak{a}$, the following are equivalent.

• $\mathfrak{a}$ is radical.
• $V(\mathfrak{a}) \subseteq V(\mathfrak{b})$ implies $\mathfrak{a \supseteq b}$ for any ideal $\mathfrak{b} \subseteq A$.

Proof. Suppose that the second property holds. Then $V(\mathfrak{a}) = V(\sqrt{\mathfrak{a}})$, so $\mathfrak{a \supseteq \sqrt{a}}$. Since $\mathfrak{a \subseteq \sqrt{a}}$, we have shown $\mathfrak{a = \sqrt{a}}$.

We showed the converse previously. Q.E.D.

The following does not appear on Atiyah-MacDonald but I had it on my algebra final exam, so I think we don’t need to be too worried. For a topological space $X$, we denote $\overline{X}$ the closure of $X$. Recall the ideal quotient: $(\mathfrak{a : b}) := \{x \in A : x\mathfrak{b} \subseteq \mathfrak{a}\}$, which is an ideal.

Closure of difference of Zariski-closed sets is “almost” Zariski-closed set of their ideal quotient. Let $\mathfrak{a}, \mathfrak{b} \subseteq A$ be ideals.

1. $\overline{V(\mathfrak{a}) \setminus V(\mathfrak{b})} \subseteq V(\mathfrak{a : b})$.
2. If $\mathfrak{a}$ is radical, then $\overline{V(\mathfrak{a}) \setminus V(\mathfrak{b})} = V(\mathfrak{a : b})$.

Proof. Let $\mathfrak{p} \in V(\mathfrak{a}) \setminus V(\mathfrak{b})$. Then $\mathfrak{p \supseteq a}$ and $\mathfrak{p \not\supseteq b}$, so we can take $y \in \mathfrak{b}$ such that $y \notin \mathfrak{p}$.

For any $x \in (\mathfrak{a : b})$, we have $x\mathfrak{b} \subseteq \mathfrak{a}$. In particular, we have $xy \in \mathfrak{a \subseteq p}$. But by primality, we have $x \in \mathfrak{p}$. This shows that $\mathfrak{(a : b) \subseteq p}$, so $\mathfrak{p} \in V(\mathfrak{a : b})$.

We have shown $V(\mathfrak{a}) \setminus V(\mathfrak{b}) \subseteq V(\mathfrak{a : b})$. Since the right-hand side is closed, we have $\overline{V(\mathfrak{a}) \setminus V(\mathfrak{b})} \subseteq V(\mathfrak{a : b})$, proving 1.

Now assume that $\mathfrak{a}$ is radical. Take any ideal $\mathfrak{s}$ such that $V(\mathfrak{s}) = \overline{V(\mathfrak{a}) \setminus V(\mathfrak{b})}$. Then $V(\mathfrak{sb}) = V(\mathfrak{s}) \cup V(\mathfrak{b}) = \overline{V(\mathfrak{a}) \setminus V(\mathfrak{b})} \cup V(\mathfrak{b}) \supseteq [V(\mathfrak{a}) \setminus V(\mathfrak{b})] \cup V(\mathfrak{b}) \supseteq V(\mathfrak{a})$. But radicality implies that $\mathfrak{a} \supseteq \mathfrak{sb}$, so $\mathfrak{s \subseteq (a : b)}$. This implies that $V(\mathfrak{a : b}) \subseteq V(\mathfrak{s}) = \overline{V(\mathfrak{a}) \setminus V(\mathfrak{b})}$. Knowing that the reverse inclusion is true by 1, we close the proof. Q.E.D.

The following did not appear on my algebra final, so we need to be careful, although the proof is simple. That is, what we have just proved is in fact another way to characterize radical ideals.

An ideal is radical precisely when closure of Zariski-closed set of it and that of any ideal is Zariski-closed set of their quotient. Let $\mathfrak{a} \subseteq A$ be an ideal. Then the following are equivalent.

• $\mathfrak{a}$ is radical.
• $V(\mathfrak{a : b}) = \overline{V(\mathfrak{a}) \setminus V(\mathfrak{b})}$ for any ideal $\mathfrak{b} \subseteq A$.

Proof. Assume the second condition. Since $V(\mathfrak{a}) = V(\mathfrak{\sqrt{a}})$, we have $V(\mathfrak{a}) \setminus V(\sqrt{\mathfrak{a}}) = \emptyset$, and its closure is again the empty set. Thus our hypothesis implies

$V(\mathfrak{a : \sqrt{a}}) \subseteq \overline{V(\mathfrak{a}) \setminus V(\mathfrak{\sqrt{a}})} = \emptyset$.

This implies that $(\mathfrak{a : \sqrt{a}}) = (1)$ (thanks to Zorn’s lemma), so $\mathfrak{\sqrt{a} = 1 \cdot \sqrt{a} \subseteq a}$, showing $\mathfrak{\sqrt{a} = a}$ (i.e., $\mathfrak{a}$ is radical).

The converse was shown before. Q.E.D.

Summary: what do we mean by ideal being “radical”? Let $\mathfrak{a} \subseteq A$ be an ideal. We have seen that the following are equivalent.

• $\mathfrak{a = \sqrt{a}}$.
• $\mathfrak{a = \sqrt{b}}$ for some ideal $\mathfrak{b} \subseteq A$.
• $\mathfrak{a}$ is an intersection of prime ideals of $A$.
• $V(\mathfrak{a}) \subseteq V(\mathfrak{b})$ implies $\mathfrak{a \supseteq b}$.
• $\overline{V(\mathfrak{a}) \setminus V(\mathfrak{b})} = V(\mathfrak{a : b})$.

# Notes on Commutative Algebra 2.

This is a “short” study note of Atiyah-MacDonald.

A trivial remark about a connected subspace. Let $X$ be a topological space that is not connected. Then $X = U \amalg V$ for some nonempty open sets $U, V \subseteq X$. Now suppose that $Y \subseteq X$ is a connected subspace. Then $Y \subseteq U$ or $Y \subseteq V$. That is, a connected subspace “cannot stay in the middle of a separation.”

If we have a set $A$ and a collection of sets $\{A_{i}\}_{i \in I}$, we say $A$ cannot stay in the middle of $\{A_{i}\}_{i=1}^{n}$ to mean the following:

if $A \subseteq \cup_{i} A_{i}$, then $A \subseteq A_{i}$ for some $i \in I$.

An ideal cannot stay in the middle of finitely many prime ideals. Given a ring $A$, let $\mathfrak{p}_{1}, \cdots, \mathfrak{p}_{n} \in \text{Spec}(A)$. For any ideal $\mathfrak{a}$, if $\mathfrak{a} \subseteq \cup_{i = 1}^{n}\mathfrak{p}_{i}$, then $\mathfrak{a} \subseteq \mathfrak{p}_{i}$ for some $i$.

Proof. Suppose that $\mathfrak{a} \not\subseteq \mathfrak{p}_{i}$ for all $i$. Then for each $i$, we can pick $x_{i} \in \mathfrak{a}$ such that $x_{i} \notin \mathfrak{p}_{i}$. We also have $x_{i} \in \mathfrak{a} \subseteq \cup_{i=1}^{n}\mathfrak{p}_{i}$ for all $i$, so far we have

• $x_{i} \notin \mathfrak{p}_{i}$ for all $i$;
• $x_{i} \in \mathfrak{p}_{1} \cup \cdots \cup \mathfrak{p}_{n}$ for all $i$.

We suspect the above phenomenon is the core of a contradiction since it is not true when $n = 1$. Thus we study the phenomenon for $n \geq 2$. But if $n = 2$, then $x_{2} \in \mathfrak{p}_{1}$ and $x_{1} \in \mathfrak{p}_{1}$, but $x_{1} + x_{2} \in \mathfrak{a} \subseteq \mathfrak{p}_{1} \cup \mathfrak{p}_{2}$. Thus $x_{1} + x_{2} \in \mathfrak{p}_{1}$ or $x_{1} + x_{2} \in \mathfrak{p}_{2}$, which leads to $x_{1} \in \mathfrak{p}_{1}$ or $x_{2} \in \mathfrak{p}_{2}$, a contradiction.

Thus, there must be $n \geq 3$ that makes the two statements above be true. Because of the second condition, we have a map $\{1, 2, \cdots, n\} \overset{f}{\longrightarrow} \{1, 2, \cdots, n\}$ such that $x_{i} \in \mathfrak{p}_{f(i)}$. We study this map $f$ as follows.

Given $m \geq 2$, if we have $f^{m}(k) = k$ for some $1 \leq k \leq n$. Then $x_{k} = x_{f^{m}(k)} \in \mathfrak{p}_{f^{m-1}(k)}$, so $x_{f(k)} \in \mathfrak{p}_{f^{m}(k)} = \mathfrak{p}_{k}$. Thus

$x_{k} \in \mathfrak{p}_{f(k)}$ and $x_{f(k)} \in \mathfrak{p}_{k}$,

and this is a contradiction since it is the case $n = 2$. Thus for arbitrarily large $m \geq 2$, we have $f^{m}(k) \neq k$ for all $1 \leq k \leq n$. This implies that the following are distinct for any $1 \leq k \leq n$:

$k, f(k), \cdots, f^{m}(k)$.

This must be a contradiction since we can take $m > n$, while everything above belongs to $\{1, 2, \cdots, n\}$. Q.E.D.

We studied a property of an ideal sitting in a finite union of prime ideals. Next, we study a property of a prime ideal containing a finite intersection of ideals.

We say a set $A$ can cut a member from the intersection of a collection of sets $\{A_{i}\}_{i \in I}$ to mean the following:

$A \supseteq \cap_{i \in I}A_{i}$ implies $A \supseteq A_{j}$ for some $j \in I$.

