Quotient Rings and Homomorphisms

Quotient Homomorphisms

In the congruence classes $\mathbb{Z}/n\mathbb{Z}$ and $F[x]/(p(x))$, we have the following natural homomorphisms:

$$\begin{array}{rl}\varphi :\mathbb{Z}& \to \mathbb{Z}/n\mathbb{Z}\\ \varphi (a)& =a+n\mathbb{Z}\end{array}$$

and

$$\begin{array}{rl}\psi :F[x]& \to F[x]/(p(x))\\ \psi (f(x))& =f(x)+(p(x)).\end{array}$$

These generalize to the following:

Proposition

Let $R$ be a commutative unitary ring and $I\subseteq R$ be an ideal. Then the map

$$\begin{array}{rl}\varphi :R& \to R/I\\ \varphi (a)& =a+I\end{array}$$

is a ring homomorphism.

The proof follows directly from proposition of congruence classes operations.

Remark

The kernel of $\varphi$ is the ideal $I$ itself. The image is the entire quotient ring $R/I$.

Kernel as Ideal

If $\varphi :R\to S$ is a ring homomorphism, we have defined kernel of $\varphi$. A notable property of the kernel is that it is an ideal of $R$.

Theorem

Let $\varphi :R\to S$ be a ring homomorphism. Then the kernel of $\varphi$ is an ideal of $R$.

Proof

Using this compact criterion for ideals, we need to show that:

1. for any $a,b\in \ker (\varphi )$, $a-b\in \ker (\varphi )$
2. for any $r\in R$, $a\in \ker (\varphi )$, $ra\in \ker (\varphi )$

The first follows from the fact that $\varphi (a-b)=\varphi (a)-\varphi (b)=0-0=0$. The second follows from the fact that $\varphi (ra)=\varphi (r)\varphi (a)=\varphi (r)0=0$. $■$

First Isomorphism Theorem

Let $\varphi :R\to S$ be a ring homomorphism. We have stated the image $\varphi (R)$ of a ring homomorphism is a subring of $S$. We just established that the $\ker (\varphi )$ is an ideal of $R$. As the above remark suggests, we should expect that the quotient ring $R/\ker (\varphi )$ is isomorphic to the image $\varphi (R)$. This is indeed true, and is known as the first isomorphism theorem.

Theorem

Let $\varphi :R\to S$ be a ring homomorphism. Let $K=\ker (\varphi )$. Then we have a well-defined map:

$$\begin{array}{rl}\psi :R/K& \to \varphi (R)\\ \psi (a+K)& =\varphi (a)\end{array}$$

is a ring isomorphism.

Proof

We first check the well-definedness of $\psi$. Suppose $a+K=b+K$, then $a-b\in K$, so $\varphi (a)-\varphi (b)=\varphi (a-b)=0$, which implies $\varphi (a)=\varphi (b)$.

Then we show $\psi$ is a homomorphism:

1. Addition: for any $a,b\in R$, we have

$$\begin{array}{rl}\psi (a+K)+\psi (b+K)& =\varphi (a)+\varphi (b)\\ & =\varphi (a+b)\\ & =\psi ((a+b)+K).\end{array}$$

2. Multiplication: for any $a,b\in R$, we have

$$\begin{array}{rl}\psi (a+K)\cdot \psi (b+K)& =\varphi (a)\cdot \varphi (b)\\ & =\varphi (ab)\\ & =\psi ((ab)+K).\end{array}$$

Finally, we show $\psi$ is a bijection.

1. Injective: Recall kernel and injectivity, it s sufficient to show $\ker (\psi )=0+K$. We have:

$$\begin{array}{rl}\ker (\psi )& =\{a+K\in R/K|\psi (a+K)=\varphi (a)=0\}\\ & =\{a+K\in R/K|a\in K\}\\ & =0+K.\end{array}$$

2. Surjective: For any $b\in \varphi (R)$, we can find $a\in R$ such that $\varphi (a)=b$. Then we can take $\psi (a+K)=\varphi (a)=b$. Thus $\psi$ is surjective. $■$