# Question on a constructive proof that space projective curves are the intersections of three hypersurfaces

$$\newcommand\P{\mathbb P} \newcommand\C{\mathcal C}$$I am a bit confused by a proof I am reading on the fact that a projective algebraic space-curve (i.e. an algebraic curve in $$\P^3(k)$$, where $$k$$ is an algebraically closed field) is the intersection of 3 hypersurfaces. I am trying to constructively create the hypersurfaces given the vanishing ideal of the curve.

The proof I am reading is from an old source (written in 1960) by Martin Kneser. Since then, many generalizations of this fact exists (e.g. Storch, Forster, Eisenbud–Evans etc.). However, I would like to understand this proof by Kneser because it is quite constructive and supposedly the simplest of all.

The text is written in German. I just don't quite understand how the proof starts.

This is how Kneser starts the proof (I will do my best by adding my interpretation and translation to the original German text):

Let $$\C$$ be the curve and suppose $$I=\langle f_1,\dotsc,f_n \rangle$$ ($$f_i$$ homogenous and nontrivial) is the vanishing ideal of the curve. And suppose that our coordinates are $$(t:x:y:z)$$ in $$\P^3$$. Without loss of generality, one may assume that the point $$P:=(1:0:0:0)$$ is in $$\C$$. Then Kneser continues by letting $$f\in k[x,y,z]$$ be a generator of the vanishing (principal) ideal of the "cone" of $$\C$$ with $$P$$ as its vertex.

First question: Is a generator of the vanishing ideal of the cone the elimination ideal $$I\cap k[x,y,z]$$?

Kneser then continues by assuming that $$g \in k[t,x,y,z]$$ is (my interpretation) a homogenous polynomial among the generators $$f_1,\dotsc,f_n$$ with minimum degree (say $$d$$) in $$t$$. One may write $$g = \sum_{i=0}^d g_i t^i$$ for (homogenous) polynomials $$g_i \in k[x,y,z]$$ such that $$g_d\not\equiv 0$$.

Then comes the most confusing part of the whole construction. Kneser states (without a proof) something like Lemma:

For any $$p\in I$$ with degree $$m$$ over $$t$$ one has the polynomial division (by $$g$$)

$$g_d^m p = qg + r$$ where the remainder $$r$$ is divisible by $$f$$.

Second question: Why must $$r$$ be divisible by $$f$$?

From the above "Lemma" Kneser concludes that $$Z(f,g) = \C\cup V$$ where $$V$$ is a subset of the vanishing of $$(f,g_d)$$. He denotes the vanishing set $$D:=Z(f,g_d)$$ and states that $$D$$ is the union of finitely many lines passing $$P$$.

I will not continue the proof here (unless someone needs to see all of it in my broken translation). I am already confused on why $$V\subset Z(f,g_d)$$ and that $$Z(f,g_d)$$ is the union of lines.

Can anyone who understands this better than me clarify my confusion? The title of the paper (which is only 2 pages) in the original German is:

Über die Darstellung algebraischer Raumkurven als Durchnitte von Flächen, Arch. Math. 11, 157–158 (1960).

• Answer to the first question is yes. The second question is answered by thinking of the degree of $r$ in $t$. By the way your notation mixes up $n$ and $d$. Aug 14, 2021 at 16:36
• Thank you for the short answers. I just corrected the typos as well. Aug 14, 2021 at 17:37
• I did some proofreading while this was on the front page. In the meantime, since there was an identified "First Question" but not explicitly identified "Second Question", I singled out what seems to be the Second Question. If I got it wrong, please feel free to change back. Oct 9, 2021 at 15:14

$$\newcommand\C{\mathcal C} \newcommand\spn[1]{\langle #1\rangle}$$Using cohomology is interesting when dealing with higher dimensions and more general results but I don't see why it would result into a shorter proof. In fact the generalization of this result of Kneser (Eisenbud and Evans) avoids any sheaf cohomology as well and the result is just as short as this specific case.

Here is Kneser's algorithm with some explanation:

The initial assumption is that, without loss of generality, $$P=(1:0:0:0)$$ is on the curve. We use the same notation $$(t:x:y:z)$$ as Kneser for the coordinates. Let $$\mathbb k$$ be your (algebraically closed) ground field. One also assumes that $$\C$$ is not a line (otherwise the algorithm won't work, but a line is trivially a set-theoretic complete intersection).

1. There exists a $$g \in I(\C)$$ with the property that:
a.) $$d > 0$$ where $$d=\deg_t(g)$$
b.) $$g_d$$ is not in $$I(\C)$$ (where $$g_d\in \mathbb k[x,y,z]$$ is the coefficient of $$t^d$$).
c.) $$d$$ is minimum with the property a.) and b.)

