I am stuck at the following exercise from Gathmann's notes on Algebraic Geometry on page 21:
Let $R = K[x_1,x_2,x_3,x_4] / \langle x_1x_4 - x_2x_3 \rangle$. Show that $R$ is an integral domain of dimension $3$.
While I know the definition of dimension (from page 17 in the above notes):
The dimension $X \in \mathbb{N} \cup \{\infty\}$ is the supremum over all $n \in \mathbb{N}$ such that there is a chain $$\emptyset \subset Y_0 \subsetneq Y_1 \subsetneq \cdots \subsetneq Y_n \subset X$$ of length $n$ of irreducible closed subsets $Y_1,\ldots,Y_n$ of $X$.
I do not understand how to apply this definition on $R$. Could you please give me a hint?
I'm assuming $K$ is a field. Thus $K[x_1,x_2,x_3,x_4]$ is a UFD and it remains to show that $\langle x_1x_4-x_2x_3 \rangle$ is prime, which just comes down to showing $x_1x_4 - x_2x_3$ is irreducible in $K[x_1,x_2,x_3,x_4]$, which I'm sure you know how to do. It follows from there that $K[x_1,x_2,x_3,x_4]/\langle x_1x_4-x_2x_3 \rangle$ is an integral domain.
To show the dimension, recall that by Krull's Hauptidealsatz, we have that $\langle x_1x_4 - x_2x_3 \rangle$ is of height $1$, we have that
$\begin{align*} \text{dim}(K[x_1,x_2,x_3,x_4)/\langle x_1x_4 - x_2x_3 \rangle) = \text{dim}(K[x_1,x_2,x_3,x_4)) - \text{height}(\langle x_1x_4 - x_2x_3 \rangle) = 4 -1 = 3 \end{align*}$
as desired.