A prime ideal can cut a member from the intersection of a finite collection of ideals. Let $\mathfrak{a}_{1}, \cdots, \mathfrak{a}_{n} \subseteq A$ be ideals and $\mathfrak{p} \in \text{Spec}(A)$. Then

$\mathfrak{p} \supseteq \cap_{i=1}^{n}\mathfrak{a}_{i}$ implies $\mathfrak{p} \supseteq \mathfrak{a}_{j}$ for some $j$.

Proof. Suppose that $\mathfrak{p} \not\supseteq \mathfrak{a}_{i}$ for all $i$. Then for each $i$, we can pick $x_{i} \in \mathfrak{a}_{i}$ such that $x_{i} \notin \mathfrak{p}$. But then $x_{1} \cdots x_{n} \in \cap_{i=1}^{n} \mathfrak{a}_{i} \subseteq \mathfrak{p}$. This is a contradiction since $x_{j} \in \mathfrak{p}$ for some $j$ by primality. Q.E.D.

As a corollary with recycled notation, notice the following:

$\mathfrak{p} = \cap_{i=1}^{n}\mathfrak{a}_{i}$ implies $\mathfrak{p} = \mathfrak{a}_{j}$ for some $j$.

Given ideals $\mathfrak{a, b} \subseteq A$, the ideal quotient of $\mathfrak{a, b}$ is defined by

$(\mathfrak{a : b}) = \{x \in A : x\mathfrak{b} \subseteq \mathfrak{a}\}$.

A direct computation shows that the ideal quotient of any ideals is an ideal. In particular, we see $\text{Ann}(\mathfrak{a}) = (0 : \mathfrak{a})$ is an ideal.

Let’s try to imbue some meanings to the definition. The ideal quotient $(\mathfrak{a} : \mathfrak{b})$ are the elements of $A$ that makes $\mathfrak{b}$ smaller than $\mathfrak{a}$ by an action.

We conjecture that $(\mathfrak{a} : \mathfrak{b})$ is an ideal version of fraction $a/b$ of positive numbers $a/b$ with the following characteristics.

1. $a$, the numerator, is large enough.
2. $b$, the denominator, is extremely small.

If our analogy is correct, the right-hand sides of the following are reasonable conjectures that corroborate the analogy.

• $a \leq a/b$ analogous to $\mathfrak{a} \subseteq (\mathfrak{a :b})$.
• $(a/b)b \leq a$ analogous to $(\mathfrak{a : b})\mathfrak{b} \subseteq \mathfrak{a}$.
• $a/b/c = a/(bc) = a/c/b$ analogous to $\mathfrak{((a: b):c) = (a:bc) = (a:(b:c))}$.
• $a/c + b/c \leq (a + b)/c$ analogous to $\mathfrak{(a:c) + (b:c)} \subseteq \mathfrak{(a + b : c)}$.

Fortunately, the first three right-hand sides above are all true and can be checked by not complicated computations. However, the fourth one does not appear in Atiyah-MacDonald, so we carefully prove it here.

Proof of the fourth property. Let $x + y \in \mathfrak{(a : c) + (b : c)}$ such that $x\mathfrak{c \subseteq \mathfrak{a}}$ and $y\mathfrak{c \subseteq \mathfrak{b}}$. Then $(x + y)\mathfrak{c} \subseteq \mathfrak{a + b}$, showing the desired inclusion. Q.E.D.

Unlike “traditional numbers,” ideals can be operated with the intersection operator. It is difficult to make conjectures without computation, but nevertheless, we inflate our analogy dictionary as follows.

• $(a_{1} \cap \cdots \cap a_{n})/b = a_{1}/b \cap \cdots \cap a_{n}/b$ analogous to $(\cap_{i=1}^{n}\mathfrak{a}_{i} : \mathfrak{b}) = \cap_{i=1}^{n}(\mathfrak{a}_{i} : \mathfrak{b})$.
• $a/(b_{1} + \cdots + b_{n}) = a/b_{1} \cap \cdots \cap a/b_{n}$ analogous to $(\mathfrak{a} : \sum_{i=1}^{n} \mathfrak{b}_{i}) = \cap_{i=1}^{n}(\mathfrak{a} : \mathfrak{b}_{i})$.

Again, both results are true and can be proved by not complicated computations. Notice that both do not violate our intuitive conjectures too much. First consider

$(a_{1} \cap \cdots \cap a_{n})/b = a_{1}/b \cap \cdots \cap a_{n}/b$.

The numerator of left-hand side gets smaller as we diversify $a_{i}$, but according to the conjecture, it cannot get too small, because the numerator is supposed to have some large enough quantity. Intuitively, the left-hand side must have a “moderate” size.

The right-hand side also gets smaller as we diversify $\mathfrak{a}_{i}$, but according to the conjecture, it cannot get too small, since each $\mathfrak{a}_{i}/b$ is huge as the denominator $b$ was supposed to be small. Again intuitively, the right-hand side must have a “moderate” size.

Now consider

$a/(b_{1} + \cdots + b_{n}) = a/b_{1} \cap \cdots \cap a/b_{n}$.

The denominator of the left-hand side gets larger as we diversify $b_{i}$, but it cannot get large at all by adding finitely many more, because we assumed that the denominator is extremely small. Thus the fraction cannot get small at all, so it must be “relatively large.”

Just as its counter part, the right-hand side gets smaller as we diversify $b_{i}$, but again, it cannot get small at all because each $a/b_{i}$ is huge. Thus the right-hand side is also “relatively large.”

For usual numbers, there are square root, cubic root, and so on, but for ideals, all $n$th roots for $n \geq 1$ are lumped together as “radical.”

For an ideal $\mathfrak{a} \subseteq A$, its radical is defined by

$\sqrt{\mathfrak{a}} := \{x \in A : x^{n} \in \mathfrak{a} \text{ for some } n \geq 1\}$.

We can immediately see that the name ‘nilradical’ makes a perfect sense:

$\mathfrak{N} = \sqrt{0}$.

Any radical is an ideal since it is the preimage of the nilradical $\mathfrak{N}_{A/\mathfrak{a}}$ under the projection $A \twoheadrightarrow A/\mathfrak{a}$.

Given an ideal $\mathfrak{a} \subseteq A$, define

$V(\mathfrak{a}) := \{\mathfrak{p} \in \text{Spec}(A) : \mathfrak{p \supseteq a}\}$.

If we write $\phi : A \twoheadrightarrow A/\mathfrak{a}$ for the projection, then

$V(\mathfrak{a}) = \{\phi^{-1}(\mathfrak{q}) : \mathfrak{q} \in \text{Spec}(A/\mathfrak{a})\}$.

We call $V(\mathfrak{a})$ the Zariski-closed set of $\mathfrak{a}$.

Recall that any preimage operator commutes with any intersection operator.

The radical of an ideal is the intersection of its Zariski-closed set. Given an ideal $\mathfrak{a} \subseteq A$, we have

$\sqrt{\mathfrak{a}} = \cap_{\mathfrak{p} \in V(\mathfrak{a})}\mathfrak{p}$.

Proof. Denote $\phi : A \twoheadrightarrow A/\mathfrak{a}$ for the projection. Then

$\sqrt{\mathfrak{a}} = \phi^{-1}(\mathfrak{N}_{A/\mathfrak{a}}) = \phi^{-1}(\cap_{\mathfrak{q} \in \text{Spec}(A/\mathfrak{a})}\mathfrak{q}) = \cap_{\mathfrak{q} \in \text{Spec}(A/\mathfrak{a})}\phi^{-1}(\mathfrak{q}) = \cap_{\mathfrak{p} \in V(\mathfrak{a})}\mathfrak{p}$. Q.E.D.

Zariski topology (Ex 1.15 in AM). We describe a way to topologize $\text{Spec}(A)$. Let

$\mathcal{T} := \{V(\mathfrak{a}) : \mathfrak{a} \subseteq A \text{ is an ideal}\}$.

We generalize Zariski-closed set for any subset $E \subseteq A$ as $V(E) := \{\mathfrak{p} \in \text{Spec}(A) : \mathfrak{p} \supseteq E\}$, but notice that $V(E) = V((E))$, so the operator $V$ does not see any difference between a subset and the ideal generated by it of a ring $A$, so we do not add any more members to $\mathcal{T}$ by this generalization.

A trivial but nice remark: $\mathfrak{p} \in V(f) := V((f))$ if and only if $f \in \mathfrak{p}$.

Zarski-closed set does not see any difference between an ideal and its radical. $V(\mathfrak{a}) = V(\sqrt{\mathfrak{a}})$.

Proof. Let $\mathfrak{p} \in \text{Spec}(A)$. It is enough to show $\mathfrak{p} \supseteq \mathfrak{a}$ if and only if $\mathfrak{p} \supseteq \sqrt{\mathfrak{a}}$. But we are done by seeing that the following are equivalent:

• $x \in \mathfrak{a}$ implies $x \in \mathfrak{p}$;
• $x^{n} \in \mathfrak{a}$ for some $n \geq 1$ implies $x^{n} \in \mathfrak{p}$;
• $x^{n} \in \mathfrak{a}$ for some $n \geq 1$ implies $x \in \mathfrak{p}$;
• $x \in \sqrt{\mathfrak{a}}$ implies $x \in \mathfrak{p}$. Q.E.D.

Zariski-closed sets satisfy axioms for closed sets. In fact, the fact that $\mathcal{T}$ form a topology of closed sets, which we call Zariski topology, on $\text{Spec}(A)$, is a corollary of the following stronger phenomenon.

