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  • 0 posts edited
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  • 107 votes cast
Apr
25
accepted $K/\mathbb{Q}$ contains a complex number - is complex conjugation in $\text{Aut}(K/\mathbb{Q})$?
Apr
23
comment Example of a field extension $K/F$ such that $K$ is the splitting field of some (non-separable) polynomial in $F$, but not Galois over $F$?
Yes, I was mistaken in that comment, I was indeed referring to why we require it in the characterization of Galois. Thanks for the recommendations.
Apr
23
comment Example of a field extension $K/F$ such that $K$ is the splitting field of some (non-separable) polynomial in $F$, but not Galois over $F$?
I'm not familiar with that in particular, is there a good resource on it? I don't think its discussed in Dummit and Foote, but I can check.
Apr
23
revised Example of a field extension $K/F$ such that $K$ is the splitting field of some (non-separable) polynomial in $F$, but not Galois over $F$?
added 23 characters in body
Apr
23
comment Example of a field extension $K/F$ such that $K$ is the splitting field of some (non-separable) polynomial in $F$, but not Galois over $F$?
$\mathbb{Q}(i)$ is the splitting field of $(x^2+1)^2$ over $\mathbb{Q}$, a non-separable polynomial. However, clearly, $\mathbb{Q}(i)$ is also the splitting field of $(x^2+1)$, a separable polynomial. So, $\mathbb{Q}(i)$ is Galois. So, I don't see how your second claim holds.
Apr
23
asked Example of a field extension $K/F$ such that $K$ is the splitting field of some (non-separable) polynomial in $F$, but not Galois over $F$?
Apr
22
comment $K/\mathbb{Q}$ contains a complex number - is complex conjugation in $\text{Aut}(K/\mathbb{Q})$?
@QiaochuYuan I would like to show the claim is true if $K$ is Galois, but I'm not quite seeing how from your comment. Could you expand your argument a bit more or maybe even post an answer? Thanks!
Apr
22
comment $K/\mathbb{Q}$ contains a complex number - is complex conjugation in $\text{Aut}(K/\mathbb{Q})$?
Ah, sure, I suppose in general $(a+ib)^{-1} = \frac{1}{a^2+b^2}(a - ib)$, but $a^2+b^2$ may not necessarily be rational or even in the field itself...
Apr
22
asked $K/\mathbb{Q}$ contains a complex number - is complex conjugation in $\text{Aut}(K/\mathbb{Q})$?
Apr
17
accepted Subgroups of $\mathbb{Z}_p^n$
Apr
16
awarded  Yearling
Apr
16
asked Subgroups of $\mathbb{Z}_p^n$
Apr
3
comment Separable polynomials are the product of the minimal polynomials of their roots?
Specifically, the word distinct clears up the meaning a whole lot.
Apr
3
comment Separable polynomials are the product of the minimal polynomials of their roots?
Thanks, this clears things up immensely!
Apr
3
accepted Separable polynomials are the product of the minimal polynomials of their roots?
Apr
1
asked Separable polynomials are the product of the minimal polynomials of their roots?
Mar
25
asked Does this inequality involving the sum of components of a vector and its norm have a name?
Mar
7
comment Components of a bounded vector must be bounded
I'm familiar with orthonormal bases, but we haven't explicitly discussed them. I think it should be able to be done without assuming/taking an orthonormal basis.
Mar
7
asked Components of a bounded vector must be bounded
Feb
25
accepted Elegant proof that maximum of sums is, at most, sum of maximums