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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
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
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
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!
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.
Feb
25
comment Elegant proof that maximum of sums is, at most, sum of maximums
Of course, don't know how I missed something so simple. Perhaps the question is too basic and I was trying to overcomplicate things.
Feb
17
comment More elegant way to make the argument that $\mathbb{Q}[x,y]/\langle x,y \rangle \cong \mathbb{Q}$
Ah, I really like this. It sort of "cuts out the middleman" by setting up a smart map directly to the target space. Thanks!
Feb
1
comment Could I do this to an infinite series?
Typically, we define $\sum_{k=1}^{\infty} x_k$ as $\lim_{n \rightarrow \infty} \sum_{k=1}^n x_k$ if and only if the limit exists, so this doesn't quite work. It would be correct to say that the partial sums are equal to the resulting expression, i.e. $\sum_{k=1}^n k + \sum_{k=1}^n k = \sum_{k=1}^n 2k$, but neither of these sums converge so we can't really talk about their limits.
Jan
17
comment Finding at least 2 elements in a set that satisfys an equation
For each $n$, temporarily consider the additive group modulo $n$, that is $\mathbb{Z}_n$. By "$-1$", we merely mean the inverse of the element $1$. In $\mathbb{Z}_5$, we have $-1 = 4$, because $1 + 4 = 0$ (and in general, $-1 = n-1$). Now, try out squaring this additive $-1$ in the multiplicative group $U(n)$ (first verify that it's always in there), and show that it works.
Jan
17
comment Finding at least 2 elements in a set that satisfys an equation
Namely, $1$ and $-1$ are the two elements in $\mathbb{Z}$ under usual multiplication that square to give the identity. So, what are the analogues of $1$ and $-1$ in the finite group $U(n)$?
Jan
17
comment Finding at least 2 elements in a set that satisfys an equation
Okay, so the group operation is multiplication mod $n$. Then actually $U(n) = (\mathbb{Z}_n)^\times$ as a group. Got it.
Jan
17
comment Finding at least 2 elements in a set that satisfys an equation
What group structure are you giving $U(n)$ when we determine if there are two elements $x,y \in U(n)$ that satisfy $x^2 = y^2 = 1$? If we're working in the group $\mathbb{Z}$, the only element that satisfies $x^2 = 1$ is 1.
Oct
19
comment Using Pumping Lemma to show a language is not regular
This is correct, you could probably be a little more concise on the ending. Just say that the fact that $xyyz$ is not in the language means the language does not satisfy the pumping lemma. So our assumption that the language is regular must be incorrect. It also couldn't hurt to write out $xyyz$ to be more explicit, that is, $xyyz = 0^{p+k}1^p 1^p$ for some $k \geq 1$, noting that $2(p+k) > 2p$.
Nov
24
comment Solutions of $a^{2} - 2b^{2} \equiv 0$ mod $p$
Thank you very much! The terminology really helps when researching this!
Apr
11
comment Why is proof of the [topological] closed graph theorem incorrect?
Thanks; I see now there's nothing guaranteeing that $N$ can be be written as $U' \times V'$ with $U'$ definitely a subset of $U$. Can I fix this approach or am I off altogether? I could use some hints going forward because I'm pretty stumped.
Oct
14
comment How can I better understand manipulating “operators” in mathematical relations?
Thanks. Okay, I understand that you can do this algebra of functions, but if I just have some operator or map $A : U \rightarrow V$ and I have an element $u \in U$, then when I write $Au$ I mean the result of applying operator $A$ with input $u$? Is writing the operator next to a member of its input type "implied application"?
Jul
8
comment What “is” a matrix in the context of a vector space?
I think I understand. Since the space of ALL possible linear maps from V to W is isomorphic with the set of ALL m x n matrices, we can work with matrices and perform arithmetic on them and interpret our results in terms of linear mappings because of the isomorphism. So would it be reasonable then to view the "multiplication" of a vector by a matrix just an expression of the matrix's corresponsing map being applied to the vector? I think I'm splitting hairs at this point but I do feel more comfortable thanks to your post.