Mathematics - 2018-05-14 (page 2 of 2)

Mike Miller

22:01

here's a way to reparse your old argument: let $g$ be a polynomial in $n$ complex variables. for any choices $a_2, \dots, a_n \in \Bbb C$, $g(z, a_2, \cdots, a_n)$ is a complex polynomial of a single variable. if it's identically zero then any $(z, a_2, \cdots, a_n)$ is a solution; otherwise there exists at least one solution $(z_1, a_2, \cdots, a_n)$ by FTA

you're looking at the function on $\Bbb C^{n+1}$ by looking at the "slices" by a single $\Bbb C$, of which there are $\Bbb C^n$ from plugging into the other coordinates

Leaky Nun

well I did choose 1 explicitly

@MikeMiller intersecting with a line :P

Mike Miller

yep

Leaky Nun

so you're saying that in fact $V(f)$ intersects every line?

Mike Miller

I'm pointing out that you can use this as a way to see that there are solutions no matter what other things you plug in, and you can sort of see the variety as built up out of these slices

yeah

Leaky Nun

that's interesting

I've never thought about that before

I mean, $xy-1$ certainly doesn't intersect the line $x=0$...

Mike Miller

22:03

rip

Leaky Nun

aha, your argument is flawed, it can be a constant

Mike Miller

you could get a nonzero constant

yeah

the picture I'm describing of scanning $V(f)$ by looking at its various intersections, is in other words understanding by the fibers of $V(f) \to \Bbb C^n$

Leaky Nun

how to fix your argument?

Mike Miller

the easy answer is "exercise", since I don't know but have a good idea

but I'll think for a few minutes

Leaky Nun

well then currently my argument is the only argument that works lol

Mike Miller

22:07

the objection you wanted to state above was "what if I get a constant nonzero polynomial when you substitute 1 to the other values"

Leaky Nun

right

I'm proud of my proof

Mike Miller

I feel like this is a contest for you, which makes me less inclined to bother thinking about it, and I wasn't interested in making you feel worse about your proof

Leaky Nun

@MikeMiller sorry...

I'm really looking for a more elementary proof

sorry if my tone makes you think otherwise

Mike Miller

the easy way to fix the original argument, using $0$ instead of $1$ to simplify notation, is to pick a variable $z_i$ that actually shows up in the monomial expansion of the polynomial, and say "plug in $0$ to everything else"

Leaky Nun

wouldn't work for $xy-1$

Mike Miller

22:11

irritating, this isn't hard

sorry about my incompetence

yeah that was the wrong thing to say

Leaky Nun

I was just... excited about my new tools

Mike Miller

if $g(z, z_2, \cdots, z_n)$ is constant - and the same constant - for all $(z_2, \cdots, z_n)$, then $g$ itself is the constant polynomial. this is the case you want to avoid

I realize. Sorry for making you feel bad.

loch

your proof says more generally that if you have a bunch (not just one!) of polynomials that don't generate the ideal $(1)$ then there is a point lying in their zero locus

which i think should be harder if you want to do this using elementary methods

Leaky Nun

aha, someone generalized my proof!

indeed "a bunch" (finite) is sufficient, since our ring is noetherian

I was about to say "you can even have infinitely many polynomials"

Mike Miller

you were right, I don't see a clean argument here. here is an argument that works.
1) If $f$ is not constant, there will either be a line $(z, a_2, \cdots, a_n)$ on which it restricts to a nonconstant function, or it will restrict to a different constant on 2 different lines.
2) In the first case, FTA.
3) In the second case, pick a line that intersects both of your previous two parallel lines. $g$ is not constant on this line. FTA.

Leaky Nun

22:27

I think you implicitly used "constant polynomial iff constant function"

which to me isn't immediately obvious

Mike Miller

ok

MatheinBoulomenos

So here's an exercise: proof that if $k$ is an infinite field, then the obvious function $k[x_1, \dots, x_n] \to \operatorname{Hom}_{Set}(k^n,k)$ is injective

Leaky Nun

@MatheinBoulomenos is that relevant?

