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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).

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2 Answers 2

up vote 25 down vote accepted

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.

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Yes, each of the symmetric polynomials in the conjugates of $\alpha$ yields a nice map and this generalizes the norm and trace; but I'm looking for more of a uniqueness theorem, along the lines of a result like "if $K/F$ is finite and algebraic and $\phi:K\rightarrow F$ is multiplicative (or additive) and does ... to elements of $F$ and ... (etc, etc) then $\phi$ is the norm (or trace)." –  Zev Chonoles Dec 3 '09 at 1:08
    
Thanks for the helpful explanation - I thought about this off and on for a while, and never could quite distill what I meant by "nice", but your result has very much clarified things. –  Zev Chonoles Dec 3 '09 at 1:46
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If all you have is a finite extension of fields, then isn't the issue not so much to do with "niceness" than the fact that there are very few ways to define anything canonical at all associated with the situation, other than using the char poly of multiplication? You can choose a basis for K/F and then you get lots of functions like "project onto the i'th coordinate" but none of these are canonical. Given an endomorphism of a f.d. vector space the char poly is the natural invariant because it's basis-independent. What I'm trying to say is that it's not so much niceness, as all we have! –  Kevin Buzzard Dec 3 '09 at 9:10
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I just found this result in Swinnerton-Dyer's A Brief Guide to Algebraic Number Theory - it's a bit closer to what I was looking for: Thm. Let $K\supset k$ be number fields. Then every $k$-linear map $K\rightarrow k$ is given by $\alpha\mapsto Tr_k^K(\beta\alpha)$ for some $\beta\in K$. Proof: Set $\phi_\beta=(\alpha\mapsto Tr_k^K(\beta\alpha))$. The $k$-linear map from $K$ to the dual space of $K$, given by $\beta\mapsto\phi_\beta$, has trivial kernel. The two spaces involved have the same dimension as $k$-vector spaces, so the map is an isomorphism. –  Zev Chonoles Jan 14 '10 at 19:36
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Zev: Your description of a property of the trace map on number fields is true for any finite separable extension of fields $K/k$: the trace pairing $K \times K \rightarrow k$ where $(\alpha,\beta) \mapsto {\rm Tr}_{K/k}(\alpha\beta)$ identifies $K$ with its $k$-dual space. This is definitely false if $K/k$ is inseparable, since in that case the trace mapping from $K$ to $k$ is identically 0. –  KConrad Jul 11 '12 at 18:46

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 ${\rm N}_{K/k}$.

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