# Showing an ideal is prime in polynomial ring

Let $k=\mathbb{C}$ and let $J$ the ideal $(xw-yz,y^{3}-x^{2}z,z^{3}-yw^{2},y^{2}w-xz^{2})$. I want to see why $J$ is a prime ideal in $k[x,y,z,w]$. I know that $Z(J)$ (the zero set of J) is exactly the image of the $4$-fold Veronese embedding from $\mathbb{P}^{1}$ to $\mathbb{P}^{3}$, i.e the map given by $[s : t] \mapsto [s^{4}: s^{3}t : st^{3}: t^{4}]$. What I tried: if we consider the following ring homomomorphism:

$k[x,y,z,w] \rightarrow k[s,t]$ given by $x \mapsto s^{4}$, $y \mapsto s^{3}t$, $z \mapsto st^{3}$ and $w \mapsto t^{4}$, then one can check that $J$ is contained in the kernel of this ring homomorphism. Now if we can show the the other inclusion we are done because then $k[x,y,z,w]/J$ embeds a subring of $k[s,t]$ and hence $J$ is prime. However I don't see the other inclusion, can you please help? perhaps there's an easier way.

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Your ideal is generated by particularly nice elements; they impose relations like $xw = yz$ which can be interpreted as relations in the commutative monoid of monomials (to get back to the ring you just take the monoid ring). Figure out what the quotient monoid is and you're done. –  Qiaochu Yuan May 8 '12 at 4:11
Since you know $Z(J)$ is the image of the Varonese embedding, $J$ is an irreducible ideal, as $\mathbb{P}^1$ is irreducible. The hard part now is to justify why $J$ is radical... –  Michael Kasa May 8 '12 at 5:50
@Michael Kasa: yeah I tried that but don't see why $J$ is radical. Any idea? –  user6495 May 8 '12 at 6:36
Isn't the Veronese embedding of degree $4$ given by $\mathbb{P}^1 \to \mathbb{P}^4, [s:t] \mapsto [s^4 : s^3 t : s^2 t^2 : s t^3 : t^4]$? –  Martin Brandenburg May 15 '12 at 18:47