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visits member for 2 years, 5 months
seen Apr 10 at 1:24

Jul
2
awarded  Curious
Apr
10
comment Confused on a proof that $\langle X,1-Y\rangle$ is not principal
Thanks! Is it known that the units of the quotient ring are the same as the units of $\mathbb{Q}[X,Y]$, i.e., just $\mathbb{Q}^\times$? I was working it out, I could only determine that the units are the polynomials $g$ such that there exists $h$, such that $gh-1$ vanishes on the unit circle.
Apr
10
revised Confused on a proof that $\langle X,1-Y\rangle$ is not principal
added 286 characters in body
Apr
9
comment Confused on a proof that $\langle X,1-Y\rangle$ is not principal
May I ask a follow up? If $\langle f^2\rangle=\langle 1-Y\rangle$, doesn't that only imply $f(X,Y)^2g(X,Y)=1-Y$ for some $g(X,Y)$? How can you in fact assume $f(X,Y)^2=1-Y$?
Apr
8
comment Confused on a proof that $\langle X,1-Y\rangle$ is not principal
Thank you Alex.
Apr
8
asked Confused on a proof that $\langle X,1-Y\rangle$ is not principal
Feb
24
comment Equivalent definition of purely inseparable field extension concerning extensions of morphisms.
Thanks Martin. My hypothesis includes that $K/F$ is finite, hence algebraic. So is it correct to conclude that your third paragraph in fact proves the claim in greater generality? Supposing $K/F$ is finite from the start, one could just jump to the last three sentences of your answer, right?
Feb
24
awarded  Critic
Feb
24
asked Equivalent definition of purely inseparable field extension concerning extensions of morphisms.
Feb
6
awarded  Yearling
Jan
12
accepted If $p\in R[X_1,\dots,X_n]$ is irreducible, is it still irreducible in $R[X_1,\dots,X_n,\dots,X_N]$?
Jan
11
revised If $p\in R[X_1,\dots,X_n]$ is irreducible, is it still irreducible in $R[X_1,\dots,X_n,\dots,X_N]$?
edited body
Jan
11
asked If $p\in R[X_1,\dots,X_n]$ is irreducible, is it still irreducible in $R[X_1,\dots,X_n,\dots,X_N]$?
Nov
6
comment When is $ 0\to\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/nm\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}\to 0$ split?
Thanks for answering, I thought the obvious choice would be the right choice, but I guess I was wrong!
Nov
6
revised When is $ 0\to\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/nm\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}\to 0$ split?
[Edit removed during grace period]
Nov
6
comment When is $ 0\to\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/nm\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}\to 0$ split?
@JyrkiLahtonen Thanks. I'm not in any course actually, maybe I should remove the homological algebra tag. Just one thing, why do we need $x\equiv 0\pmod{n}$ for $s$ to be well defined? Is it something like: if $k\equiv j\pmod{m}$, then $s$ is well defined iff $kx\equiv jx\pmod{mn}$, iff $mn\mid (k-j)x$. Are we then supposed to just choose $k$ and $j$ which forces $n\mid x$?
Nov
6
comment When is $ 0\to\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/nm\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}\to 0$ split?
@JyrkiLahtonen Isn't then number of solutions to $dx\equiv 0\pmod{mn}$ just $\gcd(d,mn)=d$? Sorry, I don't see how either of those comments relate to my question. I'm just curious if a necessary and sufficient condition exists on $m$ and $n$ for the sequence to be split.
Nov
6
revised When is $ 0\to\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/nm\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}\to 0$ split?
edited title
Nov
6
comment When is $ 0\to\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/nm\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}\to 0$ split?
I know that they're equivalent by the Splitting Lemma. I'm curious about a condition on $n$ and $m$ that ensures the sequence is split.
Nov
6
comment When is $ 0\to\mathbb{Z}/n\mathbb{Z}\to\mathbb{Z}/nm\mathbb{Z}\to\mathbb{Z}/m\mathbb{Z}\to 0$ split?
@zibadawatimmy Thanks, but is that a necessary and sufficient condition for the sequence to be split? That's what I can't tell.