Mathematics

Alessandro Codenotti

9:02 PM

Hi @Dami

Daminark

Yo

Mary Star

Hello!!

Let $V$ be a vector space with with a 5-element basis $B=\{b_1, \ldots , b_5\}$ and let $v_1:=b_1+b_2$, $v_2:=b_2+b_4$ and $\displaystyle{v_3:=\sum_{i=1}^5(-1)^ib_i}$.

I want to determine all subsets of $B\cup \{v_1, v_2, v_3\}$ that form a basis of $V$.

Are the desired subsets the following ones?

$B$, $\{v_1, b_2, b_3, b_4, b_5\}$, $\{v_1, b_1, b_3, b_4, b_5\}$, $\{v_2, b_1, b_2, b_3, b_5\}$, $\{v_2, b_1, b_3, b_4, b_5\}$, $\{v_3, b_1, b_2, b_3, b_4\}$, $\{v_3, b_1, b_2, b_3, b_5\}$, $\{v_3, b_1, b_2, b_4, b_5\}$, $\{v_3, b_1, b_3, b_4, b_5\}$, $\{v_3, b_2, b_3, b_4, b_5\}$

Balarka Sen

\o

Alessandro Codenotti

Hi @Balarka long time no see! What kind of math have you been thinking about lately?

Balarka Sen

Various

Alessandro Codenotti

9:12 PM

Fair enough, I've been doing mostly AG lately but variety is important

9

Balarka Sen

I've been a little unproductive since I had endsems this week. But still thinking about different things.

Alessandro Codenotti

9:22 PM

I see, so you're in a break between semesters now?

Balarka Sen

Almost, the day after tomorrow I have a CS exam and I'm done

Alessandro Codenotti

Nice, when will the next semester begin?

Balarka Sen

The first week of January

Alessandro Codenotti

That's a long break! Have you already decided which courses to take in the next semester?

Balarka Sen

We don't have electives until third year

Alessandro Codenotti

9:29 PM

Ah, I see, that's pretty much how it was in my uni too apart from a single second year course. So what will you do next semester?

Balarka Sen

linear algebra, analysis-II, probability-II and physics

Leaky Nun

@AlessandroCodenotti so basically $k[X^2,X^3]$ is the polynomials where the coefficient of $X$ is zero right. and then $(X^2,X^3)$ is a prime ideal just by the fact that the quotient is $k$. so if you localize there you get a local ring with exactly two prime ideals if we include the zero ideal... but I don't know what you mean by "compute $k[X^2,X^3]_{(X^2,X^3)}$"

Alessandro Codenotti

The issue was deciding whether it is integrally closed mostly

Leaky Nun

oh well, I think it still doesn't have $X$

but lemme think about it

Alessandro Codenotti

It is not integrally closed because $x^2=y^3$ looks bad in the origin

Leaky Nun

9:41 PM

if we want $X(a+fX^2) = b+gX^2$ then comparing coefficients of $X$ gives contradiction

so indeed $X$ isn't in the localization

so it isn't integrally closed

right?

Alessandro Codenotti

yeah I agree $X$ is not in the localization

Leaky Nun

nice

and what's the relation between that ring and that point?

the ring at that point is $(k[X,Y]/(X^2-Y^3))_{(X,Y)}$ right

Alessandro Codenotti

The coordinate ring of $V(x^2-y^3)$ is isomorphic to $k[t^2,t^3]$

Leaky Nun

heh...

just by taking $X=t^3$ and $Y=t^2$?

Alessandro Codenotti

Yeah

Leaky Nun

9:44 PM

interesting

Alessandro Codenotti

So $(X,Y)\mapsto(T^2,T^3)$

Leaky Nun

you flipped them but ok

maybe you meant $y^2=x^3$ in the first place

Alessandro Codenotti

eh, doesn't really make a lot of difference

Leaky Nun

and what does integrally closed have to do with anything?

Alessandro Codenotti

I probably swapped them 12 times

Leaky Nun

9:45 PM

what does integral closure look like geometrically?

Alessandro Codenotti

Well that's the definition, a variety is normal at a point $p$ if it's local ring $\mathcal O_p$ is integrally closed. The variety is normal if it is normal everywhere

Leaky Nun

well I mean, why would that correspond to the point being bad

Alessandro Codenotti

Turns out that for an affine variety $X$ checking whether $k[X]$ is integrally closed or $k[X]_p$ is integrally closed for every $p$ is equivalent. So that localization was actually superfluous

Leaky Nun

because somehow the coordinate ring $k[t^2,t^3]$ managed to be an integral domain, i.e. both irreducible and reduced... right

so what is actually bad about that point?

loch

it's singular

Leaky Nun

9:48 PM

maybe I should compute the tangent space instead

ok, but why would that correspond to not being integrally closed?

Alessandro Codenotti

singular means that the tangent space has issues, what's the relationship between being normal and singular though?

loch

turns out in general, a variety being normal (local rings = integrally closed) is equivalent to the variety being "regular in codimension one" and S2

regular in codimension one essentially means that your singular locus has codimension $\ge 2$

so for curves you just get that your curve is smooth

Leaky Nun

what is S2?

loch

Serre's criterion for normality

In algebra, Serre's criterion for normality, introduced by Jean-Pierre Serre, gives necessary and sufficient conditions for a commutative Noetherian ring A to be a normal ring. The criterion involves the following two conditions for A: R k : A p {\displaystyle R_{k}:A_{\mathfrak {p}}} is a regular local ring for any prime ideal p ...

Leaky Nun

wat

Alessandro Codenotti

9:51 PM

@LeakyNun This is probably not what you meant, but if you take an affine variety $X$, take $k[X]'$ the integral closure of $K[X]\subseteq k(X)$ and then take the variety corresponding to $k[X]'$ this is the normalisation of $X$

Leaky Nun

and how much more machinery is hidden in your "turns out"? @loch

loch

so uh
for S2 specifically, I think it's equivalent to satisfying "hartogs"

Leaky Nun

@TedShifrin do you have any comments from your diff.geom pov :P

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$Mathematics$ Mathematics