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Let $M_n(\mathbb C)$ be the algebra of $n\times n$ complex matrices. The coefficients of the characteristic polynomial $\det(\lambda I-A)=\sum f_i(A)\lambda^i$ are polynomials in the entries of $A\in M_n(\mathbb C)$, and are conjugation invariant. Moreover, every conjugation invariant polynomial function $F:M_n(\mathbb C)\to \mathbb C$ is of the form $P(f_0,f_1,\ldots, f_{n-1})$ for $P\in \mathbb C[x_0,\ldots,x_{n-1}]$.

Here is a proof (hover mouse over to view):

Given a group action on a space, any continuous invariant function must be constant on the closure of each orbit. Because the closure of every conjugacy class contains a diagonal matrix, the entries of which are the eigenvalues of the matrices in the conjugacy class, invariant functions must be polynomials in the eigenvalues. Given any permutation $\sigma\in S_n$, we have that $\operatorname{diag}(a_1,\ldots, a_n)$ is conjugate to $\operatorname{diag}(a_{\sigma(1)},\ldots, a_{\sigma(n)})$, and hence an invariant function must be a symmetric function in the eigenvalues. The coefficient $f_i(A)$ is up to a sign the $(n-i)$th elementary symmetric function in the eigenvalues of $A$, and since the elementary symmetric functions generate the ring of all symmetric functions, the result follows.

Is there a proof that doesn't require reducing the problem to eigenvalues and symmetric functions? Perhaps more important, can the result be extended to non-algebraically closed fields or other base rings where this particular proof fails because we don't have diagonalization? If not, what additional conjugation-invariant polynomial functions are there over $\mathbb R$, $\mathbb Q$, or $\mathbb Z$?

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    [Here](https://math.stackexchange.com/a/181252/374722) is a link for Qiaochu Yuan's proof.2018-09-01

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Let $K$ be an infinite field and $f:M_n(K)\to K$ a polynomial map which is constant on the conjugacy classes.

Claim. $f$ is a polynomial in the coefficients of the characteristic polynomial.

We can (and will) assume that $K$ is algebraically closed.

Observation. If a polynomial map $g:M_n(K)\to K$ vanishes on the diagonalizable matrices, then $g=0$.

Proof of the observation. We have $dg=0$, where $d$ is the discriminant of the characteristic polynomial. As $K$ is infinite, this implies $g=0$. QED

Proof of the claim.

By the observation, $f$ is determined by its restriction to the diagonal matrices.

This restriction is invariant by permutation of the diagonal entries.

Thus, this restriction is a polynomial in the elementary symmetric polynomials in the diagonal entries.

But these elementary symmetric polynomials are the coefficients of the characteristic polynomial.

Using again the observation, we see that $f$ itself is a polynomial in the coefficients of the characteristic polynomial. QED

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    Dear @Aaron: You're welcome. Thanks for your comment. I agree with what you say. I'll take a look at Keith Conrad notes. (By the way, he's an MSE user. See, below this [question](http://math.stackexchange.com/questions/38781/complex-galois-representations-are-finite/38816#38816), the comment to starting with `@KCd: Dear Keith`. I'm sure he'd have interesting things to say on your questions.) For the observation, if $d(A)\neq0$, then $A$, having $n$ distinct eigenvalues, is diagonalizable, and we have $g(A)=0$.2011-11-02