Timeline for Are the field norm and trace the unique "nice" maps between fields?
Current License: CC BY-SA 2.5
14 events
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| Jul 11, 2012 at 23:42 | comment | added | Gerry Myerson | "For simplicity let $F$ have characteristic 0." Was that a pun (based on finite extensions in characteristic zero being simple)? | |
| Jul 11, 2012 at 22:32 | comment | added | George Lowther | @KConrad, @Zev: If the extension is inseparable then you can let $E\subset K$ be the elements which are separable over $k$, let $p$ be the characteristic, and choose $r$ such that $K^{p^r}\subseteq E$. Then, use the linear map $\alpha\mapsto Tr_{E/k}(\alpha^{p^r})$ instead. I'm not sure how nice or canonical that would be though. | |
| Jul 11, 2012 at 18:46 | comment | added | KConrad | 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. | |
| Jan 14, 2010 at 19:36 | comment | added | Zev Chonoles | 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. | |
| Dec 3, 2009 at 17:19 | comment | added | Greg Kuperberg | Canonical and polynomial are two reasonable interpretations of the word "nice", especially if you have them both together. Note that there are other canonical maps that are not polynomials. | |
| Dec 3, 2009 at 9:10 | comment | added | Kevin Buzzard | 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! | |
| Dec 3, 2009 at 1:46 | comment | added | Zev Chonoles | 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. | |
| Dec 3, 2009 at 1:38 | history | edited | Greg Kuperberg | CC BY-SA 2.5 |
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| Dec 3, 2009 at 1:35 | vote | accept | Zev Chonoles | ||
| Dec 3, 2009 at 1:31 | history | edited | Greg Kuperberg | CC BY-SA 2.5 |
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| Dec 3, 2009 at 1:25 | history | edited | Greg Kuperberg | CC BY-SA 2.5 |
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| Dec 3, 2009 at 1:08 | comment | added | Zev Chonoles | 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)." | |
| Dec 3, 2009 at 1:06 | history | edited | Greg Kuperberg | CC BY-SA 2.5 |
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| Dec 3, 2009 at 0:53 | history | answered | Greg Kuperberg | CC BY-SA 2.5 |