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This question has been asked and discussed here: Show the set of points $(t^3, t^4, t^5)$ is closed in $\mathbb A^{3}$ and here: Neat way to find the kernel of a ring homomorphism.

However, I came across a solution online posted here: http://hartshornesolutions.blogspot.com/2015/06/chapter-1-exercise-111-variety-that-is.html, it seems really neat and I followed it, but I could not understand the final several lines.

So, our curve is $$Y:=\{(t^{3},t^{4},t^{5}):t\in k\},\ \text{where}\ k\ \text{is an algebraically closed field.}$$ It is fairly easy to guess and to prove that $$Y=Z(zx-y^{2},yz-x^{3}, yx^{2}-z^{2}).$$ Denote $\mathfrak{a}:=\langle zx-y^{2}, yz-x^{3}, yx^{2}-z^{2}\rangle$, and we want to show that $I(Y)=\mathfrak{a}$. It is clear that $I(Y)\supseteq \mathfrak{a}$, and we need to show the inverse inclusion.


The solution I referred above basically goes the following way:

Let $g\in k[x,y,z]$ be such that $g(t^{3},t^{4},t^{5})=0$. We want to prove that there exists $h_{1},h_{2},h_{3}\in k[x,y,z]$ such that $$g(x,y,z)=(zx-y^{2})h_{1}+(yz-x^{3})h_{2}+(yx^{2}-z^{2})h_{3}.$$

Suppose that $g$ has degree $n$, then we can write $g$ as $$g(x,y,z)=\sum_{i=0}^{n}\sum_{j=0}^{n}\sum_{k=0}^{n}a_{i,j,k}x^{i}y^{j}z^{k},$$ so that $$g(t^{3},t^{4},t^{5})=\sum_{i=0}^{n}\sum_{j=0}^{n}\sum_{k=0}^{n}a_{i,j,k}t^{3i+4j+5k}.$$

Now, note that the set of all possible values of $3i+4j+5k$ is a subset of $\{0,1,\dots, 12n\}$. For each $s\in \{0,\dots, 12n\}$, if $3i+4j+5k=s$ has no solution, then we define $b_{s}:=0$. If $3i+4j+5k=s$ has solutions, then we define $$b_{s}:=\sum_{3i+4j+5k=s}a_{i,j,k}.$$ Using the $b_{s}$, we can then write $g(t^{3},t^{4},t^{5})$ into $$g(t^{3},t^{4},t^{5})=\sum_{s=1}^{12n}b_{s}t^{s}.$$

As $g(t^{3},t^{4},t^{5})=0$, it follows that $b_{s}=0$ for all $s\in\{0,\dots, 12n\}.$ This means that for any $s\in \{0,\dots, 12n\}$ such that $3i+4j+5k=s$ has solutions, the following sum is zero: $$\sum_{3i+4j+5k=s}a_{i,j,k}=0\ \ \ \ \ (*).$$

Now we compute the solution of $3i+4j+5k=s$ for $s\in\{0,1,2,\dots, 10\}$, and the apply $(*)$ to get the information of $a_{i,j,k}$ and hence of $x^{i}y^{j}z^{k}$.

For $s=0$, the only solution is $i=j=k=0$, so $a_{0,0,0}=0$. Hence, $g$ does not have constant term.

For $s=1,2$, we do not have a solution.

For $s=3$, the only solution is $(1,0,0)$, and thus $a_{1,0,0}=0$. Hence, $g$ does not have term proportional to $x$.

For $s=4$, the only solution is $(0,1,0)$, and thus $a_{0,1,0}=0$. Hence, $g$ does not have term proportional to $y$.

For $s=5$, the only solution is $(0,0,1)$, and thus $a_{0,0,1}=0$. Hence, $g$ does not have term proportional to $z$.

For $s=6$, the only solution is $(2,0,0)$, and thus $a_{2,0,0}=0$. Hence, $g$ does not have term proportional to $x^{2}$.

For $s=7$, the only solution is $(1,1,0)$, and thus $a_{1,1,0}=0$. Hence, $g$ does not have term proportional to $xy$.

For $s=8$, we have two solutions, $(1,1,0)$ and $(0,2,0)$, and thus $a_{1,1,0}+a_{0,2,0}=0$, which means that $xy$ and $z^{2}$ have opposite coefficients.

For $s=9$, we have two solutions, $(0,1,1)$ and $(3,0,0)$, and thus $a_{0,1,1}+a_{3,0,0}=0$, which means that $yz$ and $x^{3}$ have opposite coefficients.

