Show that $\langle X_1X_4-X_2X_3 \rangle$ is irreducible in $\mathbb{Q}[X_1,X_2,X_3,X_4]$.

Question

In lecture we did the following example.

Show that $X_1X_4-X_2X_3$ is irreducible in $\mathbb{Q}[X_1,X_2,X_3,X_4]$.

We wrote down that if $X_1X_4-X_2X_3 = a\cdot b$ for some $a,b \in \mathbb{Q}[X_1,X_2,X_3,X_4] \setminus \mathbb{Q}$ then w.l.o.g. we may assume that $a$ has degree $1$ in $X_1$, i.e. $\deg_{X_1}(a) = 1$. Then it follows that $b \mid X_4$ and $b \mid X_2X_3$. However, this is a contradiction since the obviously the only common divisors of $X_4$ and $X_2X_3$ are units.

However, I do not understand why $\deg_{X_1}(a) = 1$ implies $b \mid X_4$ and $b \mid X_2X_3$. Could you please explain this to me?

Answer 1

This is an instance of the following theorem:

In a polynomial ring $R[x]$, if $a,b \in R[x]$ and $\deg(b) = 0$ (so $b\in R$), then $b$ divides (in $R$) each of the coefficients of $ab$.

If you replace $x$ with $X_1$ and $R$ with $\mathbb{Q}[X_2, X_3, X_4]$, then you get exactly your statement. In particular, the only fact that needs shown at all is that $\deg_{X_1}(b) = 0$.

In that case, we have $a = cX_1 + d$ where $b,c,d \in \mathbb{Q}[X_2,X_3,X_4]$ and $$bcX_1 + bd = ab = X_1X_4 - X_2X_3$$