7.3 Ring Homomorphisms and Quotient Rings

Definition. Let $R$ and $S$ be rings. A ring homomorphism is a map $f:R \to S$ such that

1. $f(a + b) = f(a) + f(b)$ for all $a, b \in R$ and
2. $f(ab) = f(a)f(b)$ for all $a, b \in R$. The kernel of $f$ is the preimage of $0$ in $S$. A bijective ring homomorphism is an isomorphism.

Examples.

1. The canonical quotient map $f:\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$ is a ring homomorphism. Notice that the kernel is the set of even integers and the fiber of $1$ is the set of odd integers.
2. More generally, any canonical quotient map $R \to R/I$ where $I$ is an ideal (we will ignore the fact that we haven’t defined an ideal yet) is clearly a homomorphism.

Any ring homomorphism gives rise to two natural subrings. The proof is obvious so we omit the proof.

Proposition. Let $f:R \to S$ be a ring homomorphism. Then the image of $f$ is a subring of $S$ and the kernel of $f$ is a subring of $R$.

Definition. Let $R$ be a ring and $I$ be a subset of $R$. We say that $I$ is a left ideal of $R$ provided that $rx + sy \in I$ for all $r,s\in R$ and $x, y\in I$. We say that $I$ is a right ideal of $R$ provided that $xr + ys \in I$ for all $r,s\in R$ and $x, y\in I$. If $I$ is both a left and right ideal, it is called a two-sided ideal (or, for short, an ideal).

The notions of left, right, and two-sided ideals are specifically to account for technicalities in noncommutative rings. If $R$ is commutative, we need not consider ideals so carefully.

Proposition. If $R$ is commutative, then every left or right ideal is a two-sided ideal.

This result is, of course, obvious so we omit the proof.

The whole purpose of considering ideals is to add relations between the ring elements (i.e. modding out). With the obvious choices of addition and multiplication, one can verify that $R/I$ is a ring.

Definition. Let $I$ be an ideal of $R$. Then the ring $R/I$ with the obvious choice of addition and multiplication is called the quotient ring of $R$ by (or modulo) $I$.

Examples

1. Clearly $n\mathbb{Z}$ is an ideal of $\mathbb{Z}$. Furthermore, $\mathbb{Z}$ modulo $n\mathbb{Z}$ is just the quotient ring $\mathbb{Z}/n\mathbb{Z}$.
2. The subset $I \subseteq \mathbb{Z}[x]$ of polynomials of degree at least $2$, together with the zero polynomial, is an ideal of $\mathbb{Z}[x]$. It is clear that the representatives are just the polynomials of degree at most $1$. Notice that this ring has zero divisors since $x \cdot x = 0$ in $\mathbb{Z}[x]/I$.
3. If $R$ is the ring of all functions from $X$ to a ring $A$, then the evaluation map $\phi_c:R \to A$ given by $\phi_c(f) = f(c)$ with $c \in X$ is a ring homomorphism. The kernel of $\phi_c$ is the set of functions such that $c$ is a zero. Notice that this means that the representatives are given by all possible choices of a value for $f(c)$. This shows that $R/\ker \phi_c \cong A$.
4. The map $R[x] \to R$ given by $p(x) \mapsto p(c)$ is a ring homomorphism. The kernel of this homomorphism is the set of polynomials with coefficients in $R$ that have $c$ as a root.
5. Let $J$ be an ideal of $R$ and $n \ge 2$. Then $M_n(J)$ is the kernel of the ring homomorphism $M_n(R) \to M_n(R/J)$ given by taking each entry modulo $J$. Thus, $M_n(J)$ is a two-sided ideal.
6. Let $R$ be a commutative unital ring and $G = {g_1, \ldots, g_n}$ be a finite group. The map $f:RG \to R$ given by $\sum_{i=1}^n a_i g_i \mapsto \sum_{i=1}^n a_i$ is clearly ring homomorphism. This particular homomorphism is called the augmentation map and its kernel is called the augmentation ideal. By definition, the augmentation ideal is the set of elements of $RG$ whose coefficients sum to $0$. Clearly, any element of the ring can be written in this way and representatives of $RG/\ker f$ are given by the elements of $R$. So $RG/\ker f \cong R$.
7. One-sided ideals in $M_n(R)$ (with $n \ge 1$ and $R$ nontrivial) can easily be found by letting $L_j \subseteq M_n(R)$ consist of the matrices where nonzero entries may only occur in column $j$. Clearly, $L_j$ is closed under subtraction and the definition of matrix multiplication ensures $L_j$ absorbs products on the left. It is easy to see that $L_j$ fails to absorb products on the right. An analogous construction where $R_j \subseteq M_n(R)$ consists of the matrices where nonzero entries may only occur in row $j$ gives us a right ideal but not a left ideal.

Theorem (First isomorphism theorem for rings). Let $f:R \to S$ be a ring homomorphism. Then $R/\ker f \cong f(R)$.

