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Let field $\mathbb F$ be either $\mathbb R$ or $\mathbb C$ and $M_n(\mathbb F)$ all $n \times n$ matrixes.

We denote by $I_n(\mathbb F)$ the space of all functions:

$P : M_n(\mathbb F) \rightarrow \mathbb F$

which are polynomial (in the sense that $P(A)$ is a polynomial in the entries of $A$), and which are invariant under the conjugation, i.e. $P(gAg^{-1}) = P(A)$ for all $g \in GL_n(\mathbb F)$.

For each $p \ge 0$, we define $\Sigma_p \in I_n(\mathbb F)$ as $\Sigma_p(A)=Tr(A^p)$.

Then do we have the isomorphism of algebras $I_n(\mathbb F) \cong \mathbb F[\Sigma_0,\Sigma_1,\cdots,\Sigma_n]$?

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    You don't want an isomorphism of algebras. The RHS is clearly a subalgebra of the LHS and you want to $k$now whether it is the entire algebra; that is, you want an equality. An abstract isomorphism is a weaker statement.2012-08-11

1 Answers 1

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Yes. This is almost true in general, but not quite for two reasons: one involving the size of the field and the other involving characteristic.

The correct statement over an infinite field $F$ is that the algebra of invariant polynomials is generated by the elementary symmetric functions of the eigenvalues, or more concretely by the coefficients of the characteristic polynomial $\det (I - At)$. To see this, pass WLOG to the algebraic closure $\bar{F}$ (which we can do because $P$ is polynomial). The identity $P(gAg^{-1}) = P(A)$ is a polynomial identity in the coefficients of $A$, the coefficients of $g$, and $\det g$, and since $F$ is an infinite field two polynomials which are equal when all possible elements of $F$ are plugged in are equal identically. Consequently, this identity continues to hold over $\bar{F}$. (This part of the argument can be ignored over $\mathbb{C}$.)

Now suppose that $M$ is a diagonalizable matrix. Then $P(M)$, being conjugation-invariant, is necessarily a polynomial, in fact a symmetric polynomial, in the eigenvalues of $M$. Since the diagonalizable matrices over $\bar{F}$ are Zariski-dense (as they contain the matrices with distinct eigenvalues, which is a Zariski-open condition), it follows that $P$ must be a symmetric polynomial in the eigenvalues of $M$ identically. Hence $P$ is a polynomial in the elementary symmetric polynomials with coefficients in $\bar{F}$, but using the fact that $P$ is defined over $F$ and letting $M$ be a suitable collection of companion matrices shows that $P$ is a polynomial in the elementary symmetric polynomials with coefficients in $F$.

The above argument does not work over a finite field because the identity $P(gAg^{-1}) = P(A)$ is no longer guaranteed to continue to hold over the algebraic closure, although I don't know a counterexample in this setting.

The result you want is true in characteristic zero or characteristic greater than $n$ by Newton's identities but false in small positive characteristic because using Newton's identities will require dividing by zero. For example, in characteristic $2$, unlike in higher characteristic, it is not possible to express $e_2 = \sum_{i < j} \lambda_i \lambda_j$

in terms of $p_1 = \sum_i \lambda_i$

and $p_2 = \sum_i \lambda_i^2 = (\sum_i \lambda_i)^2 = p_1^2$

(where $\lambda_i$ are the eigenvalues).

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    Nice proof. Two comments: 1) I think we can remove the use of $\overline F$ by noting that diagonalizable matrices include those with pairwise distinct eigenvalues, that are those which do not annihilate the discriminant of their characteristic polynomial, so they are dense. 2) Is your explanation that similarity is polynomial sound? As you said it, it seems that the determinant is in the denominator, giving a rational function; one possible alternative is to use that two matrices are similar iff they have same invariant factors ($n$ polynomial identities).2018-09-19