1st. $V(0) = \text{Spec}(A)$ and $V(1) = \emptyset$.

2nd. For any family of subsets $\{E_{\alpha}\}_{\alpha \in I}$,

$V(\cup_{\alpha \in I}E_{\alpha}) = \cap_{\alpha \in I}V(E_{\alpha})$.

3rd. For any ideals $\mathfrak{a, b} \subseteq A$,

$V(\mathfrak{a}) \cup V(\mathfrak{b}) = V(\mathfrak{ab}) = V(\mathfrak{a \cap b})$.

Proof. (1st.) Any prime ideal contains $0$, so $V(0) = \text{Spec}(A)$, and no prime ideal contains $1$, so $V(1) = \emptyset$.

(2nd.) For any $\mathfrak{p} \in \text{Spec}(A)$, the following are equivalent:

• $\mathfrak{p} \in V(\cup_{\alpha \in I}E_{\alpha})$;
• $\mathfrak{p} \supseteq \cup_{\alpha \in I}E_{\alpha}$;
• $\mathfrak{p} \supseteq E_{\alpha}$ for all $\alpha \in I$;
• $\mathfrak{p} \in V(E_{\alpha})$ for all $\alpha \in I$;
• $\mathfrak{p} \in \cap_{\alpha \in I}V(E_{\alpha})$.

(3rd.) Let $\mathfrak{p} \in \text{Spec}(A)$ be arbitrary.

Suppose that $\mathfrak{p} \supseteq \mathfrak{a \cap b}$. Since any prime ideal can cut a member from the intersection of a finite collection of ideals, we have $\mathfrak{p} \supseteq \mathfrak{a}$ or $\mathfrak{p} \supseteq \mathfrak{b}$. This shows $V(\mathfrak{a} \cap \mathfrak{b}) \subseteq V(\mathfrak{a}) \cup V(\mathfrak{b})$.

Suppose that $\mathfrak{p \supseteq a}$ or $\mathfrak{p \supseteq b}$. Then $\mathfrak{p} \supseteq \mathfrak{ab}$ element-wise, so $V(\mathfrak{a}) \cup V(\mathfrak{b}) \subseteq V(\mathfrak{ab})$.

Suppose that $\mathfrak{p \supseteq ab}$. For any $x \in \mathfrak{a} \cap \mathfrak{b}$, we have $x^{2} \in \mathfrak{ab \subseteq p}$, which implies $x \in \mathfrak{p}$. Thus $\mathfrak{p} \supseteq \mathfrak{a \cap b}$, which shows $V(\mathfrak{ab}) \subseteq V(\mathfrak{a \cap b})$.

We achieved $V(\mathfrak{a \cap b}) \subseteq V(\mathfrak{a}) \cup V(\mathfrak{b}) \subseteq V(\mathfrak{ab}) \subseteq V(\mathfrak{a \cap b})$, so the inclusions must be equalities. Q.E.D.

Now, we see some properties of radical.

Any ideal is in its radical. Let $\mathfrak{a} \subseteq A$ be an ideal. Then

$\mathfrak{a \subseteq \sqrt{a}}$.

Proof. We have $\mathfrak{a} \subseteq \cap_{\mathfrak{p} \in V(\mathfrak{a})}\mathfrak{p} = \sqrt{\mathfrak{a}}$. Q.E.D.

Taking more than one radical does not give anything new. For any ideal $\mathfrak{a} \subseteq A$, we have $\mathfrak{\sqrt{\sqrt{a}} = \sqrt{a}}$.

Proof. We have

$\mathfrak{\sqrt{\sqrt{a}}} = \cap_{\mathfrak{p} \in V(\sqrt{\mathfrak{a}})}\mathfrak{p} = \cap_{\mathfrak{p} \in V(\mathfrak{a})}\mathfrak{p} = \sqrt{\mathfrak{a}}$. Q.E.D

Radical operator is compatible with product inside and intersection in and out. For an ideal $\mathfrak{a} \subseteq A$, we have

$\mathfrak{\sqrt{ab} = \sqrt{a \cap b} = \sqrt{a} \cap \sqrt{b}}$.

Proof. We have

• $\sqrt{\mathfrak{ab}} = \cap_{\mathfrak{p} \in V(\sqrt{\mathfrak{ab}})}\mathfrak{p} = \cap_{\mathfrak{p} \in V(\mathfrak{ab})}\mathfrak{p}$;
• $\sqrt{\mathfrak{a \cap b}} = \cap_{\mathfrak{p} \in V(\mathfrak{a \cap b})}\mathfrak{p}$;
• $\mathfrak{\sqrt{a} \cap \sqrt{b}} = \cap_{\mathfrak{p} \in V(\mathfrak{\sqrt{a} \cap \sqrt{b}})} \mathfrak{p} = \cap_{\mathfrak{p} \in V(\mathfrak{\sqrt{a}}) \cup V({\sqrt{b}})} \mathfrak{p} = \cap_{\mathfrak{p} \in V(\mathfrak{a}) \cup V(\mathfrak{b})}\mathfrak{p}$.

But we know $V(\mathfrak{ab}) = V(\mathfrak{a \cap b}) = V(\mathfrak{a}) \cup V(\mathfrak{b})$, so the right-hand sides above are all equal. Q.E.D.

Only radical of the unit ideal is the unit ideal (and vice versa). Let $\mathfrak{a} \subseteq A$ be an arbitrary ideal. We have

$\sqrt{\mathfrak{a}} = (1)$ if and only if $\mathfrak{a} = (1)$.

Proof. We see the following are equivalent.

• $(1) = \sqrt{\mathfrak{a}}$.
• $(1) = \cap_{\mathfrak{p} \in V(\mathfrak{a})}\mathfrak{p}$.
• $V(\mathfrak{a}) = \emptyset$.
• $\mathfrak{a} = (1)$. Q.E.D.

Before the next property, we prove twp more properties of Zariski closet sets. Despite being elementary, they do not appear in Atiyah-MacDonald, so we need to be extra careful for the proofs. The difficult part of the proof recycles the proof of Proposition 1.7 in Atiyah-MacDonald.

Inclusion of ideals give reverse inclusion of Zariski-closed sets (but not necessarily vice versa). Let $\mathfrak{a, b} \subseteq A$ be ideals. Then $\mathfrak{a} \subseteq \mathfrak{b}$ implies $V(\mathfrak{a}) \supseteq V(\mathfrak{b})$.

Proof.  If $\mathfrak{a \subseteq b}$ for any prime ideal $\mathfrak{p \supseteq a}$, we have $\mathfrak{p \supseteq a \supseteq b}$. Q.E.D.

Zariski closed sets cannot tell the difference between the sum of ideals and the sum of their radicals. Let $\mathfrak{a, b} \subseteq A$ be ideals. Then

$V(\mathfrak{a + b}) = V(\mathfrak{\sqrt{a} + \sqrt{b}})$.

Proof. Since $\mathfrak{a + b \subseteq \sqrt{a} + \sqrt{b}}$, we have $V(\mathfrak{a + b}) \supseteq V(\mathfrak{ \sqrt{a} + \sqrt{b} })$.

Now take any prime ideal $\mathfrak{p \supseteq a + b}$. Take any $x + y \in \mathfrak{\sqrt{a} + \sqrt{b}}$ where $x \in \mathfrak{a}$ and $y \in \mathfrak{b}$. Take $m, n \geq 1$ such that $x^{m} \in \mathfrak{a}$ and $y^{n} \in \mathfrak{b}$. Then $x^{r}y^{s}$ appears in each term of $(x + y)^{m + n -1}$ for some $0 \leq r, s \leq m + n - 1$ and $r + s = m + n - 1$.

Since $r < m$ and $s < n$ gives a contradiction, we see $r \geq m$ or $s \geq n$, so each $x^{r}y^{s} \in \mathfrak{a + b}$. This implies that $(x + y)^{m+n-1} \in \mathfrak{a + b} \subseteq \mathfrak{p}$. By primality, we get $x + y \in \mathfrak{p}$, showing $\mathfrak{\sqrt{a} + \sqrt{b} \subseteq p}$. Thus, we conclude $V(\mathfrak{a + b}) \subseteq V(\mathfrak{\sqrt{a} + \sqrt{b}})$. Q.E.D.

A radical cannot tell the difference between the sum of ideals and the sum of their radicals. Let $\mathfrak{a, b} \subseteq A$ be ideals. Then

$\mathfrak{\sqrt{a + b} = \sqrt{\sqrt{a} + \sqrt{b}}}$.

Proof. We have

$\sqrt{\mathfrak{a + b}} = \cap_{\mathfrak{p} \in V(\mathfrak{a + b})}\mathfrak{p} = \cap_{\mathfrak{p} \in V(\mathfrak{\sqrt{a} + \sqrt{b}})}\mathfrak{p} = \mathfrak{\sqrt{\sqrt{a} + \sqrt{b}}}$. Q.E.D.

Radical of any prime power is the base prime. For any $\mathfrak{p} \in \text{Spec}(A)$, we have

$\sqrt{\mathfrak{p}^{n}} = \mathfrak{p}$ for all $n \geq 1$.

Proof. First, we study the easiest case as follows:

$\mathfrak{\sqrt{p}} = \cap_{\mathfrak{q} \in V(\mathfrak{p})}\mathfrak{q} = \mathfrak{p}$.

Now, for any $n \geq 1$, we have

$\sqrt{\mathfrak{p}^{n}} = \cap_{i=1}^{n}\sqrt{\mathfrak{p}} = \sqrt{\mathfrak{p}} =\mathfrak{p}$. Q.E.D.