Proof. Let $g:R/\ker f \to f(R)$ be defined by $\overline{r} \mapsto f(r)$. First, we show that this map is well-defined. Suppose $\overline{r} = \overline{s}$. Then $f(r - s) = 0$ implies $f(r) = f(s)$ as desired. For injectivity, we note that if $f(r) = 0$, then $r \in \ker f$ so $\overline{r} = 0$. For surjectivity, we note that $g(\overline{r}) = f(r)$ for any choice of $r$. Finally, $g$ being a ring homomorphism follows from $f$ being a ring homomorphism.

Theorem (Second isomorphism theorem for rings). Let $A$ be a subring and $B$ be an ideal of $R$. Then $A + B$ is a subring of $R$, $A\cap B$ is an ideal of $A$ and $(A + B)/B \cong A/(A \cap B)$.

Proof. The fact that $A + B$ is a subring of $R$ and $A \cap B$ is an ideal is obvious. Let $f:A \to (A + B)/B$. This map is surjective since any $\overline{a + b} = \overline{a}$ in $(A + B)/B$ and $f(a) = \overline{a}$. Clearly, this map is also a ring homomorphism. Furthermore, if $f(a) = 0$, then $a \in B$ so $r\in A\cap B$. Conversely, $a \in A\cap B$ implies $a \in B$ and $f(a) = 0$. So $\ker f = A \cap B$. Invoking the first isomorphism theorem implies the desired result.

Theorem (Third isomorphism theorem for rings). Let $I$ and $J$ be ideals of $R$ with $I \subseteq J$. Then $J/I$ is an ideal of $R/I$ and $(R/I)/(J/I) \cong R/J$.

Proof. First, we show that $J/I$ is an ideal of $R/I$. By definition $J/I$ consists of elements $j + I$ for some $j \in J$. Clearly, $J/I$ is closed under subtraction and, if $r + I \in R/I$, we have $(r + I)(j + I) = rj + I$. Since $J$ is an ideal of $R$, $rj\in J$ so $rj + I \in J/I$. Thus, $J/I$ is an ideal of $R/I$.

Let $f:R/I \to R/J$ be given by $r + I \mapsto r + J$. Clearly, $f$ is a ring homomorphism that is also surjective. If $f(r + I) = J$, then $r\in J$ and $r + I \in J/I$ by definition. If $r + I\in J/I$, then $r\in J$ and $f(r + I) = J$. Thus, $\ker f = J/I$ and the first isomorphism theorem implies the result.

Theorem (Correspondence theorem for rings). Let $I$ be an ideal of $R$. The correspondence $A \leftrightarrow A/I$ is an inclusion-preserving bijection between the set of subrings $A$ of $R$ that contain $I$ and the set of subrings of $R/I$. Furthermore, $A$ is an ideal of $R$ if and only if $A/I$ is an ideal of $R/I$.

Proof. First, we show that the existence of the bijection. If $A/I = B/I$ for subrings $A, B$ of $R$ containing $I$. For any $a \in A$, there is some $b \in B$ such that $a + I = b + I$. Thus, $a - b \in I$. Since $B$ contains $I$, it follows that $a - b \in B$. Thus, $a = (a - b) + b \in B$. An identical argument shows that $B \subseteq A$ so $A = B$. For surjectivity, it suffices to show that any subring $S$ of $R/I$ is of the form $A/I$ for some subring of $R$ containing $I$. Let $A$ be the preimage of $S$ under $R \to R/I$. Closure under addition follows by letting $a, b \in A$ and noting $\overline{a} + \overline{b} = \overline{a + b} \in S$ and so $a + b \in A$. An identical argument shows closure under products. So $A$ is a subring of $R$. Since $S$ contains $I$ as an element, $A$ contains $I$. Thus, the correspondence is, in fact, a correspondence as claimed. The inclusion preservation of the correspondence is immediate: if $A \subseteq B$, then clearly $A/I \subseteq B/I$.

For the final claim, we note that if $J$ is an ideal of $R$ containing $I$, we proved earlier that $J/I$ must be an ideal of $R/I$. For the converse direction, if $J/I$ is an ideal, then $\overline{r}\overline{j}=\overline{rj} \in J/I$ for any $r \in R$ and $j \in J$. So then $rj$ is in the preimage of $\overline{rj}$ under the quotient map $R \to R/I$. And, so, $rj \in J$. Thus, $J$ is absorbs products and we are done.

Definition. Let $I$ and J$ be ideals of $R$.

1. The sum $I + J$ of $I$ and $J$ is given by ${a + b \mid a \in I, b \in J}.$
2. The product $IJ$ of $I$ and $J$ is the ideal consisting of all finite sums of elements of the form $ab$ with $a\in I$ and $b \in J$.
3. For any $n\ge 1$, the $n$th power $I^n$ of $I$ is the set of all finite sums of elements of the form $a_1a_2\cdots a_n$ with each $a_i\in I$.

Clearly, each of the sets above are ideals. Notice that ${ab \mid a \in I, b\in J}$ can fail to be an ideal as it can fail to contain sums of such elements.

Exercises

Problem 2. Prove that the rings $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ are not isomorphic.

Proof. Suppose that $\mathbb{Z}[x]$ and $\mathbb{Q}[x]$ were isomorphic. Then there is some isomorphism $f:\mathbb{Q}[x] \to \mathbb{Z}[x]$.