Height of irreducible polynomials

Question

I started to read about dimension of varieties. The encountered the algebraic version of it, Krull Dimension. To get intuition I am trying to calculate heights of primes. I found that Atiyah Macdonald and Wikipedia mostly only have theoretical results.

So I took $f \in \mathbb{C}$ an irreducible polynomial. I am trying to find its height.

I am looking for a prime ideal $\mathcal{P}$ such that $ 0 \subset \mathcal{P} \subset (f)$.

By the definition of $\mathcal{P}$, $$\mathcal{P} = S\times f.$$

Now, one can check that $S$ is an ideal. Also because we are assuming that $\mathcal{P}$ is strictly in $(f)$ so $f $ is not in $\mathcal{P}$ and since $\mathcal{P}$ is prime we must have $S \subset \mathcal{P}$.

But then $S = \mathcal{P} = \mathcal{P} \times f$ But this is not possible. So there is no such $\mathcal{P}$ and so height of $f$ is 1.

My Question is, Is my work correct? I am bit skeptic because I have never seen this result and also this is the first time I am finding heights.

I had a second question. It is related but different. If the Krull Dimension of a ring is finite then is the ring Noetherian? This is the first question that came to my mind when I read the definition of Krull dimension in terms of increasing chains of primes. It can also be phrased like: If all chains of primes terminate and infact are bounded above uniformly then does that imply that all chains of ANY Ideals will also terminate? (They will not have a uniform bound obviously, we can look at integers). By bound here I mean bound on the length of the chains.

This seems like it can be true, as primes do seem to control the behaviour of ideals in general. For instance we have the equvialent definition of Noetherian rings in terms of primes being finitely generated being enough to imply all ideals are finitely generated!

This question seems to ambituous for me to currently get started on so any help is much appreciated.

Answer 1

For your first question, your proof is absolutely right, though it's worth maybe elaborating on why the equality $\mathcal{P}=\mathcal{P}(f)$ is a contradiction. (Of course, this is because $\mathcal{P}$ is non-zero; as such, we can choose some $g\in\mathcal{P}\setminus\{0\}$ of minimal possible degree. But then $g\in\mathcal{P}(f)$, so there exists $h\in\mathcal{P}\setminus\{0\}$ with $g=hf$. Since $f$ is irreducible, $\deg f>0$, so $\deg h<\deg g$, thus contradicting minimality of the degree of $g$. However, note that this argument won't work for more general rings; for example, the prime ideal $\mathbb{C}\times\{0\}$ of $\mathbb{C}\times\mathbb{C}$ is actually idempotent, ie equal to its own square.) Additionally, a very minor point: we normally don't refer to the height of an element, but rather to the height of an ideal. Besides this, your proof looks right to me!

As an aside, you can actually vastly generalize the statement you have just proved: if $R$ is any (commutative) Noetherian ring, $f\in R$ is an element, and $\mathcal{P}\vartriangleleft R$ is any minimal prime ideal over $(f)$, then the height of $\mathcal{P}$ is at most one. This is the so-called principal ideal theorem, and it is a fundamental result in dimension theory. Note that the theorem does not require that $\mathcal{P}$ itself be principal! Only that $\mathcal{P}$ be a minimal prime over some principal ideal.

For your second question, the answer is no. For instance, let $F$ be your favorite field, and let $B=F[x_n:n\in\mathbb{N}]$ be the polynomial ring over $F$ in infinitely many variables. Now, let $A$ be the quotient of $B$ by the ideal $\langle x_n^2:n\in\mathbb{N}\rangle$. For convenience, denote the image of each $x_n$ in $A$ as $a_n$. The ring $A$ is non-Noetherian, since the ideal $\mathcal{P}=\langle a_n:n\in\mathbb{N}\rangle$ is not finitely generated. (Why not?) But $\dim A$ is finite, and in fact $\mathcal{P}$ is the unique prime ideal of $A$! (And hence $\dim A=0$.) To see this, suppose $\mathcal{Q}$ is a prime ideal of $A$. For each $n\in\mathbb{N}$, we have $a_n^2=0$, hence $a_n^2\in\mathcal{Q}$, hence, since $\mathcal{Q}$ is prime, $a_n\in\mathcal{Q}$. Thus $\mathcal{Q}\supseteq\mathcal{P}$. On the other hand, $A\big/\mathcal{P}\cong F$, so $\mathcal{P}$ is maximal, thus forcing $\mathcal{P}=\mathcal{Q}$, as desired.