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Let $\mathbf R$ denote the real numbers, let's take a finite number of points in $\mathbf R^2$ and let's take the ideal $I$ of all the polynomials that vanish on this points. Using the Hilbert basis theorem we know that $I$ is finitely generated. I want to know if there exists an element in this ideal that is an irreducible polynomial.

Clearly I can suppose that all the finite generators are not irreducible, otherwise it's done. Using this, how can I find such a polynomial?

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Take $F=\sum_{i+j\le N} a_{ij}x_1^jx_2^j$. To pass through the points imposes a finite set of conditions on the coefficients $a_{ij}$. Taking $N$ sufficiently large you can find such an $F$ which is also irreducible. –  J.C. Ottem Apr 5 '12 at 20:05
But how can I prove that it´s irreducible? –  Daniel Apr 5 '12 at 20:09
Let ${\cal P}_N$ be the space of degree-$N$ polynomials. The reducible polynomials in ${\cal P}_N$ form the union of subvarieties each of codimension at least $N-O(1)$ [see e.g. my answer to]. The polynomials vanishing on a finite set $S$ of points constitute a subspace of ${\cal P}_N$ of codimension at most $\#(S)$. Therefore, once $N > \#(S) = O(1)$ most polynomials in that subspace are irreducible, QED. –  Noam D. Elkies Apr 5 '12 at 21:06
I don´t understand your answer, and your link is deleted. –  Daniel Apr 5 '12 at 23:59
@Daniel: Try (I see that if you click on what I wrote the final "]" gets appended to the URL). What I wrote has a slight typo: should be $N > \#(S) + O(1)$, not $\#(S) = O(1)$. Basically, once $N$ is large enough (bigger than the size of the finite subset plus a bit), there aren't enough reducible polynomials of degree $N$ to fill the space of degree-$N$ polynomials vanishing on $S$. (If $N < \#(S)$ the claim can fail, because all the points in $S$ might lie on one line $l$, and then a degree-$N$ polynomial vanishing on $N$ would vanish identically on $l$.) –  Noam D. Elkies Apr 6 '12 at 0:08

1 Answer 1

To give an explicit answer, choose a system of coordinates $x,y$ such that no two points have the same $x$ coordinate. This is possible since the slopes of lines that pass through pairs of points in the set are only finitely many of all the slopes. Then use basic algebra or the Chinese remainder theorem to find a polynomial equation $y=f(x)$ that passes through all your points. $y-f(x)$ is irreducible so you're done.

(Proof: its degree in $\mathbb R[x][y]$ is one so it must be the product of a linear and a constant term, but no nontrivial constant terms divide it.)

Edit: A secondary question might be: what is the worst-case scenario lowest-degree polynomial that accomplishes this goal? This method shows that, with $n$ points, there is always a degree $n-1$ irreducible polynomial. Sometimes, there is no degree $n-2$ polynomial: Take $n-1$ points on a line, and one off it. Then an irreducible polynomial vanishing at all $n$ points cannot vanish identically on the line, but vanishes at $n-1$ points on it, so has degree at least $n-1$.

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Ah and I forgot something, not only must cancel on this points, and be irreducible, also has only that roots , and no more! –  Daniel Apr 6 '12 at 5:46
I would never have thought there was such a beautifully elementary solution: congratulations, Will ! –  Georges Elencwajg Apr 6 '12 at 7:15
@Will Sawin Your polynomials has other roots right? ( Not just the finite points, What can I do to find a polynomial that only has that finite points and it´s also irreducible)? –  Daniel Apr 6 '12 at 15:31
@J.C. Ottem I have a question, how can I put in the coefficients the conditions of being irreducible? –  Daniel Apr 6 '12 at 15:32
@Daniel: Let $y,f(x)$ be as before and $g(x)$ vanish at the $x$ coordinates of the set of points, then $(y-f(x))^2+g(x)^2$ is irreducible in $\mathbb R[x,y]$, because it splits in $\mathbb C[x,y]$ into $(y-f(x)+ig(x))(y-f(x)-ig(x))$, neither of which is in $\mathbb R[x,y]$ –  Will Sawin Apr 7 '12 at 0:03

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