8.2: Ring Homomorphisms
- Page ID
- 100764
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)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 \(R\cong S\).
- 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.
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 \(\displaystyle \phi\left(\begin{pmatrix}a & b\\ 0 & c\end{pmatrix}\right)=a\).
[thm:kernelsubring] Let \(\phi:R\to S\) be a ring homomorphism.
- \(\phi(R)\) is a subring of \(S\).
- \(\ker(\phi)\) is a subring of \(R\).
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.
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 \(R\).