Finally, for $s=10$, we have two solutions, $(2,1,0)$ and $(0,0,2)$, and thus $a_{2,1,0}+a_{0,0,2}=0$. This means that $x^{2}y$ and $z^{2}$ have opposite coefficients.

Hence, we rule out the following terms in $g(x,y,z)$: constant term, $x,y,z$, $x^{2}$ and $xy$. And the following pairs must have opposite coefficients (with the same absolute value): $$xy\ \text{and}\ z^{2},\ yz\ \text{and}\ x^{3},\ x^{2}y\ \text{and}\ z^{2}.$$


Then, the solution concluded that $zx-y^{2}, yz-x^{3}, yx^{2}-z^{2}$ lie in $I(Y)$ and thus generate it.

Why does this conclusion follow from above computations? What I just showed is only that $$g(x,y,z)=ay^{2}+b(xy-z^{2})+c(yz-x^{3})+d(x^{2}y-z^{2})+h(x,y,z),$$ where $h(x,y,z)$ does not contain constant, $x,y,z$, $x^{2}$, $xy$ and any summands before it. Right?

This whole process seems like some elimination theory. I am not familiar with elimination ideal and related stuff, but if there exists a theorem that can directly conclude like what the solution suggested, I am happy to accept it.

Thank you!

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I will first explain why $zx-y^2,yz-x^3,yx^2-z^2$ generate $I(Y)$, then explain why $I(Y)$ cannot be gererated by two elements.

First let $\theta[x,y,z]\to k[t]$ be identity on $k$ and $x\mapsto t^3$, $y\mapsto t^4$, $z\mapsto t^5$. You can verify $I(Y)=\ker\theta$.

We already know $(zx-y^2,yz-x^3,yx^2-z^2)\subseteq I(Y)$. Take $f\in I(Y)$, we want to show $f\in(zx-y^2,yz-x^3,yx^2-z^2)$. We can "interchange" $zx$ with $y^2$ (similarly, $yz$ with $x^3$, $yx^2$ with $z^2$) for each term of $f$. For example, if $x^3y^2z$ is a term in $f$, then since $x^3y^2z=(x^3-yz)y^2z+(yz)y^2z$, we get $y^3z^2$ as a result of the interchange (note the degree has decreased by 1), and our goal is to do more interchanges so that this term, combined with other terms in $f$, becomes $0$.

Now fix a monomial $g$ in $f$, and we apply the "interchanges" in this rule: whenever possible, replace $y^2$ by $xz$, replace $x^3$ by $yz$, replace $yx^2$ by $z^2$. This algorithm will finally stop because we are decreasing the degree of $g$ in all but $1$ type of interchange. Finally g becomes (up to multiplication by a constant) one of the possible five forms: $z^nx^2$, $z^nxy$, $z^nx$, $z^ny$, $z^n$. Applying this algorithm to each term of $f$, we get $f=g+h$ where $g\in (zx-y^2,yz-x^3,yx^2-z^2)$, $h\in I(Y)$ has monomials of these 5 forms. Now something nice happens: if we compute degrees of these monomials after substitution $x\mapsto t^3, y\mapsto t^4, z\mapsto t^5$, we get $5n+6$, $5n+7$, $5n+3$, $5n+4$, $5n$, which are all different modulo $5$. Using this observation and the fact that $\theta(h)=0$, we get $h=0$. So $I(Y)=(zx-y^2,yz-x^3,yx^2-z^2)$.

Next, FSOC, suppose $I(Y)$ is generated by two elements $f$ and $g$. Then any element $h\in I(Y)$ is equal to $af+bg$ for some $a,b\in A$. Note that multiplication of any nonconstant term of $a$ with any term of $f$ increases the "t-degree" of that term of $f$ by at least $3$, and the lowest nontrivial "t-degree" term of $h$ has "t-degree" $8$, as shown by your calculation, where "t-degree" means degree in $t$ after substitutions $x\mapsto t^3, y\mapsto t^4, z\mapsto t^5$. This means that if we define a $k$-linear map $\phi:I(Y)\to k^3$ by $\phi(f)$=(coefficient of $zx-y^2$,coefficient of $yz-x^3$,coefficient of $x^2y-z^2$),then $dim_k(im(\phi))=2$. But $zx-y^2,yz-x^3,yx^2-z^2$ are in $I(Y)$ and span of their image is the whole space $k^3$, so we see contradiction.

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