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21h
comment If a polynomial $g$ divides $f$ and $f'$, then $g^2$ divides $f$?
Dear @GeorgesElencwajg, it is not unduly critical to ask that an answer address the question that was actually asked. And there absolutely should not be room on this site for rude personal comments.
22h
comment If a polynomial $g$ divides $f$ and $f'$, then $g^2$ divides $f$?
I don't need to change your mind, I just need there to be a record of why it's a bad answer, and there is. :)
23h
comment If a polynomial $g$ divides $f$ and $f'$, then $g^2$ divides $f$?
And it's not up to you to decide how others comment on your help, or to make rude personal comments. For shame.
23h
comment If a polynomial $g$ divides $f$ and $f'$, then $g^2$ divides $f$?
And god forbid you take a second to explain why, by pointing out where his original method has a gap. Because as you said, if the original method is correct then why do it differently?
23h
comment If a polynomial $g$ divides $f$ and $f'$, then $g^2$ divides $f$?
You don't have to encourage it to point out what's wrong with it, and figuring out your mistakes is never a waste of time. For shame.
23h
comment If a polynomial $g$ divides $f$ and $f'$, then $g^2$ divides $f$?
Not when the question is "have I made a mistake". By not addressing that you just do an end run around their learning process and they don't benefit. It's the laziest way possible to pretend to give help.
23h
comment If a polynomial $g$ divides $f$ and $f'$, then $g^2$ divides $f$?
You did not answer his question.
23h
answered If a polynomial $g$ divides $f$ and $f'$, then $g^2$ divides $f$?
23h
comment Show that the union of cosets formed by a subgroup of the quotient group G/N where N is a normal subgroup of group G
It's obvious that $A$ is nonempty so you have to show that $A$ is closed under inverses and multiplication. What have you tried?
Apr
12
asked Adjoint action of a Lie algebra in MAGMA
Apr
9
comment Free modules and ideals
It's not true that the ideal $(x_1, x_2) \subseteq \mathbb C[x_1, x_2]$ is a free $\mathbb C[x_1, x_2]$-module. The fact that it's not true does generalize to an arbitrary number of variables.
Apr
6
answered Determining the necessary values for a matrix' coefficients to achieve a certain rank.
Apr
5
accepted Does the exponential map respect module actions?
Apr
1
comment Prove if a matrix is diagonalizable with eigenvalues all plus or minus one then the matrix is its inverse
What have you tried?
Mar
27
answered An identity in Ring of characteristic $p$ prime
Mar
26
comment Schemes to the rescue?
Yes, that is one thing you can do. Another is to take the generic point of the pure quaternions (which is a well defined object now that they've been schemified) and square it. This is now a well defined element which should give you the equations for the conic that you want.
Mar
26
comment A morphism from $\mathbb P^1_\mathbb C$ to $\mathbb P^1_\mathbb C$
What is the difference between $\mathbb P^1_{\mathbb C}$ and $\mathbb P^1(\mathbb C)$?
Mar
26
comment Elements of a free module written uniquely as a linear combination of basis elements
The answer is yes by the way, we just need to know which definition you're using in order to explain why the answer is yes.
Mar
26
comment Elements of a free module written uniquely as a linear combination of basis elements
There are multiple ways to define "basis", which are you using?
Mar
26
comment Help in this proof in Lang's Algebra book
@user42912: As I say in the last paragraph, it's only impossible if you assume $f_1$ has positive degree. If you allow $f_1$ to be constant then let $(f_1, \ldots, f_n) = (g_1, \ldots, g_n) = (1, 0, \ldots, 0)$. Then we obviously have $\sum_ig_if_i = 1$ and $f_2, \ldots, f_n \in \mathfrak m[x]$. If you read the other answer carefully you'll see that you only get a contradiction if you assume that $f_1g_1$ has positive degree.