- #1
Palindrom
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I'm sure whoever is familiar with this subject has already seen this several times. I've seen it several times myself, and I even remember proving it in detail a couple of years ago, but now I'm stuck.
I'm quoting what my professor did in class.
Given some separable extension L/K, say for simplicity char(K)=0 and forget separability issues, we know that there are exactly n=[L:K] K embeddings of L into some algebraic closure C of L. For a in L we define its trace and norm (with respect to the extension L/K) respectively as the sum and the product of the n embeddings' actions on a.
All good.
Now the proposition that's bugging me is the following one: had we defined a linear operator on L by T_a(x)=ax, then the trace and the norm of a are exactly that trace and determinant of T_a.
I'm trying to show that the characteristic polynomial of T_a is exactly (x-a_1)...(x-a_n), where a_1,...,a_n are the images of a under the K embeddings of L into C. While this is a very nice idea, I'm failing miserably.
Help?
I'm quoting what my professor did in class.
Given some separable extension L/K, say for simplicity char(K)=0 and forget separability issues, we know that there are exactly n=[L:K] K embeddings of L into some algebraic closure C of L. For a in L we define its trace and norm (with respect to the extension L/K) respectively as the sum and the product of the n embeddings' actions on a.
All good.
Now the proposition that's bugging me is the following one: had we defined a linear operator on L by T_a(x)=ax, then the trace and the norm of a are exactly that trace and determinant of T_a.
I'm trying to show that the characteristic polynomial of T_a is exactly (x-a_1)...(x-a_n), where a_1,...,a_n are the images of a under the K embeddings of L into C. While this is a very nice idea, I'm failing miserably.
Help?
