Show that $\text{deg}(M_n(\mathbb{K})) = n$, where $\mathbb{K}$ is a field. Definition: Let $A$ be a ring and $Z=Z(A)$ its center. We say that $t \in A$ is algebraic over $Z$ if there exist $z_0,z_1, \ldots , z_n \in Z$ such that
$$z_0+z_1t+ \cdots + z_n t^n = 0 \quad \text{and} \quad z_n \neq 0.$$
Moreover, in this case we say that $t$ is algebraic of degree at most $n$. The degree of algebraicity is the smallest $n$ that satisfies this property; it will be denoted by $\text{deg}(t)$.
Show that $\text{deg}(M_n(\mathbb{K})) = n$, where $\mathbb{K}$ is a field.
Comments: I'm trying to use the Cayley-Hamilton's Theorem.
 A: A very simple computation shows that the center of $A=M_n(K)$ is the subring of scalar matrices, that is, those matrices which are diagonal and constant along the diagonal. It follows then that $Z=Z(A)$ is isomorphic to $K$ as a ring. Let $\phi:K\to Z$ be the isomorphism which maps each element of $K$ to the corresponding scalar matrix.
Let now $X$ be an element of $A$. As you observed, the Cayley—Hamilton theorem gives us a polynomial  $f$ with coefficients in $K$ such that $f(X)=0$ and the degree of $f$ is exactly $n$. From $f$ we can construct a new polynomial $\tilde f$, which is the result of applying $\phi$ to the coefficients of $f$. You can show very easily that $\tilde f(X)$ is zero.
From this, it follows that the degree of $A$ over $Z$ is at most $n$. let us show that it is equal to $n$. To do this, it is enough to show that there is a matrix $X\in A$ such that no polynomial of degree strictly smaller that $n$ vanishes on $X$. You should check that the permutation matrix $$\begin{pmatrix}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\1&0&0&0&0\end{pmatrix}$$ has degree $5$ in $M_5(K)$ and then generalize this to arbitrary $n$.
