$\newcommand\P{\mathbb P} \newcommand\C{\mathcal C}$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](https://doi.org/10.1007/BF01236924) (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*](https://doi.org/10.1007/BF01236924),
Arch. Math. 11, 157–158 (1960).



Question on a constructive proof that space projective curves are the intersections of three hypersurfaces
$\newcommand\C{\mathcal C} \newcommand\spn[1]{\langle #1\rangle}$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](https://mathoverflow.net/questions/401720/question-on-a-constructive-proof-of-space-projective-curve-are-the-intersection#comment1026831_401720) 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 $\sqrt{\spn{f,g,h}} = I(\C)$.


