Is this algebra a simple ring?

Question

Is the algebra $ \mathcal{M}_{3} ( \mathbb{C} [X,Y,Z] ) $ a simple ring (a simple algebra)?

$ \mathcal{M}_{3} ( \mathbb{C} [X,Y,Z] ) $ is the matrix algebra over $ \mathbb{C} [X,Y,Z] $. Thanks a lot for your help.

Answer 1

If $R$ is a ring, then there is a bijection between the set $\mathcal{I}(R)$ of (bilateral) ideals of $R$ and the set $\mathcal I(M_3(R))$ of ideals of $M_3(R)$. In particular, if $R$ is not simple, then neither is $M_3(R)$.

The bijection is the "obvious" one: if $I$ is an ideal in $R$, then the set $M_3(I)$ of matrices of $M_3(R)$ with all its entries in $I$ is an ideal of $M_3(I)$, and this correspondence gives us a function $$I\in\mathcal I(R)\mapsto M_3(I)\in\mathcal I(M_3(R))$$ and you should prove that this is a bijection.

Later. You originally asked whether $M_3(\mathbb C[X,Y,Z])$ was a simple ring, and now you edited the question to ask instead if it is a simple algebra. The answer to both questions is the same: if $R$ is a commutative ring and $A$ is a unitary $R$-algebra, then a bilateral ideal of $A$ viewed as a ring is the same thing as a bilateral ideal of $A$ viewed as an $R$-algebra.