Showing an ideal is prime in polynomial ring

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 $u=0$.