Let $V$ be a Zariski-closed subset of $\mathbb{A}^n_k$, where $k$ is an algebraically closed field. Assume that $V$ may be defined by polynomials of degree at most $d$ (or to put it otherwise $V$ is an intersection of hypersurfaces of degree at most $d$). My question is the following: is it also possible to define the irreducible components of $V$ by polynomials of degree at most $d$?

This is true if $V$ is an hypersurface (the irreducible components are defined by the factors of a polynomial defining $V$), so the answer is positive when $n$ is at most 2. I have managed to prove it in a few other cases but not much and I would appreciate any advice.

There exists algorithms to compute the irreducible components. I have checked a few of them but they could let the degree of generators grow. Any algorithm using Gröbner basis for instance will not fit. For the same reasons, trying to prove the results using projections is probably hopeless, since they may increase the degree of the generators.

A few more remarks:

  • I am not sure how relevant the fact that $k$ is algebraically closed is, but I suspect there could be very non-trivial arithmetic issues otherwise, even when $V$ is 0-dimensional.

  • I do not mind to work in the projective space instead of the affine space (it implies the result anyway and makes it easier to deal with degrees).

  • I do not mind to get the irreducible components only as a set (i.e. the ideal up to a radical).

  • I tried to make a few computations but found it rather hard. If you know a way to compute the least integer $d$ such that a given Zariski-closed subset may be defined by polynomials of degree $d$, I would also be glad to know.


3 Answers 3


Here's a counterexample with $n=d=3$.
Let $C$ be the rational curve $\lbrace (x,y,z) = (t,t^4,t^6) \rbrace$. Then the space $S$ of cubics that vanish on $C$ is the span of $\lbrace x^2 y - z, x^2 z - y^2, y^3 - z^2 \rbrace$. But all such cubics vanish also on the line $y=z=0$. Therefore we can take $V$ to be the zero-locus of any two-dimensional subspace of $S$, and $C$ will not be cut out by the space $S$ of cubics vanishing on $C$ (even though for this example we defined $V$ by a proper subspace of $S$).


Edit: the first version of this was completely wrong, I hope this one works.

Take a line $L$ in $\mathbb P^3$, and two general surfaces of degree $d > 3$ passing through $L$; it is not hard to see that the only line contained in $S_1$ is $L$. Their intersection is the union of $L$ with an irreducible curve $C$ of degree $d^2-1$. Now, suppose that $S$ is a surface of degree $d$ containing $C$; then the intersection of $S$ with $S_1$ must be the union of $C$ and a line, which must coincide with $L$. So every surface of degree $d$ that contains $C$ also contains $L$, and this means that we can't cut $C$ with surfaces of degree $d$.

[Edit]: this also works for $d = 3$. A smooth cubic surface famously contains 27 lines, but they define different classes in the Picard group.

  • $\begingroup$ But the component is not required to be a complete intersection of hypersurfaces of degree at most $d$. If $d=2$ then your construction gives a twisted cubic curve $C$, which is indeed cut out by quadrics, though of course not as a complete intersection. $\endgroup$ Mar 28, 2012 at 23:51
  • $\begingroup$ To Noam: oops, sorry, of course you are right. $\endgroup$
    – Angelo
    Mar 29, 2012 at 0:25
  • $\begingroup$ Thank you for your answer. I guess $S_1$ refers to one of the general surfaces you choose at the beginning. $\endgroup$ Mar 29, 2012 at 6:29
  • $\begingroup$ I would like to add that I am sorry I cannot accept your answer too. I chose Elkies' since it was a direct and concrete answer to my question (but I believe yours could also be made concrete). I hope you do not mind. Actually I like your answer a lot even if I was rather disappointed not to have found it myself, since this was really in my line of thoughts. Indeed, I was trying to find a counterexample by trying to find a curve of degree smaller than $d^2$ in $\mathbb{P}^3$ such that the space of surfaces of degree d is a line. And that is exactly what you did. Still a long way to go... $\endgroup$ Mar 29, 2012 at 9:39
  • $\begingroup$ Dear Jérôme, of course I don't mind. $\endgroup$
    – Angelo
    Mar 29, 2012 at 13:53

I am no expert, so might be misunderstanding the question (AND the answer), but a seemingly relevant fact is proved on page 251 of this survey. (Danilov, Algebraic varieties and Schemes, Encyclopaedia of Mathematics), where it is attributed to Fulton.

  • $\begingroup$ I think you refer to Bézout's theorem. In my case, it would say that if I intersect s hypersurfaces of degree d, the sum of the degrees of the irreducible components is at most $d^s$. This is certainly relevant and Angelo actually used it in his answer. On the other hand, it is not so easy to use it directly since I do not want to bound s. $\endgroup$ Mar 29, 2012 at 6:56

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