Subrings and Ideals

When discussing groups, we explored subgroups and saw how useful they help us analyse (and even connect) algebraic objects. Could we say the same for subrings?

Let $R$ be any ring, not necessarily commutative.

Definition 1. We call $S \subseteq R$ a sub-ring of $R$ if $S$ is a ring.

Lemma 1. Fix $\emptyset \neq S \subseteq R$ . Then $S$ is a sub-ring of $R$ if and only if the following conditions hold:

For any $x,y \in S$ , $x-y \in S$ .
For any $x,y \in S$ , $xy \in S$ .

Proof. The first criterion holds if and only if $S$ is an additive subgroup of $R$ .

$(\Rightarrow)$ If $S$ is a sub-ring of $R$ , then it must be an additive subgroup of $R$ , and the first criterion holds. It must also be closed under multiplication, so the second criterion holds.

$(\Leftarrow)$ By the first criterion, $S$ is an additive subgroup of $R$ , and is thus Abelian. By the second criterion, $S$ is closed under multiplication. Since $S$ inherits the ring properties of $R$ , we have that $S$ is a sub-ring of $R$ .

Example 1. For any $k \in \mathbb N^+$ , $k\mathbb Z$ is a sub-ring of $\mathbb Z$ .

Remark 1. If we defined rings to require the multiplicative identity, then Lemma 1 no longer holds, and sub-rings become rather useless objects. This deficiency hints at us to re-think of algebraic structures more useful than the sub-ring.

Theorem 1. Let $R, S$ be rings. A map $\phi : R \to S$ is called a ring homomorphism if for any $x,y \in R$ ,

$\phi(r + s) = \phi(r) + \phi(s),\quad \phi(r \cdot s) = \phi(r) \cdot \phi(s).$

The following hold:

For any subring $K$ of $R$ , $\phi(K)$ is a subring of $S$ .
For any subring $L$ of $S$ , $\phi^{-1}(L)$ is a subring of $R$ .

In particular, $\ker(\phi) := \phi^{-1}(\{0\})$ and $\phi(L)$ have ring structures. If $\phi$ is bijective, we call $\phi$ a ring isomorphism, and we say that $R, S$ are isomorphic as rings, denoted $R \cong S$ .

Proof. Since $\phi : (R, +) \to (S, +)$ is a group homomorphism, $\phi(K)$ forms an Abelian subgroup of $S$ . Closure of $\phi(K)$ under multiplication follows from $K$ being closed under multiplication and $\phi$ being a ring homomorphism. The case for $\phi^{-1}(L)$ is similar.

Recall that given a group homomorphism $\phi : (G,+) \to (H,+)$ , there exists a canonical isomorphism $\tilde \phi : (G/{\ker(\phi)}, +) \to (\phi(G), +)$ such that

$\phi = \tilde{\phi} \circ \pi_{\phi},\quad \pi_{\phi}(x) := x +\ker(\phi).$

We call this result the first isomorphism theorem for groups. Do we get a similar property for rings?

Theorem 2. Let $\phi : R \to S$ be a ring homomorphism. Then there exists a canonical ring isomorphism $\tilde \phi : R/{\ker(\phi)} \to \phi(R)$ such that

$\phi = \tilde{\phi} \circ \pi_{\phi},\quad \pi_{\phi}(x) := x +\ker(\phi).$

So yes, the first isomorphism theorem holds for rings.

Proof. Since $\phi$ is a group homomorphism under addition, the first isomorphism theorem holds for groups. In particular, $(R/{\ker(\phi)}, +)$ forms a group, and there exists a bijection $f := \tilde{\phi} : R/{\ker(\phi)} \to \phi(R)$ .

Denote $[x] \equiv x + \ker(\phi)$ for brevity. Define the ring structure on $R/{\ker(\phi)}$ using the ring structure on $\phi(R)$ : define multiplication by

$[x] \cdot [y] := f^{-1}(f([x]) \cdot f([y])).$

Then it is not hard to verify the ring structure on $R/{\ker(\phi)}$ and hence, $f$ becomes a ring homomorphism trivially. Furthermore,

$[x] \cdot [y] = [xy]$

by the following argument:

$\begin{aligned} u \in f^{-1}( f([x]) \cdot f([y]) ) \quad &\iff \quad f([u]) = f([x]) \cdot f([y]) \\ &\iff \quad \phi(u) = \phi(x \cdot y) \\ &\iff \quad u \in [x \cdot y]. \end{aligned}$

Now the group operation

$(x + \ker(\phi)) + (y + \ker(\phi)) = (x+y) + \ker(\phi)$

works because $\ker(\phi) \leq R$ forms a normal subgroup. We have defined

$(x + \ker(\phi)) \cdot (y + \ker(\phi)) := x \cdot y + \ker(\phi),$

which works by Theorem 2.

