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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.2012-05-08
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    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...2012-05-08
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    @Michael Kasa: yeah I tried that but don't see why $J$ is radical. Any idea?2012-05-08
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    Isn't the Veronese embedding of degree $4$ given by $\mathbb{P}^1 \to \mathbb{P}^4, [s:t] \mapsto [s^4 : s^3 t : s^2 t^2 : s t^3 : t^4]$?2012-05-15

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