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Nov
1
comment Proving that the maximal abelian extension contains all abelian extensions
I see and am disturbed by your mention of a union of Galois groups, which however are not contained in a larger group. I do feel strongly that all of this should be done without talking about the Galois group(s), but just about the fact that any two $K$-automorphisms, of whatever extension field, commute. Feel free to e-mail me, if you like. I can send you the longer treatment, which I discarded from here.
Nov
1
comment Compare the areas of a square with perimeter $12$ and of a parallelogram with perimeter $16$
The second figure isn’t a triangle. But to comment on WLOG’s well-taken answer, the maximum area of a parallelogram of perimeter $16$ is achieved by a square of side $4$, area $16$. You can take that square and flatten it down to a rhombus of side $4$ that’s arbitrarily thin, so will have arbitrarily small area.
Nov
1
answered Prove the equality of left and right coset of a group rigorously
Nov
1
comment Prove the equality of left and right coset of a group rigorously
To show that $g_2H=Hg_1$, I would start with a $g_2h\in g_2H$, and show that it’s writable as $h'g_1$ for some other $h'\in H$. This will show that $g_2H\subset Hg_1$, and the reverse inclusion will be similar. I think your proof can easily be modified to do this, but to me, the logic is less clear in your current version than it might be. For my argument, see below.
Oct
31
answered Proving that the maximal abelian extension contains all abelian extensions
Oct
31
comment Using the compound-angle formula, determine the exact value.
You mean $11/6=1+5/6$? or something deeper?
Oct
31
answered Show these two polynomials are irreducible over $\mathbb{Q}$
Oct
31
comment Modular arithmetic for a beginner
Factoring 713, @A.P.? You can see at a glance that it’s not divisible by any prime less than $17$. For that prime, add $713+17=730$, $73+17=90$, so not divisible by $17$; for $19$, add $57$ to get $760$, so not divisible by $19$. Subtract $23$ from $713$ and get $690=30\cdot23$, so there’s your factorization. No pencil.
Oct
30
comment If $X$ is compact metric space, then $X$ is bounded
I would have done similarly, but have noted that $X\times X$ is compact as well, so that $d(X\times X)$, which is the set of all distances between two points of $X$, is a compact subset of $\Bbb R^{\ge0}$, and thus bounded.
Oct
30
comment Can a circle be specified by an arbitrary triangle when one of its sides and the angle opposite to it is known?
Yes, this is a consequence of a standard result in plane geometry. If nobody else explains in an answer, I’ll do that.
Oct
30
answered Prove that $\sqrt[3]5 - \sqrt[4]3$ is Irrational
Oct
29
comment cyclotomic polynomial $\Phi_{2n}(x)$
Sure: if $\zeta$ is a primitive $m$-th root of unity, where $m$ is odd, then $-\zeta$ is ia primitive $2m$-th root of unity.
Oct
29
comment Application of Tower Law
and by one of the basic theorems of Galois Theory, the order of the Galois group is the field extension degree of the Galois extension $\Sigma\supset\Bbb Q$.
Oct
29
comment Why is $x-(x-1)=1$ instead of $x-x-1=-1$ as in $a*(b*c)=a*b*c$?
@AkivaWeinberger, right on. My position is that in mathematics, nothing is true. Until it’s proved.
Oct
29
comment Why is $x-(x-1)=1$ instead of $x-x-1=-1$ as in $a*(b*c)=a*b*c$?
In most cases, a question of the form “Why isn’t ...” will be beside the point. Rather, you should be asking “Why is ...”. For multiplication, you can answer the question “why is $a(bc)=(ab)c)$?” in a provisional way by noticing that it seems always to be true when you try numbers for $a$, $b$, and $c$. And you answer it in a definitive way by proving it. First for the integers, then for the rationals, then for the reals.
Oct
29
comment Galois group. $K$-automorphisms take adjoined roots to other roots of minimal polynomial or take roots of $f$ to other roots of $f$
Aha. So you’ve started with a normal extension and generated it with finitely many elements, then multiplied their various minimal polynomials, and that’s $f$.
Oct
29
awarded  abstract-algebra
Oct
29
answered Showing $\sqrt{5} \in \mathbb{Q}(w)$ where $w = e^{2\pi i/5}$
Oct
28
comment Why is $\lim_{x \to \infty} (\sqrt{9x^2+x}-3x)=\frac{1}{6}$?
This is the method I would have used, even though the other is more efficient.
Oct
28
answered Irreducible or reducible polynomials in $\mathbb Z_3[X]$ of degree 0,1,2,3