Introduction to Ring Theory (5)

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:

$\begin{aligned}\{ S : I\subseteq S \subseteq R, S\text{ subring of } R\} &\leftrightarrow\{ T : T \text{ subring of } R/I\} \\ \{ J : I\subseteq J \subseteq R, J\text{ ideal of } R\} &\leftrightarrow \{ K : K \text{ ideal of } R/I\}\end{aligned}$

which preserve inclusion: i.e. $S\subseteq S'$ on the LHS if and only if $T \subseteq T'$ 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 $I \subseteq A \subseteq R$ 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 $B\subseteq R/I$ is a subring or ideal, then let the corresponding A on the LHS be $A = f^{-1}(B)$ . 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 $J_i \supseteq I$ on the LHS, indexed by i. Each ideal then corresponds to $J_i/I$ on the RHS.

Intersection. From the lattice structure on the LHS, $J = \cap_i J_i$ is the largest ideal (containing I) such that $J\subseteq J_i$ for each i. Translating it to the RHS, the corresponding J/I is also the largest ideal which contained in all $J_i/I$ . Hence $J/I = \cap_i J_i/I$ . 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 $J_i$ .
Product. See the following example (the gist is that even if $I_1, I_2\supseteq I$ , their intersection may not contain I).

Examples

Let 12Z be an ideal in Z, which gives the quotient ring Z/12. Now the ideals of Z which contain 12Z correspond to the positive divisors of 12, so we have Z, 2Z, 3Z, 4Z, 6Z, 12Z. 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 $4\mathbf{Z}\cap 6\mathbf{Z} = 2\mathbf{Z}$ which corresponds to <4> ∩ <6> = <2> on the RHS.
Product on the LHS gives 4Z × 6Z = 24Z which doesn’t contain 12Z, 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 $f^{-1}(\mathbf{R}) = \mathbf{R} + \left<y\right> \subset \mathbf{R}[x,y]$ , which has basis given by 1, together with x^myⁿ, where m ≥ 0 and n > 0.
- Consider the subring R[x²] 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^myⁿ, where n > 0, or (m, n) = (2k, 0) for some non-negative integer k.

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 $x\in R-\{0\}$ . 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 $a, b\in R, ab\in I \implies a\in I \text{ or } b\in I$ . We call such an I a prime ideal.

R/I is a field if and only if there is no ideal J of R, $I\subsetneq J\subsetneq R$ , 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

$(a+I)(b+I) = 0+I \implies a+I=0+I \text{ or } b+I=0+I$ .

Unravelling the definition, this is equivalent to saying $ab\in I \implies a\in I\text{ or }b\in I$ .

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, $I\subseteq J \subseteq R$ are J=I or J=R. ♦

Examples

Consider Z again. Its only ideals are nZ, where n ≥ 0.
1. If n=0, then {0} is a prime ideal which is not maximal.
2. If n=1, then <1> = Z is neither prime nor maximal, by definition.
3. 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 nZ is prime iff nZ is maximal iff n is prime.
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 $f(x,y) = (x-a)k_1(x,y) + (y-b)k_2(x,y) + c\in I+c$ 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:

$R/(I\cap J) \hookrightarrow (R/I) \times (R/J).$

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 $a-b\in I$ . 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 $x\in R$ 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 be coprime. [ This is motivated by the fact that in Z, ideals mZ + nZ = Z iff (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 $x\in I, y\in J$ , x+y = 1. Taking this expression mod I and J, we get x ≡ 1 (mod J) and y ≡ 1 (mod I). In a nutshell:

$x\equiv 0 \pmod I, \ x\equiv 1\pmod J,\quad y\equiv 1\pmod I, \ y\equiv 0\pmod J.$

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 $x\in I, y\in J$ , x+y=1.

By the above theorem, we get an isomorphism

$\mathbf{Z}/35 \cong \mathbf{Z}/I \times \mathbf{Z}/J \cong \mathbf{Z}/7 \times \mathbf{Z}/5,$

which is the classical Chinese Remainder Theorem for number theory. Furthermore, the above proof gives us a concrete method of constructing a solution to $n\equiv a\pmod 7$ , $n\equiv b\pmod 5$ : just take n = ay+bx = 15a + 21b.