Hartshorne Exercise 1.11* of a curve that is not a local complete intersection.

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!

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.