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by monfrere
3268 days ago
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OK, thanks. Your answer has allowed me to follow the paper a little further, though I think your answer may be inconsistent with the other two answers I got. Also I haven't been able to show myself that if the x_i are the roots of an irreducible polynomial, then Q(x_1,...,x_n) is symmetric with respect to x_1,...,x_n, in the sense claimed in the paper without proof. |
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No worries.
> Your answer has allowed me to follow the paper a little further, though I think your answer may be inconsistent with the other two answers I got.
I was trying to be helpful.
That said, in lieu of the irreducibility criterion, you could stay in the original context where all the variables a_i, x_i, and alpha_i are indeterminates so that all the extensions are higher-ranked rational function fields over Q.
So Q(a1,a2) is a field in 2 indeterminates and Q(x1,x2) is an extension of that.
They are isomorphic to subfields of C, but we don't think of them as subfields of C because there don't come with canonical embeddings. You have to choose the algebraically independent transcendentals.
> Also I haven't been able to show myself that if the x_i are the roots of an irreducible polynomial, then Q(x_1,...,x_n) is symmetric with respect to x_1,...,x_n, in the sense claimed in the paper without proof.
There's no claim. The paper's just defining what it means for a field extension to be "symmetric w.r.t." to the adjoined elements.