Ideal generated by $2 \times 2$ minors is prime

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Let $A$ be a domain and let $I$ be the ideal in $B=A[x_1,\dots,x_n,y_1,\dots,y_n]$ generated by all the $2 \times 2$ minors of the matrix $$\begin{pmatrix} x_1 & x_2 & \dots & x_n\\ y_1 & y_2 & \dots & y_n \end{pmatrix},$$ i.e. $I = (x_iy_j-x_jy_i : 1\le i<j\le n)$. I tried to show that $I$ is prime by showing that $B/I$ is a domain. Unfortunately, I have no idea how to prove this. Maybe someone can give me a hint?

This is problem 7.7 in the textbook Binomial Ideals by Herzog, Hibi, Ohsugi.

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It’s much easier to answer this question if you know some geometry (apologies if you want an answer using strictly algebra, but I think it’s more instructive).

View the ideal $ I $ as a homogeneous ideal of $ R[ y_1, \ldots, y_n] $ where $ R $ is $ \mathbb{C} [x_1, \ldots, x_n] $. Then the equations define the blow-up of $ \mathbb{A}^n $ at the origin, as a closed subscheme of $ \mathbb{P}^{n-1} \times \mathbb{A}^n $. This blow-up is also the total space of the tautological line bundle of $ \mathbb{P}^{n-1} $ and hence, irreducible and reduced, that is, a variety itself. So the ideal $ I $ is a homogenous prime ideal, hence is a prime ideal if you forget the grading.