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Let $R=M(2,\mathbb Q)$, the ring of all $2\times 2$ matrices with rational entries.

I have a function $f:A \rightarrow B$, where $A$ is the subring of upper triangular matrices and $B$ is the subring of lower triangular matrices.

So $A$ is the set of all $2\times 2$ upper triangular matrices, i.e. $$x = \begin{pmatrix} a & b \\ 0 & c \\ \end{pmatrix}$$ and $B$ is the set of all $2\times 2$ lower triangular matrices, i.e. $$y = \begin{pmatrix} d & 0 \\ e & f \\ \end{pmatrix}.$$

I need to show that the function $f$ is a ring homomorphism and it is bijective, but I cannot seem to be able to. I'm trying to show that $f(xy)=f(x)f(y)$ but I cannot seem to get it to work.

Any help would be great.

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  • $\begingroup$ are you sure that this is the given function? I think that your function maps upper triangular matrices to lower triangular matrices. Is this correct? Or were you asked to define an isomorphism yourself? $\endgroup$ – Student Feb 11 '17 at 15:36
  • $\begingroup$ yes that's exactly what the function does. i need to prove that it is a ring homomorphism and show that A is isomorphic to B. $\endgroup$ – ptsgeeg Feb 11 '17 at 15:39
  • $\begingroup$ My guess is that they just denote $A$ to be the subring of upper triangular matrices and $B$ the subring of lower triangular matrices. This means that $A$ itself is not a matrix, but that the upper triangular matrix you wrote down is an element of $A$... otherwise I don't think this question makes sense... $\endgroup$ – Student Feb 11 '17 at 15:40
  • $\begingroup$ Did you got a description of the images of upper triangular matrices under $X$ or do you have to find it yourself? $\endgroup$ – Student Feb 11 '17 at 16:04
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The point is, you have not given your intended map $f$.

I will give it for you $$ f \left( \begin{bmatrix} a & c \\ 0 & b \\ \end{bmatrix} \right) = \begin{bmatrix} b & 0 \\ c & a \\ \end{bmatrix}. $$ Now just compute to see $f(x y) = f(x) f(y)$.

Alternatively, save some time and effort by noting that $$ f \left( \begin{bmatrix} a & c \\ 0 & b \\ \end{bmatrix} \right) = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \cdot \begin{bmatrix} a & c \\ 0 & b \\ \end{bmatrix} \cdot \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} $$ and $$ \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}^{2} = I. $$


Explicitly, $$f \left( \begin{bmatrix} a & c \\ 0 & b \\ \end{bmatrix} \right) f \left( \begin{bmatrix} d & e \\ 0 & f \\ \end{bmatrix} \right) = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \cdot \begin{bmatrix} a & c \\ 0 & b \\ \end{bmatrix} \cdot \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \cdot \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \cdot \begin{bmatrix} d & e \\ 0 & f \\ \end{bmatrix} \cdot \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} =\\= \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} \cdot \begin{bmatrix} a & c \\ 0 & b \\ \end{bmatrix} \cdot\begin{bmatrix} d & e \\ 0 & f \\ \end{bmatrix} \cdot \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix} = f \left( \begin{bmatrix} a & c \\ 0 & b \\ \end{bmatrix} \cdot \begin{bmatrix} d & e \\ 0 & f \\ \end{bmatrix}\right).$$

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  • $\begingroup$ I may be being very silly but I cannot get f(xy)=f(x)f(y) $\endgroup$ – ptsgeeg Feb 11 '17 at 21:05
  • $\begingroup$ for $f(xy)$ where $x=(a,c,0,b)$ and $y=(d,e,0,f)$ I get that $f(xy)$ has $-cf$ in the top left but $f(x)f(y)$ has $cf$ in the top left entry $\endgroup$ – ptsgeeg Feb 11 '17 at 21:32
  • $\begingroup$ Please see my edit. $\endgroup$ – Andreas Caranti Feb 11 '17 at 22:43
  • $\begingroup$ thank you for your help. It makes sense. I have a question though, how did you know to define the function f(x) as you did above? $\endgroup$ – ptsgeeg Feb 12 '17 at 13:16
  • $\begingroup$ Probably the deepish reason is that the two subrings are (Borel subalgebras)[en.wikipedia.org/wiki/Borel_subgroup], and so they are conjugate (this would be my alternate answer). More elementarily, start with transposing, see that multiplication is not quite right, and fix it by exchanging the diagonal elements. $\endgroup$ – Andreas Caranti Feb 12 '17 at 13:33

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