To prove that an ideal cannot be generated by two elements

Question

Let $k$ be an algebraically closed field and let $\ Y\subset \mathbb{A}^n(k)$ be the curve given parametrically by $x=t^3, y=t^4,z=t^5$

I want to show

(i) $I(Y)$ is a prime ideal of height 2

(ii) $I(Y)$ cannot be generated by 2 elements

Since $I(Y)$ is kernel of the homomorphism sending $x\mapsto t^3,y\mapsto t^4,z\mapsto t^5$ into the integral domain $k[t]$, $I(Y)$ must be prime. And since we can consider $k\subset k[x,y,z]/I(Y)\subset k[t]$, by considering transcendence degree we have $\dim(k[x,y]/I(Y))=1$ and hence the height of $I(Y)$ equals 2.

Now I want to prove the second statement. I guess $I(Y)=\left<x^4-y^3,x^5-z^3,y^5-z^4\right>$, but I don't know how to prove it (although one inclusion is obvious), and I don't know how to use this "result" to prove (ii).

Any help or hints are appreciated, thank you!

Have you checked your `guess' that $I(Y)$ is generated by the three elements you write down? Hint: They do not. Once you find the right equations, it is only a matter of applying Nakayama's lemma appropriately for (ii). — Mohan, May 11 '16 at 15:19
Now, your equations are correct. Now apply Nakayama's lemma at the origin to get (ii). — Mohan, May 11 '16 at 16:22
@Mohan, I really don't know how to use Nakayama's lemma to conclude (ii). Could you elaborate on that? — Garcez, May 11 '16 at 18:24
Show that modulo $(x,y,z)I(Y)$, these three elements are $k$-linearly independent. — Mohan, May 11 '16 at 20:10
@Mohan Sorry, I still don't know how to use Nakayama lemma, can you give more hints? — chan kifung, May 11 '16 at 23:56
@chankifung It is not necessary to use Nakayama's lemma; see this answer. — user26857, May 12 '16 at 05:34
@Mohan Could you write an answer about this? It seems not to be trivial. — Aolong Li, Nov 07 '17 at 08:04

score 1 · Accepted Answer · edited Nov 25 '16 at 16:33

As you wrote on the comment section, $I(Y) = \left<x^2y - z^2, y^2 - xz, x^3 -yz\right>$.

You can prove it using the following result: Given a monomial order in $k[x,y,z]$, and $g,f_1,f_2,...,f_k \in k[x,y,z]$, there are $q_1,...,q_k,r \in k[x,y,z]$ such that $g = q_1f_1 + \cdots + q_kf_k + r$ and the leading monomials of the $f_i$ do not divide any of the monomials of $r$.
PS: this writing is not unique!

Using the equality above, assume that $I(Y) = \left<p,q\right>$, where $ p,q \in k[x,y,z]$.

Note that the only polynomial of degree 2 in $I(Y)$ is $y^2-xz$ and the only ones with degree 3 are $x^2y-z^2, x^3-yz$ and $\alpha (y^2-xz)$ where $\alpha$ has degree 1. Also, there are no polynomials of degree 1 in $I(Y)$. Thus we can assume that $q = y^2-xz$ and that $p$ has degree 3. We can also assume that $p$ has no monomials of degree 2 other than $z^2$ and $yz$ (why?).

Therefore, we have shown that the $k$-subspace generated by $x^2y-z^2, x^3-yz$ and $y^2-xz$ is also generated by $p$ and $q$ which is not possible since the former are linear independent over $k$. Thus $I(Y)$ cannot be generated by two elements.

Why can you conclude $x^2y-z^2$ and$x^3-yz$ can be generated linearly by $p$ and $q$? — chan kifung, May 11 '16 at 23:07

To prove that an ideal cannot be generated by two elements

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