Simple proof that $(p_1x_1-q_1y_1,...,p_nx_n-q_ny_n)$ is a prime ideal

Question

Is there a nice way to show that $$(p_1x_1-q_1y_1,\ldots ,p_nx_n-q_ny_n) \subseteq \mathbb{Z}[x_1,...,x_n,y_1,...,y_n]$$ is a prime ideal for coprime non-zero integers $p_i,q_i\,(i=1,...,n)$ ?

I have a proof, but I think there must be a simpler reasoning.

My proof goes: Denote the polynomial ring by $R$ and the ideal by $I$. Suppose we already know that $R/I$ is flat as $\mathbb{Z}$-module. Tensoring the inclusion $\mathbb{Z} \hookrightarrow \mathbb{Q}$ over $\mathbb{Z}$ with $R/I$ gives an embedding of rings $$R/I \hookrightarrow R/I\otimes_{\mathbb{Z}}\mathbb{Q}=\frac{\mathbb{Q}[x_i,y_i|i=1,...,n]}{(p_ix_i-q_iy_i|i=1,...,n)}\cong \mathbb{Q}[x_1,...,x_n]$$ Hence, as a subring of a domain, $R/I$ is itself a domain. In order to see that $R/I$ is flat, factor it as $$R/I= \frac{\mathbb{Z}[x_1,y_1]}{(p_1x_1-q_1y_1)}\otimes_\mathbb{Z}\cdots \otimes_\mathbb{Z}\frac{\mathbb{Z}[x_n,y_n]}{(p_nx_n-q_ny_n)}.$$ Since the tensor product of flat modules is again a flat module, it suffices to show that the factors are flat which is (over a pid) equivalent to being torsion-free:

For $p, q$ coprime, $px-qy$ is irreducible and thus prime in $\mathbb{Z}[x,y]$. So, if $k \in \mathbb{Z}, f \in \mathbb{Z}[x,y]$ satisfy a relation $k\cdot f \equiv 0\,(px-qy)$, px-qy divides $f$ for degree reasons. This finally shows that $\mathbb{Z}[x,y]/(px-qy)$ is torsion-free.

Answer 1

[Sorry for answering my own question]

By choosing integers $a_i,b_i$ with $a_ip_i+b_iq_i = 1$ one can define an automorphism on the polynomial ring by mapping $x_i \mapsto x'_i = p_ix_i-q_iy_i,\, y_i \mapsto y'_i = b_ix_i + a_iy_i$. This transforms the ideal in question into the prime ideal $(x'_1,...,x'_n) \subseteq \mathbb{Z}[x'_1,y'_1, ...,x'_n,y'_n]$.