Krull dimension of $\mathbb{R}[x_1, \ldots, x_n]/(F)$?

Question

Let $F \in \mathbb{R}[x_1, \ldots, x_n]$ be irreducible over $\mathbb{R}$. I am trying to understand why the Krull dimension of the ring $\mathbb{R}[x_1, \ldots, x_n]/(F)$ is $n-1$. Any comments are appreciated! Thank you!

PS In fact is it always the case that if $P$ is a prime ideal of height $k$ then the Krull dimension of the ring $\mathbb{R}[x_1, \ldots, x_n]/P$ is $n-k$?

Answer 1

Yes, it is true that for a prime ideal $P\subset\mathbb{R}[x_1,...,x_n]$ with height $k$ we have $$\dim_{\text{Krull}}(\mathbb{R}[x_1,...,x_n]/P)=n-k$$ This is not a trivial fact, specially when you consider more general commutative rings. There's a big discussion about it in the link indicated by user73985.

For the specific example of $P=(F)$ with $F$ irreducible, you will also need the following theorem:

Krull's Hauptidealsatz (principal ideal theorem): Let $A$ be a commutative noetherian ring and let $f\in A$ be neither a unit nor a zero divisor. Then every minimal prime ideal containing $f$ has height $1$.

Since $\mathbb{R}[x_1,...,x_n]$ is noetherian and $F$ is clearly not a unit and not a zero divisor, the theorem applies.

The ideal $(F)$ is prime (which follows from the fact that $\mathbb{R}[x_1,...,x_n]$ is a unique factorization domain), so obviously $(F)$ is a minimal prime containing $F$. Therefore $ht((F))=1$.

Using the formula above, $\dim_{\text{Krul}}(\mathbb{R}[x_1,...,x_n]/(F))=n-1$.