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If I have a ring R, where $R=M(2,Q)$, the set of all 2x2 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= the set of all 2x2 upper triangular matrics, e.g.$$x = \begin{pmatrix} a & b \\ 0 & c \\ \end{pmatrix}$$

and B = the set of all 2x2 lower triangular matrices, e.g.$$y = \begin{pmatrix} d & 0 \\ e & f \\ \end{pmatrix}$$

I need to show the 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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    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? – 2017-02-11
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    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. – 2017-02-11
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    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... – 2017-02-11
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    Sorry I should make my question much clearer. I'll edit it now – 2017-02-11
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    Did you got a description of the images of upper triangular matrices under $X$ or do you have to find it yourself? – 2017-02-11

1 Answers 1

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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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    I may be being very silly but I cannot get f(xy)=f(x)f(y) – 2017-02-11
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    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 – 2017-02-11
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    Please see my edit. – 2017-02-11
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    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? – 2017-02-12
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    Probably the deepish reason is that the two subrings are (Borel subalgebras)[https://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. – 2017-02-12
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    that's great. thanks for your help – 2017-02-12
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    @ptsgeeg you're welcome. – 2017-02-12