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In invariant theory the Reynold's Operator gives rise to an element invariant for that group.

For a Galois extension $K/F$ with $K=F[\alpha]$ the trace of $\alpha$ is an element of $K$. If $\alpha$ is of degree $n$, trace is of degree 1.

If I take sum of conjugates of $\alpha$ over a subgroup of index $m$ in the Galois group, and their product there seems to be asymmetry. The elements obtained are in different fields (that is their degrees are unequal.

Example: Take the $n$th cyclotomic extention over the rationals. Then the sum $\zeta+\bar \zeta$ gives rise to an element of degree $\phi(n)/2$. But their product $\zeta\bar\zeta=1$ is a rational number.

The reason $\zeta\mapsto\zeta^{-1}$ is an automorphism does not really explain what is happening.

I have three questions (at least for $F$ the field of rational numbers).

(1) Does this happen only in Galois extensions having complex conjugation as non-trivial automorphism?

(2) Are there examples where product over the (relative) conjugates giving rise to a higher degree element than the sum?

(3) Is there a classification of base fields $F$ where there the behaviour is symmetric for sum and products?

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    $\begingroup$ There is no real asymmetry. It just happens that $\zeta\bar\zeta$ is rational. But use $\alpha=1+\zeta$ instead and you get $\alpha\bar\alpha=2+\zeta+\bar\zeta$. $\endgroup$
    – Tom Goodwillie
    Commented Apr 25, 2017 at 0:51
  • $\begingroup$ Thats my point: Using N, Tr, for relative norm and trace: $N(\zeta)$ and $N(1+\zeta)$ have different degrees as algebraic numbers. But Tr($\zeta$) and Tr($1+\zeta$) have same degrees. $\endgroup$
    – P Vanchinathan
    Commented Apr 25, 2017 at 2:37
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    $\begingroup$ But you can think of another generator of the cyclotomic field such that its trace is rational. For example, in the $8$th cyclotomic field $\zeta+\zeta^2+\zeta^3$. $\endgroup$
    – Tom Goodwillie
    Commented Apr 25, 2017 at 2:50

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Basically anything can happen: fix $M/F$ of degree $m$ and $f(x) = x^n + f_{n-1} x^{n-1} + \ldots + f_0 \in M[x]$; if $f$ is irreducible (which it will be for "random" $f$), then let $\alpha$ be a root and $K=M(\alpha)$ and then $\operatorname{Norm}_{K/M}(\alpha) = \pm f_0$ and $\operatorname{Tr}_{K/M}(\alpha) = \pm f_{n-1}$. You can choose $f_0$ and $f_{n-1}$ freely (in particular they can generate whatever subfield of $M$ you like) and so the same is true for norm and trace.

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  • $\begingroup$ Is this a typo? Your $f(x)$ is having coefficients in the topmost field of the tower. Or I do not know if $M$ or $K$ is contained in the other. $\endgroup$
    – P Vanchinathan
    Commented May 18, 2017 at 0:43

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