Is $(x_1x_2+x_3x_4, 2x_1x_2+x_5x_6)$ a prime ideal? I'm studying for my Qualifying exam and I found the following question in an old question bank.

Let $\mathbb{K}$ be an algebraically closed field (char $\mathbb{K}\not=2$). Is $\mathbb{K}[x_1,x_2,x_3,x_4,x_5,x_6]/(x_1x_2+x_3x_4, 2x_1x_2+x_5x_6)$ an integral domain?

I have proven that each of the generators of the ideal is irreducible, and hence prime and I thought ideal generated by prime elements will be a prime ideal, but it turns out that's not true. TBH, I don't really know any tricks to prove an ideal is a prime ideal, other than the definition and the integral domain condition.
 A: Using the change of variables proposed by Minglingmaster, for example, you are reduced to prove that $R=K[y_1,y_2,y_4,y_6][y_3,y_5]/(y_3^2-p,y_5^2-q)$ is an integral domain, where $p,q\in K[y_1,y_2,y_4,y_6]$ are such that $p,q,p/q$ are not squares in $L=K(y_1,y_2,y_4,y_6)$.
It is then well-known (and a good exercise) to show that $1,\sqrt{p},\sqrt{q},\sqrt{pq}$ are $L$-linearly independent.
That being said, consider the ring morphism $f:K[y_1,y_2,y_4,y_6][y_3,y_5]\to L(\sqrt{p},\sqrt{q})$ , sending $P$ to $P(y_1,y_2,\sqrt{p},y_4,\sqrt{q},y_6)$.
I claim that the kernel is $I=(y_3^2-p,y_5^2-q)$. Clearly, $I$ is contained in $\ker(f)$. Now let $P\in \ker(f)$. Dividing by $y_3^2-p$, then by $y_5^2-q$, one may write $P=Q_0+Q_1y_3+Q_2y_5+Q_3y_3y_5+S$, where $Q_0,Q_1,Q_2,Q_3\in K[y_1,y_2,y_4,y_6]$ and $S\in I$.
Applying $f$, we get $Q_0+Q_1\sqrt{p}+Q_2\sqrt{q}+Q_3\sqrt{pq}=0\in L(\sqrt{p},\sqrt{q})$. By the linear independence mentioned above, $Q_0=Q_1=Q_2=Q_3=0$, meaning that $P=S\in I$.
This, $\ker(f)=I$, and $f$ induces an injective ring morphism $R\to L(\sqrt{p},\sqrt{q})$. Hence $R$ is isomorphic to a subring of a field. In particular, $R$ is an integral domain.