Pairwise coprimality of two ideals is same as coprimality of their radicals. Let $\mathfrak{a, b} \subseteq A$ be ideals. Then

$\mathfrak{a, b}$ are coprime if and only if $\mathfrak{\sqrt{a}, \sqrt{b}}$ are coprime.

Proof. We see the following are equivalent.

• $\mathfrak{a + b} = (1)$.
• $\sqrt{\mathfrak{a + b}} = (1)$.
• $\mathfrak{\sqrt{\sqrt{a} + \sqrt{b}}} = (1)$.
• $\mathfrak{\sqrt{a} + \sqrt{b}} = (1)$. Q.E.D.

The following does not appear in Atiyah-MacDonald, so we need to be careful.

Radical operator is increasing (with respect to inclusion). Let $\mathfrak{a, b} \subseteq A$ be ideals. Then

$\mathfrak{a \subseteq b}$ implies $\mathfrak{\sqrt{a} \subseteq \sqrt{b}}$.

Proof. If $\mathfrak{a \subseteq b}$, then

$\mathfrak{\sqrt{a}} = \cap_{\mathfrak{p} \in V(\mathfrak{a})}\mathfrak{p} \subseteq \cap_{\mathfrak{p} \in V(\mathfrak{b})}\mathfrak{p} = \sqrt{\mathfrak{b}}$,

since $V(\mathfrak{a}) \supseteq V(\mathfrak{b})$. Q.E.D.

# Notes on Commutative Algebra 1.

This is a “short” study note of Atiyah-MacDonald.

Conventions in the notes. The following are assumed throughout the notes, unless mentioned otherwise.

• When we say a ring, we mean a commutative ring with $1$, the multiplicative identity and $1 \neq 0$. Thus, we do not consider any zero rings.
• We exclude $0$ from zero-divisors.
• By $x \equiv y \mod \mathfrak{a}$, we mean $x - y \in \mathfrak{a}$.
• We frequently write $0$ for the zero ideal $(0)$.
• Under any ring map, $1 \mapsto 1$.
• Unless there is a special name for a theorem, we avoid calling it “Theorem,” “Lemma,” “Proposition,” etc.
• When we view a ring as an ideal in itself, we write it as $(1)$.
• If $\{X_{\alpha}\}_{\alpha \in I}$ is a collection of subsets of $A$, we define $\cap_{\alpha \in I}X_{\alpha} := \{x \in A : x \in X_{\alpha} \text{ for all } \alpha \in I\}$.

Under a ring map, the image of ideal is not generally an ideal in the target ring. However, if we consider the image of ideal is an ideal in the range ring.

Recall that for any surjection, the image of the preimage is the target (mnemonic: right inverse). Analogously, for any injection, the preimage of the image is the domain (mnemonic: left inverse).

Recall that for any surjective ring map $\phi : A \twoheadrightarrow B$, the following maps are bijections $\{\mathfrak{a} \subseteq A : \mathfrak{a} \text{ is an ideal such that } \mathfrak{a} \supseteq \ker\phi\} \leftrightarrow \{\mathfrak{b} \subseteq B : \mathfrak{b} \text{ is an ideal} \}$, that are inverses each other:

$\mathfrak{a} \mapsto \phi(\mathfrak{a})$,

$\mathfrak{b} \mapsto \phi^{-1}(\mathfrak{b})$.

Moreover, recall that above maps preserves order and primality.

However, the correspondence does not necessarily preserve maximality. Consider $\mathbb{Z} \hookrightarrow \mathbb{Q}$.

Recall that the units of $A$ form an abelian group under multiplication. We write $A^{*}$ to denote this group.

Recall that the following are equivalent ways to say that a ring $A$ is a field:

1. $A \setminus \{0\} = A^{*}$;
2. $0, (1)$ are the only ideals in $A$;
3. Any nontrivial ring map from $A$ is injective.

Recall that given a finite collection of ring maps from a ring $A$, namely $\phi_{j} : A \rightarrow B_{j}$ with $1 \leq j \leq n$, we can construct a “larger”  ring map $\phi : A \rightarrow \prod_{j=1}^{n}B_{j}$ by

$a \mapsto (\phi_{1}(a), \cdots, \phi_{n}(a))$.

Convention. If we are given only one collection of ring maps $\phi_{j} : A \rightarrow B_{j}$, we just refer to $\phi : A \rightarrow \prod_{j} B_{j}$ as the ring map to the direct product. If we are given ideals $\mathfrak{a}_{1}, \cdots, \mathfrak{a}_{n} \subseteq A$, then to interpret $A \rightarrow \prod_{j}A/\mathfrak{a}_{j}$, we assume each $\phi_{j}$ is the projection $A \twoheadrightarrow A/\mathfrak{a}_{j}$.

Moreover, this construction essentially shows that to check if $\phi : A \rightarrow \prod_{j=1}^{n}B_{j}$ is a ring map, it is enough to check if the map $A \rightarrow \prod_{j=1}^{n}B_{j} \twoheadrightarrow B_{i}$ is a ring map for all $i$.

Recall that the following are equivalent ways to say that a proper ideal $\mathfrak{p} \subseteq A$ is a prime ideal in a ring $A$.

• $xy \in \mathfrak{p}$ implies $x \in \mathfrak{p}$ or $y \in \mathfrak{p}$;
• $A/\mathfrak{p}$ is an integral domain.

We write $\text{Spec}(A)$ to mean the set of all the prime ideals of $A$.

Immediately, $A$ is an integral domain iff $0 \in \text{Spec}(A)$.

Recall that the following are equivalent ways to say that a proper ideal $\mathfrak{m} \subseteq A$ is a maximal ideal in a ring $A$.

• There is no proper ideal that contains $\mathfrak{m}$;
• $A/\mathfrak{m}$ is an field.

We write $\text{Max}(A)$ to mean the set of all the maximal ideals of $A$.

Immediately, maximal ideals are prime ideals (i.e., $\text{Max}(A) \subseteq \text{Spec}(A)$).

Existence of a maximal ideal around a proper ideal. For any ideal $\mathfrak{a} \subseteq A$, there is a maximal ideal $\mathfrak{m} \supseteq \mathfrak{a}$.

Proof. Let $A$ be a ring (recall that $A$ must have at least two elements). Let $\Sigma$ be the set of all proper ideals that contains $\mathfrak{a}$. Consider $\Sigma$ as a poset (partially ordered set) with $\subseteq$. Notice that

• $\mathfrak{a} \in \Sigma$, so $\Sigma$ is nonempty;
• any chain $(\mathfrak{a}_{\alpha})_{\alpha \in I}$ has an upperbound, namely $\cup_{\alpha} \mathfrak{a}$.

By Zorn’s lemma, there is a maximal element $\mathfrak{m} \in \Sigma$. This is a maximal ideal that contains $\mathfrak{a}$. Q.E.D.

A ring $A$ with only one maximal ideal $\mathfrak{m}$ is called a local ring (we will write is as a pair $(A, \mathfrak{m})$)with the residue field $k := A/\mathfrak{m}$.

How to find a local ring? Let $A$ be a ring. The following are equivalent.

• $(A, \mathfrak{m})$ is a local ring.
• $\mathfrak{m} \subseteq A$ is a maximal ideal and $1 + \mathfrak{m} \subseteq A^{*}$.
• $\mathfrak{m} \subseteq A$ is a proper ideal such that $A \setminus \mathfrak{m} \subseteq A^{*}$.

Proof. Suppose that $(A, \mathfrak{m})$ is a local ring. For any $x \in \mathfrak{m}$, if $1 + x \notin A^{*}$, then $(1 + x) \subseteq \mathfrak{m}$, so $1 = 1 + x - x \in \mathfrak{m}$, which is a contradiction, so $1 + x \in A^{*}$, showing $1 + \mathfrak{m} \subseteq A^{*}$.

Assume the second condition. Consider an element of the ring $x \notin \mathfrak{m}$. Then $(x) + \mathfrak{m} = (x, \mathfrak{m}) = (1)$. Thus, we can write $ax + y = 1$. Thus $ax = 1 - y \in 1 + \mathfrak{m} \subseteq A^{*}$. This implies there is $u \in A^{*}$ such that $uax = 1$, so $x \in A^{*}$.

Assume the third condition. If $\mathfrak{a} \subseteq A$ is any proper ideal, then $\mathfrak{a} \subseteq \mathfrak{m}$. In particular, any maximal ideal must be contained in $\mathfrak{m}$. Thus this must be the unique maximal ideal of $A$, so $(A, \mathfrak{m})$ is a local ring. Q.E.D.

If a ring $A$ has a finite number of maximal ideals, we say $A$ is semi-local.

In a PID, every nonzero prime ideal is maximal.

Proof. Let $(p) \neq 0$ be a prime ideal in a PID $A$. Suppose that $(p)$ is not a maximal ideal.

Take a maximal ideal $(m) \supseteq (p)$. Then $am = p \in (p)$ for some $a \in A$. Since $m \notin (p)$, the primality of $(p)$ forces $a \in (p)$, so $a = bp$ for some $b \in A$. But then $pbm = bpm = p$ implies $bm = 1$, so $(p) = (1)$, a contradiction. Q.E.D.

Given a ring $A$, denote $\mathfrak{N}$, or $\mathfrak{N}_{A}$, the set of all the nilpotent elements of $A$, and call it the nilradical of $A$.

The nilradical is the intersection of the prime spectrum. For any ring $A$, we have

$\mathfrak{N}_{A} = \cap_{\mathfrak{p} \in \text{Spec}(A)}\mathfrak{p}$.

In particular, the nilradical is an ideal.

