8.2: Ring Homomorphisms

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Definition: Ring Homomorphism

Let \(R\) and \(S\) be rings. A ring homomorphism is a map \(\phi:R\to S\) satisfying

\(\phi(a+b)=\phi(a)+\phi(b)\)
\(\phi(ab)=\phi(a)\phi(b)\)

for all \(a,b\in R\). The kernel of \(\phi\) is defined via \(\ker(\phi)=\{a\in R\mid \phi(a)=0\}\). If \(\phi\) is a bijection, then \(\phi\) is called an isomorphism, in which case, we say that \(R\) and \(S\) are isomorphic rings and write \(R\cong S\).

Example \(\PageIndex{1}\)

For \(n\in \mathbb{Z}\), define \(\phi_n:\mathbb{Z}\to \mathbb{Z}\) via \(\phi_n(x)=nx\). We see that \(\phi_n(x+y)=n(x+y)=nx+ny=\phi_n(x)+\phi_n(y)\). However, \(\phi_n(xy)=n(xy)\) while \(\phi_n(x)\phi_n(y)=(nx)(ny)=n^2xy\). It follows that \(\phi_n\) is a ring homomorphism exactly when \(n\in\{0,1\}\).
Define \(\phi:\mathbb{Q}[x]\to \mathbb{Q}\) via \(\phi(p(x))=p(0)\) (called evaluation at 0). It turns out that \(\phi\) is a ring homomorphism, where \(\ker(\phi)\) is the set of polynomials with 0 constant term.

Problem \(\PageIndex{1}\)

For each of the following, determine whether the given function is a ring homomorphism. Justify your answers.

Define \(\phi:\mathbb{Z}_4\to \mathbb{Z}_{12}\) via \(\phi(x)=3x\).
Define \(\phi:\mathbb{Z}_{10}\to \mathbb{Z}_{10}\) via \(\phi(x)=5x\).
Let \(\displaystyle S=\left\{\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\mid a, b\in \mathbb{R}\right\}\). Define \(\phi:\mathbb{C}\to S\) via \(\displaystyle \phi(a+ib)=\begin{pmatrix}a & b\\ -b & a\end{pmatrix}\).
Let \(\displaystyle T=\left\{\begin{pmatrix}a & b\\ 0 & c\end{pmatrix}\mid a, b\in \mathbb{Z}\right\}\). Define \(\phi:T\to \mathbb{Z}\) via \(\displaystyle \phi\left(\begin{pmatrix}a & b\\ 0 & c\end{pmatrix}\right)=a\).

Theorem \(\PageIndex{1}\): Kernel Subring

[thm:kernelsubring] Let \(\phi:R\to S\) be a ring homomorphism.

\(\phi(R)\) is a subring of \(S\).
\(\ker(\phi)\) is a subring of \(R\).

Problem \(\PageIndex{2}\)

Suppose \(\phi:R\to S\) is a ring homomorphism such that \(R\) is a ring with 1, call it \(1_R\). Prove that \(\phi(1_R)\) is the multiplicative identity in \(\phi(R)\) (which is a subring of \(S\)). Can you think of an example of a ring homomorphism where \(S\) has a multiplicative identity that is not equal to \(\phi(1_R)\)?

Theorem \(\PageIndex{1}\)(b) states that the kernel of a ring homomorphism is a subring. This is analogous to the kernel of a group homomorphism being a subgroup. However, recall that the kernel of a group homomorphism is also a normal subgroup. Like the situation with groups, we can say something even stronger about the kernel of a ring homomorphism. This will lead us to the notion of an ideal.

Theorem \(\PageIndex{2}\)

Let \(\phi:R\to S\) be a ring homomorphism. If \(\alpha\in\ker(\phi)\) and \(r\in R\), then \(\alpha r, r\alpha\in \ker(\phi)\). That is, \(\ker(\phi)\) is closed under multiplication by elements of \(R\).