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 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