Proof. Denote $\mathfrak{N} := \mathfrak{N}_{A}$ and $\mathfrak{N}' := \cap_{\mathfrak{p} \in \text{Spec}(A)}\mathfrak{p}$.

Let $f \in \mathfrak{N}$. Then for any prime ideal $\mathfrak{p}$, we have $f^{n} \in \mathfrak{p}$, which implies $f \in \mathfrak{p}$. Thus $f \in \mathfrak{N}'$, showing $\mathfrak{N} \subseteq \mathfrak{N}'$.

For the other inclusion, let $g \in \mathfrak{N}'$. Suppose that $g \notin \mathfrak{N}$. Then we consider $\Sigma$, the set of ideals $\mathfrak{a}$ in which the following property is satisfied:

if $n > 0$, then $g^{n} \notin \mathfrak{a}$.

Notice the following:

• $0 \in \Sigma$ (in particular, nonemtpy);
• any chain $(\mathfrak{a}_{\alpha})_{\alpha \in I}$ is bounded above by $\cup_{\alpha}\mathfrak{a}_{\alpha}$ (with inclusion being partial ordering).

Thus, we can apply Zorn’s lemma to achieve a maximal element $\mathfrak{p} \in \Sigma$. We study $\mathfrak{p}$ as follows.

If elements of the ring $x, y \notin \mathfrak{p}$, then $\mathfrak{p} + (x), \mathfrak{p} + (y)$ cannot be in $\Sigma$ by the maximality. Thus, we can take $m, n > 0$ such that

$g^{m} \in \mathfrak{p} + (x)$ and $g^{n} \in \mathfrak{p} + (y)$.

Then $g^{m+n} \in (\mathfrak{p} + (x))(\mathfrak{p} + (y)) = \mathfrak{p} + (xy)$, so $\mathfrak{p} + (xy) \notin \Sigma$, and this implies that $xy \notin \mathfrak{p}$.

From our study, we proved: $x, y \notin \mathfrak{p}$ implies $xy \notin \mathfrak{p}$. Thus we see $\mathfrak{p} \in \text{Spec}(A)$. Then $g \notin \mathfrak{N}'$, which contradicts the assumption. We showed $g \in \mathfrak{N}$, showing $\mathfrak{N}' \subseteq \mathfrak{N}$. Q.E.D.

“Removing” nonzero nilpotent elements. Given a ring $A$, the nilradical of  the quotient $A/\mathfrak{N}$ is trivial.

Proof. Consider any $\bar{a} \in A/\mathfrak{N}$ with $a \in A$. Then if $\bar{a}^{n} = \bar{0}$, then $a^{n} \in \mathfrak{N}$, so we have $m > 0$ such that $a^{nm} = 0$. This implies that $a \in \mathfrak{N}$, which means $\bar{a} = \bar{0}$. Q.E.D.

Now, for a given ring $A$, we define the Jacobson radical, which is written as $\mathfrak{R}$, or $\mathfrak{R}_{A}$, to be the intersection of the maximal spectrum. That is,

$\mathfrak{R}_{A} := \cap_{\mathfrak{m} \in \text{Max}(A)}\mathfrak{m}$.

Just like the nilradical, there is another way to descirbe the Jacobson radical.

A characterization of the Jacobson radical. Let $A$ be a ring. Then

$\mathfrak{R} = \{x \in A : 1 + (x) \subseteq A^{*}\}$.

Proof. Let $x \in \mathfrak{R}$. Take an arbitrary $y \in A$. Suppose that $1 - xy \notin A^{*}$. Then we can take $1 - xy \in \mathfrak{m} \in \text{Max}(A)$. But since $xy \in \mathfrak{m}$, we have $1 \in \mathfrak{m}$, a contradiction. Thus $1 - xy \in A^{*}$, showing $1 + (x) \subseteq A^{*}$.

Conversely, let $1 - xy \in A^{*}$ for all $y \in A$. Suppose $x \notin \mathfrak{R}$. Then we can take $x \notin \mathfrak{m} \in \text{Max}(A)$. Then $\mathfrak{m} + (x) = (\mathfrak{m}, x) = (1)$, so we can write $z + xy = 1$ for some $y \in A$. Thus $z = 1 - xy \in \mathfrak{m}$ gives a contradiction since $1 - xy$ is a unit, showing $x \in \mathfrak{R}$. Q.E.D.

A finite collection of ideals $\mathfrak{a}_{1}, \mathfrak{a}_{2}, \cdots, \mathfrak{a}_{n} \subseteq A$ are coprime if $\mathfrak{a}_{1} + \mathfrak{a}_{2} + \cdots + \mathfrak{a}_{n} = (1)$.

If $\mathfrak{a}_{i} + \mathfrak{a}_{j} = (1)$ given $i \neq j$, we say the collection of ideal is pairwise coprime.

To oppress the difficulty of notation that naturally arises, we write $a \mapsto a \mod \mathfrak{a_{i}}$ for the projection $A \twoheadrightarrow A/\mathfrak{a}_{i}$.

Recall the product of ideal is defined as $\prod \mathfrak{a}_{i} := (x_{1} \cdots x_{n})_{x_{j} \in \mathfrak{a}_{j}}$. Considering generators, we always have $\prod \mathfrak{a}_{i} \subseteq \cap \mathfrak{a}_{i}$. We can also write the elements of the product as follows:

$\prod_{i=1}^{n} \mathfrak{a}_{i} = \{\sum_{j=1}^{m}x_{1}^{(j)} \cdots x_{n}^{(j)} : m \geq 0 \text{ and } x_{i}^{j} \in \mathfrak{a}_{i}\}$.

Proof. The right-hand side is an ideal that contains the generators of the product on the left-hand side, so the left-hand side must be contained in the right-hand side. The reverse inclusion is clear. Q.E.D.

A direct corollary is as follows.

The intersection contains the product. $\prod \mathfrak{a}_{i} \subseteq \cap \mathfrak{a}_{i}$.

Product operation commutes with generating operation. Let $\mathfrak{a, b} \subseteq A$ be ideals and suppose that $\mathfrak{a} = (X)$ and $\mathfrak{b} = (Y)$ for some $X, Y \subseteq A$. Then

$(XY) = \mathfrak{ab}$,

where $XY := \{xy : x \in X \text{ and } y \in Y\}$.

Proof. Since $X \subseteq \mathfrak{a}$ and $Y \subseteq \mathfrak{b}$, we have $XY \subseteq \mathfrak{ab}$, so $(XY) \subseteq \mathfrak{ab}$.

Let $u \in \mathfrak{a} = (X)$ and $v \in \mathfrak{b} = (Y)$. Then we can write $u = u_{1}x_{1} + \cdots + u_{m}x_{m}$ and $v = v_{1}y_{1} + \cdots + v_{n}y_{n}$, where $u_{i}, v_{j} \in A$ and $x_{i} \in X$ and $y_{j} \in Y$. By a direct computation, we have $uv \in (XY)$, so $\mathfrak{ab} \subseteq (XY)$, as the right-hand side contains all the generators of the left-hand side. Q.E.D.

Since product operation commutes generating operation, we can easily prove the following.

• $\mathfrak{(ab)c} = \mathfrak{a(bc)}$;
• $\mathfrak{a}(\mathfrak{b} + \mathfrak{c}) = \mathfrak{ab + ac}$.

Proof. We have $\mathfrak{ab} = (xy)_{x \in \mathfrak{a}, y \in \mathfrak{b}}$, so $\mathfrak{(ab)c} = (xyz)_{(x, y, z) \in \mathfrak{a \times b \times c}}$. Similarly, we have $\mathfrak{a(bc)} = (xyz)_{(x, y, z) \in \mathfrak{a \times b \times c}}$, so $\mathfrak{(ab)c} = \mathfrak{a(bc)}$.

We have $\mathfrak{a}(\mathfrak{b} + \mathfrak{c}) = (x(y + z))_{(x, y, z) \in \mathfrak{a \times b \times c}} = (xy + xz)_{(x, y, z) \in \mathfrak{a \times b \times c}} \subseteq \mathfrak{ab + ac}$. But elementwise, we see $\mathfrak{ab, ac \subseteq a(b + c)}$, so $\mathfrak{ab + ac \subseteq a(b + c)}$. Q.E.D.

Pairwise coprimaity makes the product sweep the intersection. If $\mathfrak{a}_{1}, \mathfrak{a}_{2}, \cdots, \mathfrak{a}_{n}$ are finitely many coprime ideals, then

$\prod \mathfrak{a}_{i} = \cap \mathfrak{a}_{i}$.

Proof. Due to the previous remark, it is enough to show that $\cap \mathfrak{a}_{i} \subseteq \prod \mathfrak{a}_{i}$.

We first study the case $n = 2$.

Suppose we are given only two coprime ideals, namely $\mathfrak{a}, \mathfrak{b} \subseteq A$. Since $\mathfrak{a} + \mathfrak{b} = (1)$, we can take $a \in \mathfrak{a}$ and $b \in \mathfrak{b}$ such that $a + b = 1$. Then $\mathfrak{a} \cap \mathfrak{b} = (a + b)(\mathfrak{a} \cap \mathfrak{b}) = a(\mathfrak{a} \cap \mathfrak{b}) + (\mathfrak{a} \cap \mathfrak{b})b \subseteq \mathfrak{a}\mathfrak{b}$.

Our study showed that the desired statement is true for $n = 2$. It is clear that the statement is also true for $n = 1$. We try induction, so assume that for $n = 1, 2, \cdots k$, the statement is true. Suppose $n = k + 1$. Then

$\mathfrak{a}_{1} \cap \cdots \cap \mathfrak{a}_{k} \cap \mathfrak{a}_{k+1} = (\mathfrak{a}_{1} \cap \cdots \cap \mathfrak{a}_{k}) \cap \mathfrak{a}_{k+1} = (\prod_{i=1}^{k}\mathfrak{a}_{i}) \cap \mathfrak{a}_{k+1}$.

