$\newcommand\P{\mathbb P} \newcommand\C{\mathcal C}$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, Eisenbund-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 $I=\langle f_1,..,f_n \rangle$ ($f_i$ 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 $f_1,\dots,f_n$ 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 $f$

I don't quite follow why $r$ must 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).

$\newcommand\P{\mathbb P} \newcommand\C{\mathcal C}$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).

$\newcommand\P{\mathbb P} \newcommand\C{\mathcal C}$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, Eisenbund-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 $I=\langle f_1,..,f_n \rangle$ ($f_i$ 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 $f_1,\dots,f_n$ 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 $g_n\not\equiv 0$.

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 $g$)

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

I don't quite follow why $r$ must 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 $(f,g_n)$. He denotes the vanishing set $D:=Z(f,g_n)$ 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 $V\subset Z(f,g_n)$ and that $Z(f,g_n)$ 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).

$\newcommand\P{\mathbb P} \newcommand\C{\mathcal C}$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, Eisenbund-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 $I=\langle f_1,..,f_n \rangle$ ($f_i$ 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 $f_1,\dots,f_n$ 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 $f$

I don't quite follow why $r$ must 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).