Prove that $M_{n\times n}(K)$ and $P_{n^2-1}[x]$ (polynomials with degree less than or equal to $n^2-1$) are not isomorphic rings for any field $K$ and $n\ge 2$
Let $f: M_{n\times n}(K)\to P_{n^2-1}[x]$ a ring isomorphism. Let $A\in M_{n\times n}(K)$ so that $f(A)$ is not the zero polynomial: $f(AI_n)=f(A)f(I_n)$ but $AI_n=A$ hence $f(A)=f(A)f(I_n)\Rightarrow 1=f(I_n)$
But since $f$ is any ring isomorphism then I don't what polynomial is $f(I_n)$. Is this the correct approach? I would really appreciate if you can help me :)
If $d \in \mathbb{N}$, then usually $P_d(x)$ means $K[x]/(x^{d+1})$, and is called a truncated polynomial algebra. This is a ring and it should not be confused with the vector space of polynomials of degree $\leq d$. This is just (a version of) the underlying $K$-vector space. You cannot multiply polynomials of degree $\leq d$ to get polynomials of degree $\leq d$, but you can multiply in $P_d(x)$. (It is a common habit to ignore forgetful functors, but this here is one of lots examples where this is no good idea.)
In our case, $P_{n^2-1}(x) = K[x]/(x^{n^2})$. The underlying vector space has dimension $n^2$, which is also the dimension of $M_{n \times n}(K)$. Hence, the underlying vector spaces of $P_{n^2-1}(x)$ and $M_{n \times n}(K)$ are isomorphic.
But they are not isomorphic rings (when $n>1$), and there are many reasons for this. The simplest one is that $P_{n^2-1}(x)$ is commutative, and $M_{n \times n}(K)$ is not.