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 , the multiplicative identity and . Thus, we do not consider any zero rings.
- We exclude from zero-divisors.
- By , we mean .
- We frequently write for the zero ideal .
- Under any ring map, .
- 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 .
- If is a collection of subsets of , we define .
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 , the following maps are bijections , that are inverses each other:
Moreover, recall that above maps preserves order and primality.
However, the correspondence does not necessarily preserve maximality. Consider .
Recall that the units of form an abelian group under multiplication. We write to denote this group.
Recall that the following are equivalent ways to say that a ring is a field:
- are the only ideals in ;
- Any nontrivial ring map from is injective.
Recall that given a finite collection of ring maps from a ring , namely with , we can construct a “larger” ring map by
Convention. If we are given only one collection of ring maps , we just refer to as the ring map to the direct product. If we are given ideals , then to interpret , we assume each is the projection .
Moreover, this construction essentially shows that to check if is a ring map, it is enough to check if the map is a ring map for all .
Recall that the following are equivalent ways to say that a proper ideal is a prime ideal in a ring .
- implies or ;
- is an integral domain.
We write to mean the set of all the prime ideals of .
Immediately, is an integral domain iff .
Recall that the following are equivalent ways to say that a proper ideal is a maximal ideal in a ring .
- There is no proper ideal that contains ;
- is an field.
We write to mean the set of all the maximal ideals of .
Immediately, maximal ideals are prime ideals (i.e., ).
Existence of a maximal ideal around a proper ideal. For any ideal , there is a maximal ideal .
Proof. Let be a ring (recall that must have at least two elements). Let be the set of all proper ideals that contains . Consider as a poset (partially ordered set) with . Notice that
- , so is nonempty;
- any chain has an upperbound, namely .
By Zorn’s lemma, there is a maximal element . This is a maximal ideal that contains . Q.E.D.
A ring with only one maximal ideal is called a local ring (we will write is as a pair )with the residue field .
How to find a local ring? Let be a ring. The following are equivalent.
- is a local ring.
- is a maximal ideal and .
- is a proper ideal such that .
Proof. Suppose that is a local ring. For any , if , then , so , which is a contradiction, so , showing .
Assume the second condition. Consider an element of the ring . Then . Thus, we can write . Thus . This implies there is such that , so .
Assume the third condition. If is any proper ideal, then . In particular, any maximal ideal must be contained in . Thus this must be the unique maximal ideal of , so is a local ring. Q.E.D.
If a ring has a finite number of maximal ideals, we say is semi-local.
In a PID, every nonzero prime ideal is maximal.
Proof. Let be a prime ideal in a PID . Suppose that is not a maximal ideal.
Take a maximal ideal . Then for some . Since , the primality of forces , so for some . But then implies , so , a contradiction. Q.E.D.
Given a ring , denote , or , the set of all the nilpotent elements of , and call it the nilradical of .
The nilradical is the intersection of the prime spectrum. For any ring , we have
In particular, the nilradical is an ideal.
Proof. Denote and .
Let . Then for any prime ideal , we have , which implies . Thus , showing .
For the other inclusion, let . Suppose that . Then we consider , the set of ideals in which the following property is satisfied:
if , then .
Notice the following:
- (in particular, nonemtpy);
- any chain is bounded above by (with inclusion being partial ordering).
Thus, we can apply Zorn’s lemma to achieve a maximal element . We study as follows.
If elements of the ring , then cannot be in by the maximality. Thus, we can take such that
Then , so , and this implies that .
From our study, we proved: implies . Thus we see . Then , which contradicts the assumption. We showed , showing . Q.E.D.
“Removing” nonzero nilpotent elements. Given a ring , the nilradical of the quotient is trivial.
Proof. Consider any with . Then if , then , so we have such that . This implies that , which means . Q.E.D.
Now, for a given ring , we define the Jacobson radical, which is written as , or , to be the intersection of the maximal spectrum. That is,
Just like the nilradical, there is another way to descirbe the Jacobson radical.
A characterization of the Jacobson radical. Let be a ring. Then
Proof. Let . Take an arbitrary . Suppose that . Then we can take . But since , we have , a contradiction. Thus , showing .
Conversely, let for all . Suppose . Then we can take . Then , so we can write for some . Thus gives a contradiction since is a unit, showing . Q.E.D.
A finite collection of ideals are coprime if .
If given , we say the collection of ideal is pairwise coprime.
To oppress the difficulty of notation that naturally arises, we write for the projection .
Recall the product of ideal is defined as . Considering generators, we always have . We can also write the elements of the product as follows:
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. .
Product operation commutes with generating operation. Let be ideals and suppose that and for some . Then
Proof. Since and , we have , so .
Let and . Then we can write and , where and and . By a direct computation, we have , so , 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.
Proof. We have , so . Similarly, we have , so .
We have . But elementwise, we see , so . Q.E.D.
Pairwise coprimaity makes the product sweep the intersection. If are finitely many coprime ideals, then
Proof. Due to the previous remark, it is enough to show that .
We first study the case .
Suppose we are given only two coprime ideals, namely . Since , we can take and such that . Then .
Our study showed that the desired statement is true for . It is clear that the statement is also true for . We try induction, so assume that for , the statement is true. Suppose . Then
It is clear what to attempt next. We wonder if and are coprime.
Since for . Thus, we can take and such that for . This implies that
Thus , as we conjectured. We now can proceed the final step of our induction, continuing the previous computation:
The pairwise coprimality of ideals in can be neatly characterized as the surjectivity of the ring map . This is the famous Chinese remainder theorem.
The Chinese remainder lemma. A finite collection of ideals is pairwise coprime if and only if is surjective.
Proof. Suppose that is pairwise coprime. Take any , and we want to show the fiber of is not empty.
First, we study an easier case and ask if the fiber of is nonempty.
Since for all , we can take and such that . Then let and see that
From our study, we constructed in the fiber of . But now we immediately see that is in the fiber of . This finishes showing the surjectivity of .
Conversely, suppose that the ring map is surjective. Let , we need to show .
Our previous study seems suggestive, so we take in the fiber of . Then
Thus we have , as desired. Q.E.D.
Notice that , so we always have
Recall that for any ring map , surjectivity of is characterized by . Thus we can state the Chinese remainder in the following way.
The Chinese remainder theorem. For a finite collection of ideals in a ring , the following are equivalent.
Thus, if any of above is true, the product sweeps the intersection, so .
A trivial application of the Chinese remainder. Let be a field and be a separable polynomial. Then we can write over an algebraically closed field and apply the Chinese remainder to have
where the right-hand side is a direct product of fields and a -algebra.