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Prove that a given function is a ring homomorphism and describe its kernel


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.3 Exercise 7.3.7

Solution: Let $A,B \in R$ be arbitrary, with $A = \begin{bmatrix} a_1 & b_1 \\ 0 & d_1 \end{bmatrix}$ and $B = \begin{bmatrix} a_2 & b_2 \\ 0 & d_2 \end{bmatrix}$.

First we show that $\varphi$ is a ring homomorphism.\begin{align*}\varphi(A+B) =&\ \varphi \left( \begin{bmatrix} a_1 & b_1 \\ 0 & d_1 \end{bmatrix} + \begin{bmatrix} a_2 & b_2 \\ 0 & d_2 \end{bmatrix} \right)\\ =&\ \varphi \left( \begin{bmatrix} a_1 + a_2 & b_1 + b_2 \\ 0 & d_1 + d_2 \end{bmatrix} \right)\\
=&\  (a_1+a_2, d_1+d_2) = (a_1,d_1) + (a_2,d_2)\\ =&\ \varphi\left( \begin{bmatrix} a_1 & b_1 \\ 0 & d_1 \end{bmatrix} \right) + \varphi \left( \begin{bmatrix} a_2 & b_2 \\ 0 & d_2 \end{bmatrix} \right)\\=&\ \varphi(A) + \varphi(B)\end{align*} \begin{align*}\varphi(AB) =&\ \varphi \left( \begin{bmatrix} a_1 & b_1 \\ 0 & d_1 \end{bmatrix} \begin{bmatrix} a_2 & b_2 \\ 0 & d_2 \end{bmatrix} \right)\\=&\ \varphi \left( \begin{bmatrix} a_1a_2 & a_1b_2 + b_1d_2 \\ 0 & d_1d_2 \end{bmatrix} \right)\\=&\ (a_1a_2, d_1d_2) = (a_1,d_1) (a_2,d_2)\\=&\ \varphi \left( \begin{bmatrix} a_1 & b_1 \\ 0 & d_1 \end{bmatrix} \right) \cdot \varphi \left( \begin{bmatrix} a_2 & b_2 \\ 0 & d_2 \end{bmatrix} \right)\\=&\ \varphi(A) \cdot \varphi(B)\end{align*}Thus $\varphi$ is a ring homomorphism.

Let $(a,d) \in \mathbb{Z}^2$; clearly then $\varphi\left( \begin{bmatrix} a & 0 \\ 0 & d \end{bmatrix} \right) = (a,d)$. Thus $\varphi$ is surjective.

Now I claim that $$\mathsf{ker}\ \varphi = \left\{ \begin{bmatrix} 0 & b \\ 0 & 0 \end{bmatrix} \bigg| b \in \mathbb{Z} \right\}.$$ If $\begin{bmatrix} a & b \\ 0 & d \end{bmatrix} \in \mathsf{ker}\ \varphi$, then $$\varphi \left( \begin{bmatrix} a & b \\ 0 & d \end{bmatrix} \right) = (a,d) = (0,0);$$ thus $a = d = 0$. Moreover, we have $$\varphi\left( \begin{bmatrix} 0 & b \\ 0 & 0 \end{bmatrix} \right) = (0,0).$$


Linearity

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