It is clear what to attempt next. We wonder if $\prod_{i=1}^{k}\mathfrak{a}_{i}$ and $\mathfrak{a}_{k+1}$ are coprime.

Since $\mathfrak{a}_{i} \cap \mathfrak{a}_{k+1} = (1)$ for $1 \leq i \leq k$. Thus, we can take $x_{i} \in \mathfrak{a}_{i}$ and $y_{i} \in \mathfrak{a}_{k+1}$ such that $x_{i} + y_{i} = 1$ for $1 \leq i \leq k$. This implies that

$x_{1} \cdots x_{k} = (1 - y_{1}) \cdots (1 - y_{k}) \equiv 1 \mod \mathfrak{a}_{k+1}$.

Thus $\prod_{i=1}^{k}\mathfrak{a}_{i} + \mathfrak{a}_{k+1} = (1)$, as we conjectured. We now can proceed the final step of our induction, continuing the previous computation:

$\cap_{i}^{k+1}\mathfrak{a}_{i} = (\prod_{i=1}^{k}\mathfrak{a}_{i}) \cap \mathfrak{a}_{k+1} = (\prod_{i=1}^{k}\mathfrak{a}_{i})\mathfrak{a}_{k} = \prod_{i=1}^{k+1}\mathfrak{a}_{i}$. Q.E.D.

The pairwise coprimality of ideals $\{\mathfrak{a}_{i}\}_{i=1}^{n}$ in $A$ can be neatly characterized as the surjectivity of the ring map $A \rightarrow \prod_{i} A/\mathfrak{a}_{i}$. This is the famous Chinese remainder theorem.

The Chinese remainder lemma. A finite collection of ideals $\mathfrak{a}_{1}, \cdots, \mathfrak{a}_{n} \subseteq A$ is pairwise coprime if and only if $A \rightarrow \prod_{i=1}^{n}A/\mathfrak{a}_{i}$ is surjective.

Proof. Suppose that $\{\mathfrak{a}_{i}\}_{i=1}^{n}$ is pairwise coprime. Take any $y := (x_{1} \mod \mathfrak{a}_{1}, \cdots, x_{n} \mod \mathfrak{a}_{n}) \in \prod_{i=1}^{n}A/\mathfrak{a}_{i}$, and we want to show the fiber of $y \in \prod_{j} A/\mathfrak{a}_{j}$ is not empty.

First, we study an easier case and ask if the fiber of $(0, \cdots, 0, 1 \mod \mathfrak{a}_{j}, 0, \cdots, 0)$ is nonempty.

Since $\mathfrak{a}_{i} + \mathfrak{a}_{j} = (1)$ for all $i \neq j$, we can take $u_{i} \in \mathfrak{a}_{i}$ and $v_{i} \in \mathfrak{a}_{j}$ such that $u_{i} + v_{i} = 1$. Then let $y_{j} := \prod_{i \neq j}v_{i}$ and see that

• $y_{j} = \prod_{i \neq j}(1 - u_{i}) \equiv 1 \mod \mathfrak{a}_{j}$;
• $y_{j} \equiv 0 \mod \mathfrak{a}_{i}$ for any $i \neq j$.

From our study, we constructed $y_{j}$ in the fiber of $(0, \cdots, 0, 1 \mod \mathfrak{a}_{j}, 0, \cdots, 0)$. But now we immediately see that $x_{1}y_{1} + \cdots + x_{n}y_{n}$ is in the fiber of $(x_{1} \mod \mathfrak{a}_{1}, \cdots, x_{n} \mod \mathfrak{a}_{n})$. This finishes showing the surjectivity of $A \rightarrow \prod_{i=1}^{n}A/\mathfrak{a}_{i}$.

Conversely, suppose that the ring map $A \rightarrow \prod_{i=1}^{n}A/\mathfrak{a}_{i}$ is surjective. Let $i \neq j$, we need to show $\mathfrak{a}_{i} + \mathfrak{a}_{j} = (1)$.

Our previous study seems suggestive, so we take $y_{j}$ in the fiber of $(0, \cdots, 0, 1 \mod \mathfrak{a}_{j}, 0, \cdots, 0)$. Then

• $y_{j} \equiv 1 \mod \mathfrak{a}_{j}$ implies $1 - y_{j} \in \mathfrak{a}_{j}$;
• $y_{j} \equiv 0 \mod \mathfrak{a}_{i}$ implies $y_{j} \in \mathfrak{a}_{i}$.

Thus we have $1 \in \mathfrak{a}_{i} + \mathfrak{a}_{j}$, as desired. Q.E.D.

Notice that $\ker(A \rightarrow \prod_{i=1}^{n}A/\mathfrak{a}_{i}) = \cap_{i=1}^{n} \mathfrak{a}_{i}$, so we always have

$A/(\cap_{i = 1}^{n}\mathfrak{a}_{i}) \hookrightarrow \prod_{i=1}^{n}A/\mathfrak{a}_{i}$.

Recall that for any ring map $\phi : A \rightarrow B$, surjectivity of $\phi$ is characterized by $A/\ker(\phi) \simeq B$. Thus we can state the Chinese remainder in the following way.

The Chinese remainder theorem. For a finite collection of ideals $\{\mathfrak{a}_{i}\}_{i = 1}^{n}$ in a ring $A$, the following are equivalent.

• $\{\mathfrak{a}_{i}\}_{i=1}^{n}$ is pairwise coprime.
• $A \twoheadrightarrow \prod_{i=1}^{n}A/\mathfrak{a}_{i}$.
• $A /(\cap_{i=1}^{n}\mathfrak{a}_{i}) \simeq A/\mathfrak{a}_{i}$.

Thus, if any of above is true, the product sweeps the intersection, so $\prod_{i=1}^{n}\mathfrak{a}_{i} = \cap_{i=1}^{n}\mathfrak{a}_{i}$.

A trivial application of the Chinese remainder. Let $k$ be a field and $f \in k[x]$ be a separable polynomial. Then we can write $f(x) = (x - a_{1}) \cdots (x - a_{n})$ over an algebraically closed field $\bar{k}$ and apply the Chinese remainder to have

$k[x]/(f(x)) \hookrightarrow \bar{k}/(f(x)) \simeq \bar{k}/(x - a_{1}) \times \cdots \times \bar{k}/(x - a_{n}),$

where the right-hand side is a direct product of fields and a $k$-algebra.

# Notes on Covering Spaces 5.

Consider relation $\sim$ of two covering maps $p : E \rightarrow B$ and $p' : E' \rightarrow B'$ defined by existence of

$h : E \approx E'$ such that $p = p' \circ h$.

This defines equivalence relation on the class of covering maps. Thus, we say $p$ and $p'$ are equivalent if $p \sim p'$ and a homeomorphism between them is called an equivalence (of covering maps/spaces).

We say a space $X$ is locally path connected at $x \in X$ if every neighborhoood of $x$ contains a path connected neighborhood of $x$. We say $X$ is locally path connected if it is locally path connected everywhere in $X$.

It turned out that the property of locally path connected and that of path connected are independent each other (see Comb_space). We say space $X$ is nice if it is locally path connected and path connected. We say a continuous map $f : X \rightarrow Y$ is nice if $X, Y$ are nice.

We proved the following about any covering map $p : (E, e_{0}) \rightarrow (B, b_{0})$. For any $[f] \in (B, b_{0})$, TFAE:

• $[f] \in H_{0} = p_{*}(\pi_{1}(E, e_{0}))$;
• $\tilde{f} : e_{0} \sim e_{0} \in E$ (so that $[\tilde{f}] \in \pi_{1}(E, e_{0})$).

The general lifting lemma. The following is a characterization of lifting of a nice continuous map by a nice covering map.

Let $p : (E, e_{0}) \rightarrow (B, b_{0})$ be a nice covering map and $f : (Y, y_{0}) \rightarrow (B, b_{0})$ be a nice continuous map. Then TFAE:

• $f$ can be lifted to lifting $h : (Y, y_{0}) \rightarrow (E, e_{0})$ by $p$;
• $f_{*}(\pi_{1}(Y, y_{0})) \subseteq p_{*}(\pi_{1}(E, e_{0}))$.

Moreover if this happens, such $h$ is unique, so we write $\tilde{f} := h : (Y, y_{0}) \rightarrow (E, e_{0})$.

Proof. (Uniqueness) Suppose we have another lifting $h' : (Y, y_{0}) \rightarrow (E, e_{0})$. For any $y \in Y$, since $Y$ is path connected, take a path $\alpha : y_{0} \sim y \in Y$ so that we get $f \circ \alpha : e_{0} \sim f(y)$. Then $h \circ \alpha = h' \circ \alpha$ since both are liftings of $f \circ \alpha$, so $h(y) = h'(y)$.

(Easier Direction) Suppose we have the lifting $\tilde{f} : (Y, y_{0}) \rightarrow (E, e_{0})$. Then by a property of $\pi_{1}$ functor, we have

$f_{*}(\pi_{1}(Y, y_{0})) = (p \circ \tilde{f})_{*}(\pi_{1}(Y, y_{0})) = p_{*}(\tilde{f}_{*}(\pi_{1}(Y, y_{0})) \subseteq H_{0}$.

