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4h
awarded  Nice Answer
Aug
29
comment What makes it legitimate to multiply both sides?
You can apply a function to both sides of any equality. Even in settings where we aren't talking about numbers. Even when we're just talking about barren, structureless sets and elements of sets, we can apply a function to both sides of an equality. Of course, for any $u$ in an algebraic structure with a binary operation, there is a function $f(x):=x\cdot u$.
Aug
28
comment Irreducibility of $p(x)$ implies that of $p(x+c)$ only when taken over a field?
Why do you think you're missing anything? The lemma is usually proved and used in settings where we're exploring polynomial rings over a field, so that's how the lemma gets stated up front, even though as you note it is more general.
Aug
28
comment Why is $1+2+3+\cdots = 0 $?
@Did Summability methods are a thing in math, they have a rigorous meaning. In particular, the equation $1+2+\cdots=-\frac{1}{12}$ has meaning. And in this context where some divergent series can have meaning, subtracting two divergent series also has meaning. I haven't seen the numberphile video, and can't speak to whether pop-math treatments do justice to the rigorous idea of summability methods. If I were to watch the video, I imagine I would be in agreement with tired's comment.
Aug
28
comment Why is $1+2+3+\cdots = 0 $?
@Did One can subtract divergent series both of which are assigned a finite value by a given summability method. There is a different issue at play here - inserting and deleting $0$s as terms in a series is no longer valid - it does not generally preserve the value the summability method assigns to a divergent series.
Aug
24
comment Intuition behind the construction of Young Symmetrizer
Yes, the cases of symmetrizing and antisymmetrizing are comparatively obvious and intuitive - they can be discovered just playing around with algebra. Can you tell a plausible story for how we might discover general Young symmetrizers though, or give a historical account? That is the question here, after all. (Just read your last comment. I guess I'll reserve judgment until it's finished.)
Aug
22
comment I need to prove: if $a|b$ and $a|c$ then $a|(bc)$?
That $a^2$ isn't prime is irrelevant.
Aug
22
comment Intuition behind the construction of Young Symmetrizer
Assuming you're still around to read this comment, you could ask this over at MathOverflow if my bump and bounty offer don't get it any new eyes in the next week.
Aug
22
revised Intuition behind the construction of Young Symmetrizer
Bump.
Aug
22
comment $G$ finite, the number of distinct conjugates of $x$ is the index of the normalizer $N_x$ of $\{x\}$ in $G$
@GuerlandoOCs Your comment does not make sense to me. Since what you're trying to prove is a special case of orbit-stabilizer, you're talking about a specific proof essentially of the theorem, that's about as opposite of "a general proof without this theorem" as you can get.
Aug
17
comment Lie algebra vector subspace: Does $[n_1,[n_2,Y]]=[n_1,A]=B$
Well, $[n_1,[n_2,Y]]\subseteq [n_1,Y]\subseteq Y$. What are your $A$ and $B$ exactly? If they're just any two random vector subspaces with $B\subseteq A\subseteq Y$, then of course not.
Aug
15
awarded  Nice Answer
Aug
15
comment If $S\leq G$, prove that $S\unlhd G \iff \gamma (S) \leq S$ for every conjugation $\gamma$
Right. Like I said, missing some quantification. If for every $s\in S$ there is a $s'\in S$ for which $gs=s'g$ (hence $gs\in Sg$), then in particular we have $gs\in Sg$ for every $s\in S$, hence $gS\subseteq Sg$.
Aug
15
comment If $S\leq G$, prove that $S\unlhd G \iff \gamma (S) \leq S$ for every conjugation $\gamma$
How does $gs=s_1g$ imply $gS\subset Sg$? Perhaps you are missing sume quantification.
Aug
12
comment Proof a Rng cannot have exactly five non-zero divisors.
If $u$ is the unit of $M$ then $u(ux-x)=0\Rightarrow ux=x$ and similarly $xu=x$ for all $x\in R$, so in fact $u$ would be a multiplicative identity for $R$. Then you can let $M=\langle a\rangle$, define a map $\Bbb F_2[x]/(x^5-1)\to \Bbb F_2[a]$ and consider the number of units in its image (as you do)..
Aug
11
comment For polytopes, does union and linear transformation commute?
$f(X\cup Y)=f(X)\cup f(Y)$ is true for any function $f$.
Aug
9
comment Show that $\mathbb{Q}(\sqrt{2},\sqrt{3},\dots,\sqrt{p},\dots)$ is an algebraic extension of $\mathbb{Q}$, for $p$ prime.
If $x\in\bigcup_p \Bbb Q(\sqrt{2},\cdots,\sqrt{p})$ then $x\in\Bbb Q(\sqrt{2},\cdots,\sqrt{p})$ for some $p$ and so is algebraic.
Aug
9
comment Let $u,v$ be algebraic over the field $K$, such that $[K(u):K]=n$ and $[K(v):K]=m$. Show that if $\gcd(m,n)=1$, then $[K(u,v):K]=mn$
What you need to be showing is that $[K(u,v):K]$ is divisible by both $n$ and $m$. Use transitivity of degrees, and the fact that $K(u)$ and $K(v)$ are both intermediate fields.
Aug
9
comment How many sides does a circle have?
We would never use "number of $1$-cells in a CW structure imposed on the figure" to count the sides of a polygon, so why would we use it to count sides of a circle?
Aug
9
comment minimum number of leaves in a perfect binary tree
You've proven a perfect binary tree of height $h$ has $2^h$ leaves. Surely $2^h\ge h+1$ for all $h\ge0$?