Question 1. For what other subrings $I$ of $R$ would the definition

$(x + I) \cdot (y + I) = x \cdot y + I$

work?

Lemma 2. For $I, J \subseteq R$ , interpret

$I \odot J := \{x \cdot y : x \in I, y \in J\}.$

Then

$(x + \ker(\phi)) \odot (y + \ker(\phi)) \subseteq x \cdot y + \ker(\phi).$

Proof. Given $z_1, z_2 \in \ker(\phi)$ ,

$\begin{aligned} (x+z_1) \cdot (y+z_2) &= xy + xz_2 + z_1y + z_1z_2.\end{aligned}$

The terms after $xy$ belong to $\ker(\phi)$ since

$\begin{aligned} \phi(xz_2 + z_1y + z_1z_2) &= \phi(x) \cdot \phi(z_2) + \phi(z_1) \cdot \phi(y) + \phi(z_1)\cdot \phi(z_2) \\ &= \phi(x) \cdot 0 + 0 \cdot \phi(y) + 0 \cdot 0 \\ &= 0 + 0 + 0 \\ &= 0. \end{aligned}$

Denote $I = \ker(\phi)$ in hopes of uncovering some pattern. By adopting a set-theoretic lens, we notice that for $z_1,z_2 \in I$ and $x,y \in R$ , $xz_2 \in I$ and $z_1y \in I$ .

Theorem 3. Suppose the subring $I \subseteq R$ satisfies the following properties:

For any $x \in R$ and $z \in I$ , $xz \in I$ .
For any $y \in R$ and $z \in I$ , $zy \in I$ .

Then for any $x,y \in R$ , $(x + I) \odot (y + I) \subseteq xy + I$ .

If $I$ satisfies the first property, it is a left-ideal of $R$ .
If $I$ satisfies the first property, it is a right-ideal of $R$ .
If $I$ satisfies both properties, it is an ideal of $R$ .

Conversely, if

$(x + I) \odot (y + I) \subseteq xy + I$

for any $x,y \in R$ , then $I$ must be an ideal. As such, we can define

$(x + I) \cdot (y + I) := xy + I.$

Proof. Following the proof in Lemma 2, fix $z_1, z_2 \in I$ .

By the first property, $xz_2 \in I$ and $z_1 z_2 \in I$ .
By the second property, $z_1y \in I$ .
Since $I$ is a subring of $R$ , $xz_2 + z_1y + z_1 z_2 \in I$ .

Therefore,

$\begin{aligned} (x+z_1) \cdot (y+z_2) &= xy + \underbrace{ xz_2 + z_1y + z_1z_2 }_{\in I} \in xy + I.\end{aligned}$

Conversely, setting $y = 0$ yields the left-ideal property since

$(x + I) \odot I = (x + I) \odot (0 + I) \subseteq x\cdot 0 + I = I,$

which implies that for any $x \in R$ and $z \in I$ ,

$xz = (x+0) \cdot z \in (x+I) \odot I \subseteq I.$

Similarly, setting $x = 0$ yields the right-ideal property.

Corollary 1. Let $I$ be a sub-ring of $R$ . Then $R/I$ has a ring structure if and only if $I = \ker(\phi)$ for some ring homomorphism $\phi : R \to S$ to some ring $S$ .

Proof. The direction $(\Leftarrow)$ is immediate. For the direction $(\Rightarrow)$ , Theorem 3 comes from $I$ being an ideal, so that the map $\phi : R \to R/I$ defined by $\phi(r) = r + I$ is a well-defined ring homomorphism with kernel $I$ .

The first isomorphism theorem helps us prove the Chinese remainder theorem, and we will do that as a guided exercise. For now, there are many ideal-based sub-structures worth discussing in the parlance of ring theory. We will explore these new vocabularies next time.

—Joel Kindiak, 22 Jan 26, 1834H

KindiakMath

Subrings and Ideals

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Subrings and Ideals

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