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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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    To clarify: A polynomial function $F:M_n(\mathbb C)\to \mathbb C$ is a polynomial in the entries of its argument?2011-11-02
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    @Joriki: Yes. I believe this is equivalent to viewing $M_n(\mathbb C)$ and $\mathbb C$ as algebraic varieties (affine spaces) and considering regular functions between them.2011-11-02
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    Dear @joriki: I see that Aaron has already responded, but here is the (more naive) comment I prepared: Yes. You view $M_n(\mathbb C)$ as a (finite dimensional) vector space.2011-11-02
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    Dear Aaron: I answered your question as well as I could. I'm sure there will be better answers, and I'm expecting them with as much impatience as you. - I don't have the slightest idea about the conjugacy classes in $M_n(\mathbb Z)$. That's definitely too complicated form me! (Thanks for your very interesting question, which I upvoted.)2011-11-02
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    @Aaron: the proof using eigenvalues extends to non-algebraically closed fields (take the algebraic closure and extend a given polynomial function to the algebraic closure). Is this a sufficiently satisfying answer to your question?2012-04-30
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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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