Reputation
4,404
Next privilege 5,000 Rep.
Approve tag wiki edits
Badges
2 7 43
Impact
~38k people reached

Jul
26
comment Set that is not algebraic
Thanks a lot Zev! This makes more sense now. Just to paraphrase: We start with a function $f\in I(A)$ that is supposed to vanish on $A$, and we find that $f$ actually vanishes on the unit cylinder! So the Zariski closure of $A$, which is $V(I(A))$, is strictly larger than $A$, and so $A$ is not Zariski-closed.
Jul
26
comment Set that is not algebraic
I think you are right. But I still don't know how to complete your argument. So you start with $f(x, y, z)\in V(A)$, and then $f(\cos(\theta), \sin(\theta), z)=0$ (as a polynomial in $z$) for each fixed $\theta\in [0, 2\pi)$. Can you explain why this forces $f=0$ (as a polynomial in $x$, $y$, $z$)?
Jul
26
comment Set that is not algebraic
Hmmm, I think it is somewhat easier to argue as follows: Let $L\subset\mathbb{A}^{3}$ be the line given by $x=1$ and $y=0$, i.e. $L=\{(1, 0, z): z\in\mathbb{R}\}$. If $A$ were algebraic set, then $L\cap A = \{(1, 0, 2\pi k): k\in\mathbb{Z}\}$ would be an infinite proper closed subset of $L\cong \mathbb{A}^{1}$ in the Zariski topology, contradiction.
Jul
26
comment Set that is not algebraic
I am having trouble completing the argument here. So, for each fixed $\theta\in [0, 2\pi)$, we have $f(\cos(\theta), \sin(\theta), z)=0$ as a polynomial of $z$. How can we conclude that $f(x, y, z)=0$? Can't we have something like $f(x, y, z)=(x^2+y^2-1)z$?
Jul
24
comment Number of square matrices of order $n$ where each row and each column has at most one $1$
Thanks for the answer! :) I especially like the second alternative form $\sum_{k=0}^n\binom{n}k^2k!\;$, because it also gives another proof for the count: for each $0\leq k\leq n$, we pick $k$ rows and $k$ columns to fill in with $1$s (which gives a factor of $\binom{n}k^2$), and there is exactly $k!$ ways of doing this.
Jul
24
accepted Number of square matrices of order $n$ where each row and each column has at most one $1$
Jul
24
asked Number of square matrices of order $n$ where each row and each column has at most one $1$
Jul
23
comment The group $G$ has order 56. Show that it contains normal Sylow p-subgroup.
The proof looks good to me. See here for a more general result.
Jul
21
answered Prove that $xy+yz+zx \leq x^2+y^2+z^2$
Jul
20
awarded  Nice Question
Jul
15
comment Show that $V(x^2y^2+x^2-xy-y^2)$ is irreducible.
I just read your edit. Excellent! This is a very nice result.
Jul
15
comment Show that $V(x^2y^2+x^2-xy-y^2)$ is irreducible.
No problem. I think the irreducibility of $f$ does imply the irreducibility of $V(f)$ if the field is algebraically closed. Indeed, if $f$ is irreducible, then $f$ is prime (since we are in a UFD), so the ideal generated by $(f)$ is prime ideal. So $J=(f)$ is a prime ideal, in particular radical. Now applying Nullstellensatz, we get $I(V(f))=I(V(J))=\sqrt{J}=J$ is prime. But it is a theorem that $X$ is irreducible iff $I(X)$ is prime for affine algebraic set $X$ (See Proposition 3 in page 198 in Cox, Little, O'shea --- 2007 edition). In this case $X=V(f)$. Does this make sense? :)
Jul
15
comment Show that $V(x^2y^2+x^2-xy-y^2)$ is irreducible.
It is sometimes possible to have $f(x, y)$ irreducible in $\mathbb{R}[x, y]$, and yet $V ( f(x, y) )$ to be reducible! For example, $f(x, y)= (x^2 - 1)^2 + y^2$ is irreducible in $\mathbb{R}[x, y]$, but $V( f(x, y) ) $ consists of two points $(-1, 0)$ and $(1, 0)$, so is reducible. So there is something more to show! +1
Jul
15
comment Need help to show $R/I$ is not necessarily flat over $R$
Could you explain where the exactness fails in the last sequence? Isn't the map $\mathbb{Z}/3\mathbb{Z}\to \mathbb{Z}/3\mathbb{Z}$ there given by $x\mapsto 2x$, which is a bijective map?
Jul
14
accepted Can we always find $q\in\overline{\{p\}}$ with $p\notin \overline{\{q\}}$
Jul
14
comment Every quasi-compact scheme has a closed point
I just posted a follow up question concerning the parenthetical remark in your post. I think it is pretty interesting!
Jul
14
asked Can we always find $q\in\overline{\{p\}}$ with $p\notin \overline{\{q\}}$
Jul
14
accepted Every quasi-compact scheme has a closed point
Jul
14
comment Every quasi-compact scheme has a closed point
Ah, because then $P_{3}$ would be closed automatically.
Jul
14
comment Every quasi-compact scheme has a closed point
When you pick $P_{3}$ in $\overline{P_2}$, it is possible to make sure that $P_{3}\neq P_{2}$ (because $P_{2}$ is not closed in $X$). But how do you make sure that $P_{3}\neq P_{1}$? (This would follow if we can somehow show $P_{1}\notin\overline{P}_2$)