2. As you noted $$I(\C)\cap \mathbb k[x,y,z]$$ is principal, and is generated by say $$f$$. The vanishing of $$f$$ is therefore a cone with "vertex" at $$P$$ and "base" $$\C$$. Note that $$Z(f,g_d)$$ is a finite union of lines because: If $$Q\in Z(f,g_d)\backslash\{P\}$$ then the line $$PQ$$ is on the surface $$Z(g_d)$$ and on the cone $$Z(f)$$ which means that this line meets $$\C$$. We also know that $$g_d\notin I(\C)$$ so the curve $$\C$$ cuts the surface $$Z(g_d)$$ finitely many times.

3. Kneser then states that for any $$p\in I(\C)$$ we can find a $$m\ge 1$$ such that $$g_d^m p \in \spn{f,g}$$. As pointed out in the comments by Kapil, you can use degree (of $$t$$) argument to prove this.

4. Suppose now $$g(Q)=0$$ and $$f(Q)=0$$, then either $$Q\in \C$$ or $$Q\notin \C$$ and in this case there is a $$p\in I(\C)$$ such that $$p(Q)\ne 0$$. By 3. $$Q\in Z(g_d)$$. So, $$Z(f,g) \subset \C \cup Z(f,g_d).$$ In words, $$Z(f,g)$$ is contained in the union of the curve $$\C$$ and finitely many lines whose common intersection is $$P$$. This already provides an immediate and nice proof that all irreducible projective space (algebraic) curves are intersections of an analytic surface and an algebraic surface, just perturb the cone (or the surface $$Z(g)$$) so that it intersect these lines only at $$\C$$.

5. We want to now find a third algebraic surface $$Z(h')$$ (the two previous surfaces are $$Z(f)$$ and $$Z(g)$$) that avoids these finite lines $$Z(f,g_d)$$ except at $$\C$$. This can be done in several ways, Kneser himself provides an algorithm with a proof. I believe his algorithm is inefficient but didactically the easiest one to follow. I will not explain this part, but you can usually get away using a surface defined by products and sums of linear and quadratic forms avoiding the lines and having some higher multiplicity at $$P$$.

6. By Hilbert's Nullstellensatz there is therefore an $$N\in \mathbb N$$ and $$h\in I(\C)$$ such that $$h'^N - h \in \spn{g_d}.$$ You can obtain $$N$$ by just iterating through the powers of $$h'$$ and reducing via Groebner basis (ideal-membership test) of $$\spn{g_d}+I(\C)$$ and you can obtain $$h$$ as a by-product. Therefore we choose $$h$$ and get $$\sqrt{\spn{f,g,h}} = I(\C)$$.

This is not precisely an answer to your question (since you would like to understand Kneser's paper), but at the expense of using some modern technology (mainly sheaf cohomology) and being a little less constructive, I think one can get a very short proof of the result you mention.

EDIT: as abx points out in a comment below, my proof does not quite work. A fourth hypersurface containing $$C_0$$ of sufficiently high degree is needed to get rid of the potential points of intersection of $$S_3$$ with the $$C_i$$ ($$i>0$$) which do not lie on $$C_0$$.

Let $$\mathcal{C} \subset \mathbb{P}^{3}$$ be an integral projective curve (over an algebraically closed field) and denote by $$\mathcal{I}$$ its sheaf of ideal in $$\mathbb{P}^3$$. For any $$d \geq 0$$, the vector space:

$$H^0(\mathbb{P}^3, \mathcal{I}(d))$$

is the vector space of hypersurfaces of degree $$d$$ which contain $$\mathcal{C}$$. By Serre's vanishing Theorem, we know that the sheaf $$\mathcal{I}(d_1)$$ is globally generated for $$d_1\gg0$$, in particular $$H^0(\mathbb{P}^3, \mathcal{I}(d_1)) \neq 0$$ for $$d_1\gg0$$. An explicit bound for $$d_1$$ should be equivalent to the computation of the degree of your $$f$$ generating $$I \cap k[x,y,z]$$.

Let $$f_1, f_2$$ be two generic elements of $$H^0(\mathbb{P}^3, \mathcal{I}(d_1))$$. Up to increasing $$d_1$$, we can assume that the hypersurfaces $$S_1$$ and $$S_2$$ given by $$f_1 = 0$$ and $$f_2 =0$$ are integral. So in particular, we have, set-theoretically:

$$S_1 \cap S_2 = \bigcup_{i=0}^{r} \mathcal{C}_i,$$ with $$\mathcal{C}_0 = \mathcal{C}$$ and the $$\mathcal{C}_i$$ are integral projective curves that meet properly.

