0

My question rises from the theorem for the ring $\mathbb F[x]$ which suggests that if $f(\alpha)=0$ then $f(x)=g(x)(x-\alpha)$ for some $g\in \mathbb F[x]$.

Is there a similar theorem for $\mathbb F[x_1,\ldots,x_n]$?

My guess would be yes, and that it should be something like if $f(y_1,\ldots,y_n)=0$ then $f(x_1,\ldots,x_n)=g(x_1,\ldots,x_n)(x_1-y_1)\cdots(x_n-y_n)$ and that the proof should be by induction.

I understood it was wrong, thanks. Is the following right though?

could I say that for any $1\leq i\leq n$ I desire i can find $g_i$ such that $f=g_i⋅(x_i−y_i)$? that sound about right for if I consider $f$ as a one variable polynoimial, fixing all coordinated but the $i$ one

thanks

3 Answers3

3

Yes, there is a similar result for $\mathbb F[x_1,\dots,x_n]$, but different than you thought: $$f(\alpha_1,\dots,\alpha_n)=0\iff f\in(x_1-\alpha_1,\dots,x_n-\alpha_n).$$ Note that for a single variable we have $$f(\alpha)=0\iff f\in(x-\alpha)$$ which is equivalent to $x-\alpha\mid f$ as you started.

user26857
  • 53,190
  • Awesome, that's exactly the kind of answer I was looking for. Is there a refference where I could find a proof or the idea behind it? –  Oct 23 '15 at 14:06
  • @TheSpellofMath This has been asked many times before; see http://math.stackexchange.com/questions/500153/proving-that-kernels-of-evaluation-maps-are-generated-by-the-x-i-a-i – user26857 Oct 23 '15 at 14:17
  • Another topic: http://math.stackexchange.com/questions/1352606/is-a-polynomial-f-zero-at-a-1-ldots-a-n-iff-f-lies-in-the-ideal-x-a-1 – user26857 Oct 23 '15 at 16:12
  • Notice that your equivalence involves a multiply-generated ideal, whereas OP was looking for a singly-generated ideal. This was a truly excellent answer! – Lubin Oct 23 '15 at 17:54
  • @Lubin If you read the OP's comment above maybe can change your mind and withdraw this (intended to be ironic? definitely useless) comment. – user26857 Oct 24 '15 at 10:16
1

Hint : consider $f(x_1,x_2):=x_1-x_2$. Then for any $y$ we have $f(y,y)=0$... But with your proposition you would have (for $y=0$) that $(x_1-0)(x_2-0)$ divides $f(x_1,x_2)$ which is obviously not true.

Something weaker can be said... Provided that $\mathbb{F}$ is algebraically closed, you have the Nustellensatz see wikipedia.

Clément Guérin
  • 11,721
  • without assuming $\mathbb F$ is algebraically closed, could I say that for any $1\leq i \leq n$ i desire i can find $g_i$ such that $f=g_i\cdot (x_i-y_i)$? that sound about right for if I consider $f$ as a one variable polynoimial, fixing all coordinated but the $i$ one. –  Oct 23 '15 at 12:30
  • @TheSpellofMath The problem with this is that it implies that if $f$ has some zero then $f$ is reducible... This is false take $f(x,y)=x^2+y^2-1$ like in Lubin's answer or $f(x,y)=x^2-y^3$... Those polynomials have some zero but you cannot factorize them by anything... Remark, about your assertion, when you fix all coordinate but $i$ the $g_i$ you get depend (in general) of where you fixed the other coordinates... That's why you cannot factorize $f$... – Clément Guérin Oct 23 '15 at 13:52
1

No, sorry, completely wrong.

Consider the polynomial $f(x,y)=x^2+y^2-1$, describing the unit circle in the plane. That is, a zero of $f$ is a point, like $(3/5,4/5)$, on the unit circle. But you can see at a glance that $f$ is not divisible by $(x-3/5)$ nor $(y-4/5)$, indeed $f$ is irreducible, has no nonconstant factors.

Once you start looking at many variables instead of just one, suddenly Geometry rears its lovely head. In particular, a single polynomial in several variables can be expected to have infinitely many zeros, even over the ground field (as in my example), and certainly over an algebraically closed field.

Lubin
  • 65,209
  • thanks. could I say that for any $1\leq i\leqn$ i desire i can find $g_i$ such that $f=g_i\cdot(x_i−y_i)$? that sound about right for if I consider $f$ as a one variable polynoimial, fixing all coordinated but the $i$ one, although i can't find such decomposition for your example. –  Oct 23 '15 at 12:32
  • No, geometrically, a factorization of $f$ means a decomposition of its zero-set as the union of two simpler sets (which may intersect). Look at the zero-set of $f(x)=x^2-y^2$, which does factor, and so the zero-set is the union of the two lines $y=\pm x$. – Lubin Oct 23 '15 at 13:20
  • thanks, now i see a little bit more about the geomoetry of that, thanks. so that makes me look for a factor which isn't of the type of $(x_i-y_i)$ (which would give a full $n-1$ dimensional space of zeros to the polynomial which isn't necessarily true. So i thought about $f=g\cdot(\sum x_i -\sum y_i)$. is this one more reasonable? this should be a discrete set of roots, maybe even of size 1. –  Oct 23 '15 at 13:46
  • No, you really have to give this idea up. Always check against examples! – Lubin Oct 23 '15 at 17:52