Our first order of the day is to state the correspondence between the ideals and subrings of *R*/*I* and those of *R*. This is totally analogous to the case of groups.

Theorem. Let I be an ideal of R. There are 1-1 correspondences:which preserve inclusion: i.e. on the LHS if and only if on the RHS. The same holds for the ideals.

[ Correspondence for the lattices of ideals and subrings between those of *R* (containing *I*), and those of *R*/*I*. In general, an ideal can be contained in a subring, but not vice versa, unless the ideal is the whole ring *R* (why?). ]

**Sketch of Proof**.

The proof is similar to the case of groups. For *A* such that is an ideal or subring of *R*, let the corresponding element on the RHS be *A*/*I*. If *A* is an ideal (resp. subring) of *R*, then *A*/*I* is an ideal (resp. subring) of *R*/*I*.

Conversely, if is a subring or ideal, then let the corresponding *A* on the LHS be . Prove that *A* contains *I* and *A*/*I* = *B*. Now fill up the proof by showing that by composing these two arrows in either order, we get the identity map on either the LHS or RHS. ♦

**Corollary**

The above correspondence preserves intersections and sums of ideals, but not products.

**Proof**.

Suppose we have ideals on the LHS, indexed by *i*. Each ideal then corresponds to on the RHS.

*Intersection.*From the lattice structure on the LHS, is the largest ideal (containing*I*) such that for each*i*. Translating it to the RHS, the corresponding*J*/*I*is also the largest ideal which contained in all . Hence . This is yet another example of a “universal proof” we alluded to earlier.*Sum*. Similar to intersection, except now we’re dealing with the smallest ideal containing every .*Product*. See the following example (the gist is that even if , their intersection may not contain*I*).

**Examples**

- Let 12
**Z**be an ideal in**Z**, which gives the quotient ring**Z**/12. Now the ideals of**Z**which contain 12**Z**correspond to the positive divisors of 12, so we have**Z**, 2**Z**, 3**Z**, 4**Z**, 6**Z**, 12**Z**. These give corresponding ideals of**Z**/30: <1>, <2>, <3>, <5>, <6>, <10>, <15>, <30>. E.g. the ideal <5> = {0, 5, 10, 15, 20, 25} modulo 30.

- Intersection on the LHS gives which corresponds to <4> ∩ <6> = <2> on the RHS.
- Product on the LHS gives 4
**Z**× 6**Z**= 24**Z**which doesn’t contain 12**Z**, so it doesn’t correspond to anything on RHS.

- Consider the homomorphism
**R**[*x*,*y*] →**R**[*x*], where**R**[*x*,*y*] (resp.**R**[*x*]) is the ring of all polynomials in*x*,*y*(resp.*x*) with real coefficients. The homomorphism is given by*f*(*x*,*y*) →*f*(*x*, 0). It’s quite easy to show that the ideal is*I*= <*y*>, the set of all multiples of*y*. So we have an isomorphism*f*:**R**[*x*,*y*]/<*y*> →**R**[*x*].- Consider the subring
**R**of**R**[*x*]. This corresponds to the pullback , which has basis given by 1, together with*x*, where^{m}y^{n}*m*≥ 0 and*n*> 0. - Consider the subring
**R**[*x*^{2}] of**R**[*x*], which comprises of polynomials whose terms are all even powers of*x*. The pullback gives a subring of**R**[*x*,*y*] which has basis given by*x*, where^{m}y^{n}*n*> 0, or (*m*,*n*) = (2*k*, 0) for some non-negative integer*k*.

- Consider the subring

## Case of Commutative Rings

Let’s assume our *R* is commutative now. We wish to derive conditions for the ring quotient *R*/*I* to be an integral domain or a field. First, a result:

Proposition. A commutative ring R≠{0} is a field if and only if it has only two ideals: {0} and itself.

**Proof**.

We already know that a division ring has no non-trivial ideals. For the converse, suppose *R* has only two ideals {0} and itself. Let . Then <*x*> is an ideal which is non-zero, hence <*x*>=*R*. This implies 1 is contained in <*x*>, so there exists an element *y* in *R* such that *xy* = 1. ♦

It is not true that a (possibly non-commutative) ring *R* ≠ {0} is a division ring if and only if it has only two ideals. For a counter-example, the ring of *n* × *n* real matrices has no non-trivial ideals (which we’ll see in a later post). If *n* > 1, this is clearly not a division ring.

Theorem. Let R be a commutative ring and I ≠ R be an ideal.

- R/I is an integral domain if and only if . We call such an I a
prime ideal.- R/I is a field if and only if there is no ideal J of R, , i.e. no ideal is strictly contained between I and R. We call such an I a
maximal ideal.

**Proof**. For (1), saying *R*/*I* is an integral domain means

.

Unravelling the definition, this is equivalent to saying .