Now, it is sufficient to find $$d_2>0$$ such that the vanishing locus of a generic section in $$H^0(\mathbb{P}^3, \mathcal{I}(d_2))$$ does not contain any of the $$\mathcal{C}_i$$ for $$i>0$$. Since $$k$$ is infinite and any vector space over $$k$$ is thus infinte, it is sufficient to prove that given a fixed $$i \neq 0$$, there exists $$d_2(i) >0$$, such that the vanishing locus of a generic section in $$H^0(\mathbb{P}^3, \mathcal{I}(d_2(i)))$$ does not contain $$\mathcal{C}_i$$ (and then apply the result for all $$i$$ by taking the complement in $$H^0(\mathbb{P}^3, \mathcal{I}(d_3))$$ of the union of the subspaces to be avoided, where $$d_3$$ is the max of the $$d_2(i)$$).

Thus, we are left to prove that for $$d_2\gg0$$, we have: $$H^0(\mathbb{P}^3, \mathcal{I}(d_2)) \neq H^0(\mathbb{P}^3, \mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)),$$ where $$\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}$$ is the ideal sheaf of $$\mathcal{C}_0 \cup \mathcal{C}_i$$. Since the intersection $$\mathcal{C}_0 \cap \mathcal{C}_i$$ has the right codimension (i.e. $$3$$), we have an exact sequence (see Sasha's answer to Products of Ideal Sheaves and Union of irreducible Subvarieties):

$$0 \longrightarrow \mathcal{I}_{\mathcal{C}_0 \cup \mathcal{C}_i} \longrightarrow \mathcal{O}_{\mathbb{P}^3} \longrightarrow \mathcal{O}_{\mathcal{C}_0} \oplus \mathcal{O}_{\mathcal{C}_i} \longrightarrow \mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i} \longrightarrow 0.$$

Tensoring by $$\mathcal{O}_{\mathbb{P}^3}(d_2)$$ and taking Euler carateristics, we get: $$\chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{O}_{\mathbb{P}^3}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{O}_{\mathcal{C}_i}(d_2)) + \chi(\mathcal{O}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ that is:

$$\chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = \chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) - \chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)),$$ where $$\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}$$ is the ideal sheaf of $$\mathcal{C}_0 \cap \mathcal{C}_i$$ in $$\mathcal{O}_{\mathcal{C}_i}$$.

Applying Serre's vanishing theorem one more time, we find that for $$d_2\gg0$$:

• $$\chi(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2)) = H^0(\mathcal{I_{\mathcal{C}_0 \cup \mathcal{C}_i}}(d_2))$$, $$\chi(\mathcal{J}_{\mathcal{C}_0}(d_2)) = H^0(\mathcal{J}_{\mathcal{C}_0}(d_2))$$ and $$\chi(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) = H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2))$$,

• $$H^0(\mathcal{J}^{\mathcal{C}_i}_{\mathcal{C}_0 \cap \mathcal{C}_i}(d_2)) \neq 0$$.

As a consequence, for $$d_2(i)\gg0$$, the generic hypersurface of degree $$d_2(i)$$ containing $$\mathcal{C}_0$$ does not contain $$\mathcal{C}_i$$. This is true for all $$i$$ and since $$k$$ is infinite, by taking $$d_3$$ the max of the $$d_2(i)$$, we find that the generic hypersurface of degree $$d_3$$ does not contain any of the $$\mathcal{C}_i$$ for $$i \neq 0$$. Let $$S_3$$ be such a generic hypersurface, then we have, set-theoretically:

$$S_1 \cap S_2 \cap S_3 = \mathcal{C}_0.$$

• Why is it sufficient to prove that your general surface in $|\mathscr{I}(d_2)|$ doesn't contain $\mathscr{C}_i$? It could intersect $\mathscr{C}_i$ in a point not contained in $\mathscr{C}$.
– abx
Aug 16, 2021 at 7:40
• @abx : right, stupid mistake. Thanks for the comment. It seems I need a fourth hypersurface to get rid off the residual points of intersection. I will edit. Aug 18, 2021 at 16:10
• It'd be nice if the edit was a disclaimer before the main text instead of at the very end. Aug 22, 2021 at 7:34
• I edited to fix some typos and link to @‍abx's comment and the answer I think you wanted. Since "the first answer" is not well defined over time, I looked for one that seemed to fit your description, and the accepted answer does, so I linked to that. I apologise if it was not correct. In the meantime, I took @user347489's suggestion and moved the disclaimer to the front. Oct 9, 2021 at 15:25