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.
 A: One low-tech way to go from here is to write any polynomial $f$ in $k[x,y,z,w]$ as a sum of $j\in J$ and some other particularly nice $u$ and then apply the homomorphism $\varphi$ you've written down - if $f\in \ker\varphi$, then $\varphi(j)+\varphi(u)=0$, and you'll see that the only way for $\varphi(u)=0$ is if $u=0$ to begin, so $J$ is exactly the kernel of $\varphi$.
There's probably a more succinct way to write this down, but this is the way I would think through it:

*

*$xw-yz$ is in $J$, so we can assume that we've chosen $j$ and $u$ so that $u$ contains no monomials with a $yz$ in them: if $p$ is a monomial and $pyz$ is in $u$, then $pyz=-p(xw-yz)+pxw$, so we can adjust $j$ and $u$ so there are no terms with a $yz$ by repeated applications of this formula.

*$y^3-xz$ and $z^3-yw^2$ are in $J$, so via the same trick, we can assume there are no monomials in $u$ with a $y^3$ or a $z^3$ in them.

*$y^2w-xz^2$ is in $J$, so there's no monomials in $u$ with $y^2w$ in them.

So we may take $u$ to consist of a $k$-linear combination of monomials of the following type:

*

*$x^aw^b$,

*$x^aw^by$,

*$x^aw^bz$, and

*$x^aw^bz^2$
where $a,b\geq 0$.
Now let's look at what happens when we apply $\varphi$:

*

*$\varphi(x^aw^b) = s^{4a}t^{4b}$,

*$\varphi(x^aw^by) = s^{4a+3}t^{4b+1}$,

*$\varphi(x^aw^bz) = s^{4a+1}t^{4b+3}$, and

*$\varphi(x^aw^bz^2) = s^{4a+2}t^{4b+6}$.

By examining the residue classes of the exponent of $s$ modulo $4$, we see that there can be no cancellation between any of these monomials, so the only way for $\varphi(f)=\varphi(j+u)=\varphi(j)+\varphi(u)=0+\varphi(u)$ to be equal to zero is if $u=0$.
