How to show that $(y-x^2, z-x^3) \in \mathbb{C}[x,y,z]$ is irreducible or radical?

Question

In Fulton's introduction to Algebraic Geometry, there is the following exercise on page 11:

I have been struggling for a bit longer than I care to admit on this problem, and have not been able to get a handle on it. I have managed to prove that such an algebraic set is equal to $V = V(y-x^2, z-x^3)$, but I haven't managed to prove that this is the ideal of the variety. I have looked elsewhere for solutions, but have not been able to find any. However, I did come across the solution to the part a of this particular problem, and figured it would be a good idea to link it here for anyone who might stumble on this post in the future.

Irreducible components of the variety $V(X^2+Y^2-1,X^2-Z^2-1)\subset \mathbb{C}^3.$

In this particular chapter, we have covered Hilbert's Nullstellensatz,so there is an obvious route open to me that the author seems to intend: prove the ideal $\mathcal{I} = (y-x^2, z-x^3)$ is prime, in which case it is radical and therefore the ideal of the algebraic set; being prime would in turn imply that $V$ is reducible. This is an introductory course for undergraduates, so any solutions that don't use super high powered results are preferred, since this would be most pedagogically appropriate. Right now, the theorems I have at my disposal are pretty much anything you would learn in an introductory course to ring theory, and basic facts about algebraic sets, irreducibility conditions, field extensions, and a brief introduction to coordinate rings.

I would appreciate any hints to help coax me in the right direction, thanks in advance!

Answer 1

Here is a solution in a series of hints/steps, the details of which I leave to you. Some of these are a little tedious to demonstrate with absolute rigor, but the good news is that they do not use anything more than basic ring theory, as you preferred.

1. Let $R$ be a domain. Show that for any $a \in R$, the kernel of the unique $R$-algebra morphism $R[X] \to R$ sending $X$ to $a$ is $\langle X - a \rangle$. (Suggestion: use Euclidean division by $X-a$, which is possible because the leading coefficient is a unit - it's $1$!)

2. Using 1, prove that for any $a_{1}, \ldots, a_{n} \in R$, the evaluation morphism $R[X_{1}, \ldots, X_{n}] \to R$ sending $X_{i}$ to $a_{i}$ is $\langle X_{1}-a_{1}, \ldots, X_{n} - a_{n} \rangle$. Perhaps you might use induction, and the fact that $R[X_{1}, \ldots, X_{n}]$ is canonically isomorphic to $(R[X_{1}, \ldots, X_{n-1}])[X_{n}]$.

3. Conclude that the ideal $\langle X_{1} - a_{1}, \ldots, X_{n}-a_{n} \rangle \subset R[X_{1}, \ldots, X_{n}]$ is prime, as the evaluation morphism in $2$ is surjective and the image is a domain by hypothesis.

4. Apply fact 3 to your situation by choosing appropriate $R$ and $a_{i}$s. (Hint: you will have $n = 2$.)

Answer 2

From Cox, Little and O'Shea:

Proposition 3. Let $V ⊂ k^n$ be an affine variety. Then $V$ is irreducible if and only if $I(V)$ is a prime ideal.

As an example of how to use Proposition 3, let us prove that the ideal $I(V)$ of the twisted cubic is prime. Suppose that $f g ∈ I(V).$ Since the curve is parametrized by $(t, t^2, t^3),$ it follows that, for all $t ,$ $f (t, t^2, t^3)g(t, t^2, t^3) = 0.$ This implies that $f (t, t^2, t^3)$ or $g(t, t^2, t^3)$ must be the zero polynomial, so that $f$ or $g$ vanishes on $V.$ Hence, $f$ or $g$ lies in $I(V),$ proving that $I(V)$ is a prime ideal. By the proposition, the twisted cubic is an irreducible variety in ${\Bbb R}^3$. One proves that a straight line is irreducible in the same way: first parametrize it, then apply the above argument. In fact, the above argument holds much more generally.

Proposition 5. If k is an infinite field and $V ⊂ k^n$ is a variety defined parametrically $$\begin{align} x_1 &= f_1(t_1, \ldots , t_m),\\ &\vdots\\ x_n &= f_n(t_1, \ldots , t_m),\end{align}$$ where $f_1, \ldots , f_n$ are polynomials in $k[t_1, \ldots , t_m],$ then $V$ is irreducible

[...]

With a little care, the above argument extends still further to show that any variety defined by a rational parametrization is irreducible.

Answer 3

One can also use the "algebra-geometry" correspondence. $(y-x^2,z-x^3)$ is a prime ideal in $\mathbb C[x,y,z]$ if and only if the set $$ V = \{ (x,y,z) \in \mathbb A_{\mathbb C^3} : y=x^2, z = x^3\} $$

is an irreducible set in $\mathbb A_{\mathbb C}^3$. The answer of Jan-Magnis alludes that $V$ can be parameterizes as: $$ V = \{ (t,t^2,t^3) : t \in \mathbb C \} $$

Therefore, it is possible to find a bijection:

$$ \varphi : \mathbb A_{\mathbb C}^1 \to V \quad \quad t \mapsto (t,t^2,t^3) $$

$\varphi$ turns out to an isomorphism of affine algebraic sets. Since $\mathbb A_{\mathbb C}^1$ is irreducible, $V$ must be irreducible.

This argument uses some heavy machinery, but I like it because one can circumvent trying to algorithmically determine whether $(y-x^2,z-x^3)$ is a prime ideal or not.