For (2), by the proposition above, *R*/*I* is a field if and only if its only ideals are {0} and itself. By the correspondence theorem above, this is equivalent to saying the only ideals *J*, are *J*=*I* or *J*=*R*. ♦

**Examples**

- Consider
**Z**again. Its only ideals are*n***Z**, where*n*≥ 0.- If
*n*=0, then {0} is a prime ideal which is not maximal. - If
*n*=1, then <1> =**Z**is neither prime nor maximal, by definition. - For
*n*>1, whenever*n*is composite,**Z**/*n*is not an integral domain; but if*n*is prime,**Z**/*n*is a field. This shows that*n***Z**is prime iff*n***Z**is maximal iff*n*is prime.

- If
- Take the ring
**R**[*x*,*y*], set of polynomials in*x*,*y*with real coefficients. For any real numbers*a*and*b*, consider the ideal*I*= <*x*–*a*,*y*–*b*>. Replacing*x*→*x*–*a*and*y*→*y*–*b*, it’s easy to see any*f*(*x*,*y*) can be written in the form for some real*c*. Hence**R**[*x*,*y*]/*I*is isomorphic to**R**and*I*is maximal. - On the other hand, take the ideal
*I*= <*y*>. We saw earlier that**R**[*x*,*y*]/*I*is isomorphic to**R**[*x*] which is an integral domain but not a field. Hence, <*y*> is a prime ideal which is not maximal. Indeed, <*y*> is strictly contained in <*x*,*y*> so of course it’s not maximal.

## Chinese Remainder Theorem

Throughout this section, let *I* and *J* be ideals of a commutative ring *R*. Consider the natural map *R* → (*R*/*I*) × (*R*/*J*) via projecting *r* to (*r*+*I*, *r*+*J*). Clearly the kernel comprises of all *r* in both *I* and *J*, i.e. ker = *I* ∩ *J*. By the first isomorphism theorem, we get a monomorphism:

Now assume *I* ∩ *J* = {0}, i.e. we get a monomorphism *R* → (*R*/*I*) × (*R*/*J*). The question we want to devote ourselves to is:

When’s

R→ (R/I) × (R/J) an isomorphism?

To make our lives ahead easier, let us define some new notation.

Definition. If I is an ideal of R, and a, b are arbitrary elements of R, we write a ≡ b (mod I) to mean . Since I is an additive subgroup, we see that this is an equivalence relation. Further:

- if a ≡ b (mod I) and c ≡ d (mod I), then a+c ≡ b+d (mod I) and ac ≡ bd (mod I).
The proof for these properties is identical to that for the corresponding properties of modular arithmetic.

Back to the question.

For *R* → (*R*/*I*) × (*R*/*J*) to be an isomorphism, there must be an which maps to (0, 1), i.e. *x* ≡ 0 (mod *I*) and *y* ≡ 1 (mod *J*). By the same token, we get *y* such that *y* ≡ 1 (mod *I*) and *y* ≡ 0 (mod *J*). But now look at *x*+*y*. We have *x*+*y* ≡ 1 (mod *I*) and (mod *J*), so *x*+*y* ≡ 1 (mod *I*∩*J*). Since *I*∩*J* = {0}, *x*+*y*=1.

Thus the ideal *I*+*J* contains 1, and must be the whole ring *R*! This condition turns out to be sufficient too.

Definition. If I, J are ideals of R such that I+J = R, then the two ideals are said to becoprime. [ This is motivated by the fact that inZ, ideals mZ+ nZ=Ziff (m, n) = 1. ]

Chinese Remainder Theorem. If I, J are coprime ideals of R such that I ∩ J = {0}, then R → (R/I) × (R/J) is an isomorphism.

**Proof**.

It suffices to show the map is surjective. The process is essentially the reverse of the above. Since *I*+*J* contains 1, pick , *x*+*y* = 1. Taking this expression mod *I* and *J*, we get *x* ≡ 1 (mod *J*) and *y* ≡ 1 (mod *I*). In a nutshell:

Now for any *a*, *b* in *R*, the element *bx*+*ay* maps to *a* mod *I* and *b* mod *J* respectively. This shows that *R* → (*R*/*I*) × (*R*/*J*) is an isomorphism. ♦

**Example**

Consider the ring **Z**/35, with ideals *I*=<7> and *J*=<5>. Then the above conditions are satisfied:

*I*∩*J*= {0} since the only integers divisible by 7 and 5 are multiples of 35;*I*+*J*=**Z**/35 since 21 + 15 ≡ 1 (mod 35), i.e. there exist ,*x*+*y*=1.

By the above theorem, we get an isomorphism

which is the classical Chinese Remainder Theorem for number theory. Furthermore, the above proof gives us a concrete method of constructing a solution to , : just take *n* = *ay*+*bx* = 15*a* + 21*b*.