MatheinBoulomenos

sure

it says that constant polynomia iff constant function as a special case

Leaky Nun

hmm, no nullstellensatz for me :c

Mike Miller

22:30

Why can't both make you happy?

Leaky Nun

aha, I proved that before

somehow it becomes easier if you phrase it that way

ok, it isn't that the proof is really easy

MatheinBoulomenos

it's an easy inductive argument

Leaky Nun

firstly we have a lemma: if $R$ is any infinite integral domain, and $f \in R[X_1,\cdots,X_n]$, then $f=0 \iff \forall v, f(v)=0$

MatheinBoulomenos

this already implies the above statement

Leaky Nun

sure, treat it as a restatement

the base case $n=1$ is because of factor theorem, which follows from division algorithm for monic polynomials (which holds for any commutative ring, because the divisor is monic)

Now, assume $n=k+1$ and $f \in R[X_1, \cdots, X_{k+1}]$

then, treat $f \in R[X_{k+1}][X_1, \cdots, X_k]$

claim: $f(v)=0$ for every $v \in R[X_{k+1}]^k$

Mike Miller

22:33

Generally I think whenever I talk to you it seems like you're crusading to find something wrong with what I just said. Sometimes I say wrong things, like here. But if it felt like you were trying to find a mistake (as opposed to understanding) I'm just going to feel pissed off instead of encouraged to fix it.

Maybe that's my fault. In any case it seems pretty clear that neither of us got anything of value out of the past half hour so I'll make an effort to not engage in the future to avoid that situation.

Leaky Nun

...

proof: $f(v) \in R[X_{k+1}]$, resort to base case

therefore, $f=0$, by induction hypothesis, since $R[X_{k+1}]$ is also an integral domain

@MatheinBoulomenos so I generalized your thing to infinite integral domains

MatheinBoulomenos

that's just a restatement

but well done

this lemma actually comes up in the proof of the existence of normal bases

Leaky Nun

corollary: if $R$ is an integral domain, then any nonempty open set in $\Bbb A^n_R$ is dense

@MatheinBoulomenos do you see an elementary proof for the problem way above?

May 1 at 11:32, by Silent

I know that every polynomial with complex coefficients has a solution in complex field. is it true that every 'multivariate polynomial' has solution in complex field?

scroll up for my nuke proof and someone's generalization

@loch's gneralization

MatheinBoulomenos

You can use the lemma you just proved: If $f$ is a multivariate polynomal that is not constant, then there's a point where it doesn't vanish. Then prove the statement by induction. It's obvious for $n=1$. If it's true for a fixed $n$ and $f$ is a nonconstant multivariate polynomial in $n+1$ variables, write $f=\sum_{i=0}^k f_i \cdot x_{n+1}^i$, where $f_i \in k[x_1, \dots, x_n]$.
Case 1: If every $f_i$ is constant, then $f$ is a polynomial in one variable, so it has a root.
Case 2: If $f_0$ is the only noncontant polynomial among the $f_i$, then $f$ is a polynomial in $n$ variables, so it h…

Ah case 2 is nonsense, sorry

Fixed Case 2: If $f_0$ is the only nonconstant polynomial among the $f_i$, then we have $f_0=-\sum_{i=1}^k f_i x_{n+1}^i$, the RHS is a polynomial just in $x_{n+1}$, thus the RHS has a root $a$. Because of this equality, $a$ will also be a root of $f$

Or you derive a contradiction from $f_0=-\sum_{i=1}^k f_i x_{n+1}^i$, since the RHS is a polynomial in $x_{n+1}$ and the LHS is a polynomial in $x_1, \dots, x_n$

Leaky Nun

22:53

hmm...

interesting

thanks

MatheinBoulomenos

if you modify the argument a bit, you can also show that every multivariate polynomial in $\geq 2$ variables has infinitely many roots

(over an algebraically closed field)

Leaky Nun

hmm...

WeavingBird1917

23:34

Can somebody help with a weird 3 line mathematica problem?

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$Mathematics$ Mathematics