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My question is:

Show that φ$\left\{ \begin{array}{c l} T &\mbox{$\longrightarrow\mathbb{R}$} \\ \begin{pmatrix} a & 0\\ b & c \end{pmatrix} & \mbox{$\longrightarrow$ $c$} \end{array}\right.$

is a ring-homomorphism.

With T being $\left\{\begin{pmatrix} a & 0\\ b & c \end{pmatrix}\bigg|a,b,c,d \in\mathbb{R}\right\}$

I get that the definition of a Homomorphism is when a ring 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 cant seem to prove the original question. Any help will be appreciated.

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The mapping $\varphi$ associates to a lower triangular $2\times 2$-matrix its lower right coefficient.

Take two such matrices: $$T=\begin{pmatrix}a&0\\b&c\end{pmatrix},\quad T'=\begin{pmatrix}a'&0\\b'&c'\end{pmatrix},$$ compute $T+T'$, $TT'$ and check $\;\varphi(T+T')=\varphi(T)+\varphi (T')$, $\;\varphi(TT')=\varphi(T)\varphi (T')$.

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  • $\begingroup$ when stating $\;\varphi(T+T')$, are you just saying $\begin{pmatrix}a+a'&0\\b+b'&c+c'\end{pmatrix}$?? $\endgroup$ – AFraggers Nov 10 '16 at 23:57
  • $\begingroup$ Exactly. $\mbox{}$ $\endgroup$ – Bernard Nov 11 '16 at 0:02
  • $\begingroup$ If thats the case then why has the question used the function φ if it had no bearing in the addition of T and T' $\endgroup$ – AFraggers Nov 11 '16 at 0:06
  • $\begingroup$ Sorry I misunderstood your question. By $\varphi(T+T')$, I mean $\varphi\begin{pmatrix}a+a'&0\\b+b'&c+c'\end{pmatrix}$. $\endgroup$ – Bernard Nov 11 '16 at 0:13
  • $\begingroup$ Oh that makes sense now. So for checking $\varphi(T+T')$ would you start by saying $\varphi(T+T')=\varphi(\begin{pmatrix}a&0\\b&c\end{pmatrix}+\begin{pmatrix}a'&0\\b'&c'\end{pmatrix})$ which simplifies to $\begin{pmatrix}a+a'&0\\b+b'&c+c'\end{pmatrix}$ which can be rewritten as $\varphi(T) +\varphi(T')$ $\endgroup$ – AFraggers Nov 11 '16 at 0:23

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