(Harder Direction Part 1.) Let $y \in Y$. Then we take a paths $\alpha, \beta : y_{0} \sim y \in Y$ and lift $f \circ \alpha, f \circ \beta : b_{0} \sim f(y)$ to

$\alpha' : e_{0} \sim \alpha'(1)$ and $\beta' : e_{0} \sim \beta'(1)$.

Now consider a path $f \circ \bar{\beta} : f(y) \sim b_{0}$ and a loop

$f \circ (\alpha * \bar{\beta}) : b_{0} \sim b_{0}$.

Our hypothesis implies that $[f \circ (\alpha * \bar{\beta})] \in H_{0} = p_{*}(\pi_{1}(E, e_{0}))$. We now know that this implies that the lifting of $f \circ (\alpha * \bar{\beta}) = (f \circ \alpha) * (f \circ \bar{\beta})$ is a loop in $E$. But we also know how to compute the lifting. It is exactly $\alpha' * \bar{\beta'}$ by bijectivity and compatibility of lifting operator. Thus $\alpha' \simeq_{p} h * \beta'$ for some $h : e_{0} \sim e_{0}$. Thus $\alpha'(1) = \beta'(1)$. Therefore, we can define $\tilde{f} : (Y, y_{0}) \rightarrow (E, e_{0})$ by

$y \mapsto \alpha'(1)$, noting $\alpha'(1)$ is only dependent on $y \in Y$.

Notice that this map has the following property:

$p(\tilde{f}(y)) = p(\alpha'(1)) = f(\alpha(1)) = f(y)$.

Therefore, it remains to show $\tilde{f} : (Y, y_{0}) \rightarrow (E, e_{0})$.

(Harder Direction Part 2.) The rest of the proof uses local path connectedness.

Let $y_{1} \in Y$. We show continuity of $\tilde{f}$ at $y_{1}$. Take a neighborhood $N$ of $y_{1} \in Y$. Take a neighborhood $U_{0}$ of $f(y_{1}) \in B$ that is evenly covered by $p : E \rightarrow B$. Then we can slice $p^{-1}(U_{0}) \subseteq E$ disjoint open sets in $E$, and take the slice $V_{0}$ that contains $\tilde{f}(y_{1}) \in p^{-1}(U_{0})$. Note that

$p : (V_{0}, \tilde{f}(y_{1})) \approx (U_{0}, f(y_{1}))$.

Now consider $V_{0} \cap N$, which is a neighborhood of $\tilde{f}(y_{1})$. Since $B$ is locally path connected, take a path connected neighborhood $U \subseteq p(V_{0} \cap N)$ of $f(y_{1})$. Let $V := p^{-1}(U) \subseteq p^{-1}(p(V_{0} \cap N)) = V_{0} \cap N \subseteq N$. We now consider

$p : (V, \tilde{f}(y_{1})) \approx (U, f(y_{1}))$ with $V \subseteq N$ and $U$ path connected.

By continuity of $f : Y \rightarrow B$ and local connectedness of $Y$, take path connected neighborhood $W$ of $y_{1}$ such that $f(W) \subseteq U$. If we show $\tilde{f}(W) \subseteq V$, we are done since $V \subseteq N$. Consider $y \in W$, and we show $\tilde{f}(y) \in V$.

Choose path $\alpha : y_{0} \sim y_{1}$ and $\beta : y_{1} \sim y$. Let $\eta := f \circ (\alpha * \beta) : b_{0} \sim f(y)$. Then we have the lifting $\tilde{\eta} : e_{0} \sim \tilde{f}(y)$ (see our definition above). From our choice of $W$, we have $f \circ \beta : f(y_{1}) \sim f(y) \in U$ and its lifting is

$\delta := p^{-1} \circ f \circ \beta : \tilde{f}(y_{1}) \sim \tilde{f}(y) \in p^{-1}(U) = V$. Q.E.D.

Notice that the above proof does not use local path connectedness of $E$.

Characterization of equivalence of nice covering maps with same target. Let $p : (E, e_{0}) \rightarrow (B, b_{0})$ and $p' : (E', e'_{0}) \rightarrow (B, b_{0})$ be nice covering maps. Then TFAE

• $p \sim p'$;
• $H_{0} = H'_{0}$, where these are corresponding subgroups of $p$ and $p'$.

Moreover, if this happens, the equivalence $h : E \approx E'$ is unique.

Proof. (Uniqueness) Any equivalence $h : E \rightarrow E'$ is the unique lifting of $p' \circ h : E \rightarrow B$ by $p'$ according to the general lifting lemma.

(Equivalence) Now resetting the notation, suppose that $h : E \rightarrow E'$ be any continuous map (must be nice since $E, E'$ are). For this direction we can even make one of the covering map target to a different nice set $p' : (E, e_{0}) \rightarrow (B', b'_{0})$.

Without more assumption, we know that

• $p' \circ h : E \rightarrow B$ is lifted to $h : E \rightarrow E'$ by $p'$, so by GLL,
• $(p' \circ h)_{*}(\pi_{1}(E, e_{0})) \subseteq p'_{*}(\pi_{1}(E', e_{0}))$.

Also consider arbitrary map $h' : E' \rightarrow E$. By symmetry, we have

• $p \circ h' : E' \rightarrow B'$ is lifted to $h' : E' \rightarrow E$ by $p'$, so by GLL,
• $(p \circ h')_{*}(\pi_{1}(E', e_{0})) \subseteq p_{*}(\pi_{1}(E, e_{0}))$.

Thus if $h : E \rightarrow E'$ is an equivalence, thinking $h' = h^{-1}$ and recognizing $p' \circ h = p$ and $p' = p \circ h'$, we get (taking $B = B'$ if one prefers)

$H_{0} = H'_{0}$.

Conversely, suppose $H_{0} = H'_{0}$. This direction does need the same target.

Then by GLL, we have unique liftings

$\tilde{p} : E \rightarrow E'$ and $\tilde{p'} : E' \rightarrow E$.

But note that $\tilde{p} \circ tilde{p'} : E \rightarrow E$ is the lifting of $p : E \rightarrow E$. Thus it has to be equal to $id_{E}$. By symmetry, we get $\tilde{p'} \circ \tilde{p} = id_{E'}$. Q.E.D.

# Notes on Covering Spaces 4.

Read Topology by Munkres if you need a better explanation of this post.

Another notation to describe paths. We write $f : a \sim b \in X$ to mean that $f : [0, 1] \rightarrow X$ is a path from $a \in X$ to $b \in X$. For instance,

• $g \circ f : g(a) \sim g(b) \in Y$ if $g : X \rightarrow Y$ is continuous.

We review what we know about liftings by a covering map.

Bijection between lower paths and higher paths. Fix a covering map $p : (E, e_{0}, e_{1}) \rightarrow (B, b_{0}, b_{1})$. We have

• $\tilde{f} : e_{0} \sim e_{1} \leftrightarrow f : b_{0} \sim b_{1}$ by $\tilde{f} \mapsto p \circ \tilde{f}$.
• $\tilde{F} : \tilde{f} \simeq_{p} \tilde{g}$ if and only if $F : f \simeq_{p} g$ where $F = p \circ \tilde{F}$ (and here, $\tilde{F} \leftrightarrow F$).

Lifting is compatible with product and reverse. Let $f : a \sim b$ and $g : b \sim c$ be paths in some space $B$. Suppose that a (not necessarily covering) map $p : E \rightarrow B$ gives (not necessarily unique) lifting $\tilde{f} : a' \sim b'$ and $\tilde{g}: b' \sim c'$ in $E$. Then

• $\tilde{f} * \tilde{g} : a' \sim c'$ is a lifting of  $f * g : a \sim c$.
• $\bar{\tilde{f}} : b' \sim a'$ is a lifting of $\bar{f} : b \sim a$.

Proof.

• $p \circ (\tilde{f} * \tilde{g}) = (p \circ \tilde{f}) * (p \circ \tilde{g}) = f * g$.
• $(p \circ \bar{\tilde{f}})(s) = p(\tilde{f}(1-s)) = f(1-s) = \bar{f}(s)$. Q.E.D.

Given a covering map $p : (E, e_{0}) \rightarrow (B, b_{0})$, we define corresponding subgroup

$H_{0} := p_{*}(\pi_{1}(E, e_{0})) \leqslant \pi_{1}(B, b_{0})$.

We already proved that $p_{*} : \pi_{1}(E, e_{0}) \simeq H_{0}$.

If $H \leqslant G$, then $G / H := \{Hg | g \in G\}$. This is the space of right cosets, which is not necessarily a quotient group.

Let $p : (E, e_{0}) \rightarrow (B, b_{0})$ be a covering map.

Injection from right coset space to a fiber. By injectivity of lifting correspondence of a covering map, the following map is a well-defined injection:

$\pi_{1}(B, b_{0})/H_{0} \hookrightarrow p^{-1}(\{b_{0}\})$ by

$H_{0}*[f] \mapsto \phi([f]) = \tilde{f}(1)$.

Proof. Suppose $H_{0}*[f] = H_{0}*[g]$. Then $[f * \bar{g}] \in H_{0} = \text{Im}(p_{*})$, so there exists a loop $h : e_{0} \sim e_{0} \in E$ such that $[f * \bar{g}] = [p \circ h]$. Since $p \circ h : b_{0} \sim b_{0} \in B$, we have $[f] = [(p \circ h)*g]$. Note that $f \simeq_{p} (p \circ h) * g$ implies $\tilde{f} \simeq_{p} \tilde{h} * \tilde{g}$ with fixed point $e_{0}$. Thus $\tilde{f}(1) = \tilde{g}(1)$. Thus the map is well-defined.

