Are the field norm and trace the unique "nice" maps between fields? - MathOverflow

Are the field norm and trace the unique "nice" maps between fields?

2009-12-03T00:48:23Z

This seems like an obvious fact, but I'm not sure what the necessary meaning of "nice" is to get a result like this. I'm wondering if there is a theorem of the form:

For any <1> field extension $K/F$, a map from $\phi:K\rightarrow F$ that satisfies <2> is the field norm (or trace).

where <1> could be something like finite, algebraic, etc., and <2> could be anything (obviously there would be different <2>'s for norm and trace).

Answer by Greg Kuperberg for Are the field norm and trace the unique "nice" maps between fields?

2009-12-03T00:53:08Z

The field norm and trace exist when $K$ is a finite algebraic extension of $F$. In this case, an element $\alpha \in K$ can be interpreted as an $F$-linear map on $K$ by multiplication. The field norm is just the determinant of $\alpha$ as a linear map, while the trace is the trace of $\alpha$ as a linear map. This yields an evident generalization: Norm and trace are part of a family of nice maps, namely the coefficients of the characteristic polynomial of $\alpha$.

Since Zev asks for a uniqueness theorem in the comments, here is one that shows both the merits and limitations of the characteristic polynomial as an answer.

For simplicity let $F$ have characteristic 0. Let $K$ be a field extension of degree $n$ which is generic in the sense that the Galois group is $S_n$. Then any Galois-invariant polynomial in $\alpha \in K$ and its Galois conjugates, is a symmetric polynomial. The theorem is that the algebra of symmetric polynomials is generated by elementary symmetric polynomials, which are exactly the coefficients of the characteristic polynomial of $\alpha$. (This is using the fact that the eigenvalues of $\alpha$ as a map are itself and its Galois conjugates.) In particular, the trace is the unique linear such map up to a scalar; and any multiplicative polynomial of this type is a power of the norm. You can also describe the norm as the last Galois-invariant polynomial (the one of degree $n$) that provides new information.

But if the Galois group is smaller, then the ring of invariant polynomials in $\alpha$ and its Galois conjugates is larger, and any of these other invariant polynomials is also "nice". These extras are somewhat hidden by the fact that, for any Galois group, the trace is still the only linear example and the norm is still the only multiplicative example.

Well, the original question was open-ended. I think that this answer does fit one interpretation of the question, but maybe it is too standard and maybe there are also other interesting answers.

Answer by KConrad for Are the field norm and trace the unique "nice" maps between fields?

2012-07-11T18:54:11Z

Here is a nice characterization of the norm mapping on a finite extension of fields $K/k$.

If $K/k$ is any finite extension of fields with degree $n$, then the norm mapping from $K$ to $k$ is the unique function $f \colon K \rightarrow k$ satisfying the following three conditions:

1) $f(xy) = f(x)f(y)$ for all $x$ and $y$ in $K$.

2) $f(c) = c^n$ for all $c$ in $k$.

3) $f$ is a polynomial function over $k$ of degree at most $n$, by which I mean there is a basis $\{e_1,\dots,e_n\}$ of $K/k$ relative to which $f$ can be described by a polynomial: there's a polynomial $P(x_1,\dots,x_n)$ in $k[x_1,\dots,x_n]$ such that $f(\sum_{i=1}^n c_ie_i) = P(c_1,\dots,c_n)$ for all $c_1,\dots,c_n$ in $k$. (Being a polynomial function is independent of the choice of basis.)

This is due to Harley Flanders. See the following two articles of his:

The Norm Function of an Algebraic Field Extension, Pacific J. Math 3 (1953), 103--113.

The Norm Function of an Algebraic Field Extension, II, Pacific J. Math 5 (1955), 519--528.

One nice consequence of this characterization of the norm, which Flanders points out, is that it gives a slick proof of the transitivity of the norm: if $K \supset F \supset k$ then the function ${\rm N}_{F/k} \circ {\rm N}_{K/F}$ satisfies the three conditions that characterize ${\rm N}_{K/k}$ .