This question (maybe an easy one) arose when I was reading Humphrey's book "an introduction to Lie algebra and its representations".
Suppose $\mathfrak{g}$ is a complex semisimple lie algebra, $V$ is a $\mathfrak{g}-$ module, both $\mathfrak{g}$ and $V$ are finite dimensional. for $x\in\mathfrak{g}$, let $\rho(x)$ be the matrix of $x$ in $\mathfrak{sl}(V)$. $P(x)=tr[\rho^m(x)]$ be the trace of the $m$th power of $\rho(x)$, then it's fairly simple to see $P(x)$ is a polynomial function on $\mathfrak{g}$. Let $\mathfrak{g}$ act on itself by the adjoint action, then the assertion is: $P(x)$ is an invariant polynomial under this action, or we can write $P(x)\in S(\mathfrak{g})^{\mathfrak{g}}$.
This is an easy step in the proof of the Chevalley's theorem, but I found its proof quite weird: Most articles used an "polarization" way, that is, by expanding $P(x)$ to a multilinear function to handle this. And some experts(like Bernstein) even wrote "it's clearly that $P(x)$ be invariant" with out a proof.
So my question is: is there really an easy way to explain that the $P(x)$ is really $\mathfrak{g}$ invariant?