Now let $\tilde{f}(1) = \tilde{g}(1)$. To show injectivity, we need to show $H_{0}*[f] = H_{0}*[g]$. Since $\tilde{f}, \tilde{g} : e_{0} \sim \tilde{f}(1) \in E$, we have $\tilde{h} = \tilde{f} * \bar{\tilde{g}} : e_{0} \sim e_{0} \in E$, so $\tilde{h} * \tilde{g} \simeq_{p} \tilde{f}$. Then $h * g \simeq_{p} f$ by bijection and compatibility of lifting. But note that $[h] = [p \circ \tilde{h}] = p_{*}([\tilde{h}]) \in H_{0}$. Q.E.D.

Bijection from right coset space to a fiber with lifting in path connected space. If $E$ is path connected, then $\phi : \pi_{1}(E, e_{0}) \rightarrow p^{-1}(\{b_{0}\})$ is surjective, so

$\pi_{1}(B, b_{0})/H_{0}\leftrightarrow p^{-1}(\{b_{0}\})$ by same map as above.

Loop belongs to corresponding group precisely when it lifts to a loop (under covering). Given $f : b_{0} \sim b_{0} \in B$, TFAE:

• $[f] \in H_{0}$;
• $\tilde{f} : e_{0} \sim e_{0}$.

Proof. TFAE

• $[f] \in H_{0} = H_{0}*[e_{0}]$;
• $\tilde{f}(1) =\phi([f]) = \phi([e_{0}]) = \tilde{b_{0}}(1) = e_{0}$. Q.E.D.

# Notes on Covering Spaces 3.

Read Topology by Munkres if you need a better explanation of this post.

Notation. $I := [0, 1]$.

Path homotopy

Instead of saying two paths $f, g : [0, 1] \rightarrow X$ are path homotopic, we will write $f \simeq_{p} g$. If the path homotopy is $F$, we write $F : f \simeq_{p} g$.

Continuous map preserves path homoptopy. If $F : f \simeq_{p} g$ and $\psi : X \rightarrow Y$ is continuous, then $\psi \circ F : \psi \circ f \simeq_{p} \psi \circ g$.

Proof. Given path homotopy $F : I \times I \rightarrow X$ such that $F(s, 0) = f(s)$ and $F(s, 1) = g(s)$, we see $F(s, 0) \mapsto (\psi \circ f)(s)$ and $F(s, 1) \mapsto (\psi \circ g)(s)$ by $\psi : X \rightarrow Y$. Moreover

• $F(0, t) = f(0) = g(0) \Rightarrow (\psi \circ F)(0, t) = (\psi \circ f)(0) = (\psi \circ g)(0)$;
• $F(1, t) = f(1) = g(1) \Rightarrow (\psi \circ F)(1, t) = (\psi \circ f)(1) = (\psi \circ g)(1)$. Q.E.D.

Lifting of a continuous map

Given $f : X \rightarrow B$, we say a map $p : E \rightarrow B$ lifts $f$ if there is $h : X \rightarrow E$ such that $f = h \circ g$. In this case $h$ is called a lifting of $f$ pulled by $g$.

Recall: Lebesgue number lemma. Let $X$ be a compact metric space. Every open cover $\mathcal{C}$ of $X$ has a Lebesgue number $\delta > 0$ such that if $E \subseteq X$ and $\text{diam}(E) < \delta$, then there is $M_{\delta} \in \mathcal{C}$ such that $E \subseteq M_{\delta}$.

Intuitively, Lebesgue number lemma says that for any open covering of a compact metric space, there exists some non-negligible amount of overlaps among members. The proof is not terribly hard, but notice that the lemma is also intuitively clear since any open cover of compact metric space has a finite subcover. The reason metric is necessary is at least because the intuition requires the notion of distance.

A covering map lifts any path by a unique lifting. Let $p : (E, e_{0}) \rightarrow (B, b_{0})$ be a covering map. Then $p$ lifts any path $f : ([0, 1], 0) \rightarrow (B, b_{0})$ with a unique lifting $h : ([0, 1], 0) \rightarrow (E, e_{0})$. (The choice of lifting $h$ is dependent on $p$ and $f$.)

Proof. Exercise using Lebesgue number lemma.

We can generalize the fact that a path can be lifted by a covering map. In fact a continuous map on $I \times I$ can be lifted by a covering map with a lifting defined on $I \times I$. If this continuous map is a path homotopy, this lifting is a new path homotopy in a different space.

A  covering map lifts any continuous map on a box by a unique lifting. Let $p : (E, e_{0}) \rightarrow (B, b_{0})$ be a covering map. Then $p$ lifts any continuous map $F : I \times I \rightarrow (B, b_{0})$ with a unique lifting $H : (I \times I, 0) \rightarrow (E, e_{0})$. (The choice of lifting $H$ is dependent on $p$ and $F$.)

The lifting of a path homotopy by a covering map is a path homotopy. In above sentence, if $F$ is path homotopy, then $H$ is a path homotopy.

Proof. Exercise using the proof of above and generalize it.

Suppose that $B$ is covered by $p : E \rightarrow B$. Then we can pick $e_{0} \in E$ that maps to $b_{0}$. Fixing $e_{0}$ can let us define the lifting of a path $f \mapsto \tilde{f}$ starting from $e_{0}$.

A covering map can track back paths and path homotopy. Consider two paths $f, g : ([0, 1], 0, 1) \rightarrow (B, b_{0}, b_{1})$ with same end points such that $F : f \simeq_{p} g$.

Suppose that there is a covering map $p : (E, e_{0}) \rightarrow (B, b_{0})$. Then

• $\tilde{F} : \tilde{f} \simeq_{p} \tilde{g}$;

(In particular, $\tilde{f}(1) = \tilde{g}(1)$.)

Proof. Exercise using techniques one can develop from previous proofs.

Now, believing everything above, consider any space $B$.

Task. We want to understand $\pi_{1}(B, b_{0})$ for a fixed point $b_{0} \in B$.

Since each $f$ path has a unique lifting $\tilde{f} : [0, 1] \rightarrow E$ that starts at $e_{0}$, with previous statement, the following map is well-defined:

$\phi : \pi_{1}(B, b_{0}) \rightarrow p^{-1}(\{b_{0}\})$

$[f] \mapsto \tilde{f}(1)$.

We call this the lifting correspondence derived from the covering map $p$. Notice the following properties.

• If $E$ is path connected, then $\phi : \pi_{1}(B, b_{0}) \twoheadrightarrow p^{-1}(\{b_{0}\})$.
• If $E$ is simply connected, then $\phi : \pi_{1}(B, b_{0}) \hookrightarrow \hspace{-10pt} \rightarrow p^{-1}(\{b_{0}\})$.

Proof. Let $E$ be path connected. Pick any $e_{1} \in p^{-1}(\{b_{0}\}) \subseteq E$ so that $p(e_{1}) = b_{0}$. Since $E$ is path connected, there is a path $g : ([0, 1], 0, 1) \rightarrow (E, e_{0}, e_{1})$. Then $p \circ g : 0 \mapsto e_{0} \mapsto b_{0}, 1 \mapsto e_{1} \mapsto b_{0}$, so $[p \circ g] \in \pi_{1}(B, b_{0})$. By uniqueness of lifting, we see $g$ is the lifting of $p \circ g$ starting at $e_{0}$. Thus $\phi([p \circ g]) = g(1) = e_{1}$, as desired.

Assume that $E$ is simply connected. We now know that the lifting correspondence is surjective. To show injectivity, suppose $\phi([f]) = \phi([g])$. This is equivalent to $\tilde{f}(1) = \tilde{g}(1)$ where $\tilde{f}, \tilde{g} : ([0, 1], 0) \rightarrow (E, e_{0})$ are liftings of $f, g$. But notice that these liftings are paths in simply connected space $E$, so there exists a path homotopy $F : \tilde{f} \simeq_{p} \tilde{g}$. Then we have $p \circ F : f \simeq_{p} g$, since $p$ preserves path homotopy. Q.E.D.

Exercise. Show $\pi_{1}(x_{0}, S^{1}) \simeq \mathbb{Z}$ for any fixed point $x_{0} \in S^{1}$.

1. Since $S^{1}$ is path connected, we can choose any $x_{0}$ freely.
2. Show that the map $p : \mathbb{R} \rightarrow S^{1}$ given by $x \mapsto (\cos(2\pi x), \sin(2\pi x)$ is a covering map.
3. By 1., choose $x_{0} = (1, 0)$ so that $p(x_{0}) = 0$.
4. Show $p^{1}(\{0\}) = \mathbb{Z}$.
5. Show that the lifting correspondence $\phi : \pi_{1}(x_{0}, S^{1}) \rightarrow \mathbb{Z}$ is a group homomorphism.
6. Since $\mathbb{R}$ is simply connected, the lifting correspondence needs to be bijective and by 5., it is an isomorphism.

We fix a covering map $p : (E, e_{0}) \rightarrow (B, b_{0})$.

Covering map induces an injective homomorphism. That is,

$p_{*} : \pi_{1}(E, e_{0}) \hookrightarrow \pi_{1}(B, b_{0})$.

Proof. Let $[f] \in \ker(p_{*})$. Then $F : p \circ f \simeq_{p} b_{0}$ for some path homotopy $F : (I \times I, (0, 0)) \rightarrow (B, b_{0})$. But then the covering map $p$ lifts $F$ by a unique lifting $\tilde{F} : (I \times I, (0, 0)) \rightarrow (E, e_{0})$, which is a path homotopy such that $\tilde{F} :f \simeq_{p} e_{0}$ since $f$ and $e_{0}$ are liftings of $p \circ f$ and $b_{0}$. Thus $[f] = [e_{0}]$ Q.E.D.