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$\newcommand\C{\mathcal C} \newcommand\spn[1]{\langle #1\rangle}$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 $$h'^N - h \in \spn{f,g_d}.$$ You can obtain $N$ by just iterating through the powers of $h'$ and reducing via Groebner basis of $\spn{g_d}+I(\C)$ and you can obtain $h$ using Groebner basis and/or syzygies. Therefore we choose $h$ and get $\sqrt{\spn{f,g,h}} = I(\C)$.

$\newcommand\C{\mathcal C} \newcommand\spn[1]{\langle #1\rangle}$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 $\sqrt{\spn{f,g,h}} = I(\C)$.

$\newcommand\C{\mathcal C} \newcommand\spn[1]{\langle #1\rangle}$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 $$h'^N - h \in \spn{f,g_d}.$$ You can obtain $N$ by just iterating through the powers of $h'$ and reducing via Groebner basis of $\spn{g_d}+I(\C)$ and you can obtain $h$ using Groebner basis and/or syzygies. Therefore we choose $h$ and get $\sqrt{\spn{f,g,h}} = I(\C)$.

$\newcommand\C{\mathcal C} \newcommand\spn[1]{\langle #1\rangle}$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 $$h'^N - h \in \spn{f,g_d}.$$ You can obtain $N$ by just iterating through the powers of $h'$ and reducing via Groebner basis of $\spn{g_d}+I(\C)$ and you can obtain $h$ using Groebner basis and/or syzygies. Therefore we choose $h$ and get $\sqrt{\spn{f,g,h}} = I(\C)$.

$\newcommand\C{\mathcal C} \newcommand\spn[1]{\langle #1\rangle}$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 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 $$h'^N - h \in \spn{f,g_d}.$$ You can obtain $N$ by just iterating through the powers of $h'$ and reducing via Groebner basis of $\spn{g_d}+I(\C)$ and you can obtain $h$ using Groebner basis and/or syzygies. Therefore we choose $h$ and get $\sqrt{\spn{f,g,h}} = I(\C)$.

$\newcommand\C{\mathcal C} \newcommand\spn[1]{\langle #1\rangle}$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 $$h'^N - h \in \spn{f,g_d}.$$ You can obtain $N$ by just iterating through the powers of $h'$ and reducing via Groebner basis of $\spn{g_d}+I(\C)$ and you can obtain $h$ using Groebner basis and/or syzygies. Therefore we choose $h$ and get $\sqrt{\spn{f,g,h}} = I(\C)$.