Is \( A \) Isomorphic to \( \dfrac{R<X,Y>}{((X^2+1)X,(X^2+1)Y,YX)} \)?

[Fermat1]

let $x=\begin{bmatrix}i&0\\0&0 \end{bmatrix}$ and $y=\begin{bmatrix}0&1\\0&0 \end{bmatrix}$.

Define $A={{\begin{bmatrix}a&b\\0&c\\ \end{bmatrix}}where c\in\mathbb{R}}$

Show that A is isomorphic to $\dfrac{R<X,Y>}{((X^2+1)X),(X^2+1)Y,YX)}$

My work: Define $f:R<X,Y>\implies A$ by $f(X)=x$, $f(Y)=y$, and define $f$ to be an algebra homomorphism. Things to do: Show $(X^2+1)X$ etc. are in the kernel, which I have done.

Show $f$ is surjective, which I haven't been able to do. Finally I would need to show those elements are in fact the whole kernel in order to invoke the first isomorphism theorem

 

[Deveno]

I'm a bit confused...is the domain of $f$: $\Bbb R[X,Y]$, or does $R$ refer to some other ring?

Also, you state that $c \in \Bbb R$, but what about $a,b$?

If indeed all the entries of elements of $A$ are to be real, then $x$, as you have defined it, is not in $A$, so your proposed homomorphism does not have the right co-domain.

What does the original problem actually state? I feel we are missing something, here.

 

[Fermat1]

a,b can be complex numbers and yes R is the set of reals. A is a set by the way, not a matrix

 

[Deveno]

Ok, so $A$ is the set of all upper triangular matrices in $\Bbb C_{2 \times 2}$ with real entries in the 2,2 position.

Also, just to clarify, by $\Bbb R\langle X,Y\rangle$ you mean the non-commutative real polynomial ring in two indeterminates, right?

Also, you are taking:

$f(c) = \begin{bmatrix}c&0\\0&c\end{bmatrix}$, yes?

If that is so, then writing:

$a = a_0 + a_1i$
$b = b_0 + b_1i$

we have, for:

$M = \begin{bmatrix}a&b\\0&c\end{bmatrix} \in A$:

$f(c + a_1X + (c - a_0)X^2 + b_0Y + b_1XY) = M$,

which shows $f$ is surjective.

Frankly, I think that's all you need to show, because $R\langle X,Y\rangle$ is *free* in $X$ and $Y$, and $f$ evidently annihilates the set of generators of the ideal, so $A$ IS (isomorphic to) the quotient of the free $\Bbb R$-algebra on two letters modulo the ideal generated by the generating set (using the universal properties of free algebras and quotient rings).

I hope that makes sense to you.

 

[blue_raver22]

.

Thank you for sharing your work and thought process. As a scientist, it is important to provide clear and logical reasoning for our conclusions. It appears that you have made some progress in establishing an isomorphism between $A$ and the given quotient ring. However, it is important to fully prove and explain each step to ensure the validity of the result.

To show that $f$ is surjective, we need to show that for any matrix $A$ in the form $A=\begin{bmatrix}a&b\\0&c \end{bmatrix}$ with $c\in\mathbb{R}$, there exists an element $r\in R<X,Y>$ such that $f(r)=A$. In other words, we need to show that for any $a,b,c\in\mathbb{R}$, there exists $r\in R<X,Y>$ such that $f(r)=\begin{bmatrix}a&b\\0&c \end{bmatrix}$.

To do this, we can use the fact that $f(X)=x$ and $f(Y)=y$. This means that any polynomial in $X$ and $Y$ can be mapped to a corresponding matrix in $A$. Thus, we can define $r=Xa+Yb$ and we have $f(r)=f(Xa+Yb)=f(X)a+f(Y)b=ax+by=\begin{bmatrix}a&b\\0&0 \end{bmatrix}$. This shows that for any $a,b,c\in\mathbb{R}$, we can find an element $r\in R<X,Y>$ such that $f(r)=\begin{bmatrix}a&b\\0&c \end{bmatrix}$, proving that $f$ is surjective.

To show that the elements in the kernel are in fact the whole kernel, we can use the fact that the polynomials $(X^2+1)X$, $(X^2+1)Y$, and $YX$ are all in the kernel. This means that any element in the kernel can be written as a linear combination of these polynomials. Furthermore, we can also use the fact that $f$ is an algebra homomorphism to show that any polynomial in $X$ and $Y$ that evaluates to $0$ under $f$ must also be in the kernel.