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My question is:

$\varphi=\left\{ \begin{array}{c l} T &\mbox{$\longrightarrow R$} \\ \begin{pmatrix} a & 0\\ b & c \end{pmatrix} & \mbox{$\longrightarrow$ $\begin{pmatrix} a & b\\ 0 & c \end{pmatrix}$} \end{array}\right.$

Show that this function is not a ring homomorphism

I get that the definition of $f$ a ring homomorphism is when satisfies

$f(a + b) = f(a) + f(b)$ for all $a$ and $b$ in $\mathbb{R}$

and $f(ab) = f(a) f(b)$ for all $a$ and $b$ in $\mathbb{R}$

But I can't see how to prove the original question. Any help will be appreciated.

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  • $\begingroup$ Have you checked if the given map satisfies those two properties? $\endgroup$ – arctic tern Nov 14 '16 at 5:17
  • $\begingroup$ Ive tried and got it to fail on the dirst condition i stated above. Is that correct?? @arctictern $\endgroup$ – AFraggers Nov 14 '16 at 5:30
  • $\begingroup$ The map does not fail the first condition. The map is in fact additive. It's the second condition concerning multiplicativity that fails. $\endgroup$ – arctic tern Nov 14 '16 at 5:31
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I will call $f$ to your $\varphi$. Look that if you have:

$$ A= \begin{pmatrix} 1 & 0\\ 1 & 0 \end{pmatrix}, B= \begin{pmatrix} 0 & 0\\ 1 & 1 \end{pmatrix} $$

Then,

$$\begin{pmatrix} 1 & 0\\ 1 & 0 \end{pmatrix}*\begin{pmatrix} 0 & 0\\ 1 & 1 \end{pmatrix}= \begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}$$

So

$$f(A*B)=\begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix}$$

But:

$$f(A)*f(B)=\begin{pmatrix} 1 & 1\\ 0 & 0 \end{pmatrix}*\begin{pmatrix} 0 & 1\\ 0 & 1 \end{pmatrix}= \begin{pmatrix} 0 & 2\\ 0 & 0 \end{pmatrix}$$

Making a contradiction to supposing $f$ is an homomorphism.

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