Looking for alternative argument $(x^2 - y^3, y^2 - z^3)\subset k[x,y,z]$ is prime ideal.

Question

Consider $I=(x^2 - y^3, y^2 - z^3)\subset k[x,y,z]$ as an ideal with $k$ a field.

$\textbf{Q:}$ I am Looking for alternative argument to conclude $I$ is prime ideal. It is clear that I can use parametrization $k[x,y,z]\to k[t]$ by parametrization $(x,y,z)\to (t^9, t^6, t^4)$ and argue this descends to a monomorphism under quotient $k[x,y,z]/I$. Note this map is clearly not surjection. Hence this is not isomorphic to $A^1_k$ and this is already indicated by singularity at $(0,0,0)$. Can I conclude primeness of ideal $I$ by some other better argument? Say intersection,grobner basis,...

Answer 1

Lemma: Let $f:X\to Y$ be a continuous map of topological spaces. If $X$ is irreducible, then $f(X)$ is irreducible.

Proof: If $f(X)$ had a decomposition in to two nontrivial proper closed subsets $A\cup B$, then $f^{-1}(A)\cup f^{-1}(B)$ would be a decomposition of $X$ in to two nontrivial proper closed subsets, contradicting irreducibility of $X$.

We apply this to the situation at hand in the following fashion: $V(I)$ is the image of the morphism $\Bbb A^1\to\Bbb A^3$ given by $t\mapsto (t^9,t^6,t^4)$, so $V(I)$ is irreducible. Thus $I$ must be prime.

Answer 2

KReiser has given a really nice answer, but I just wanted to add that you're already essentially there with what you've written in your question. You've observed that the map $(x,y,z)\mapsto (t^9,t^6,t^4)$ induces an injection $k[x,y,z]/I \hookrightarrow k[t]$. The latter ring is a domain, and any subring of a domain is a domain, thus $I$ must be prime.