# What is the “prime ideal of the polynomial ring of an operator”?

I am unable to lookup a piece of terminology I have bumped into. I have been watching AGITTOC: Algebraic geometry in the time of Covid; Pseudolecture 3, where a user claimed that the reason the spectrum of a ring is called as the "prime spectrum" is through the analogy from physics:

(Name hidden for privacy): spectrum of light -> eigenvalues of the hamiltonian operator -> prime ideal of the polynomial ring of the operator

I'm failing at finding references to "prime ideal of the polynomial ring of the operator". Could someone please point me to the mathematics they are referring to?

• They're talking about the prime spectrum of a ring - $Spec(\mathbb C[T])$ - and probably just left out a word. The first result when one searches your question seems describes exactly the history mentioned: mathoverflow.net/questions/9125/… – user208649 Jul 12 '20 at 7:47
• I don't grok the answer on MO, unfortunately. How does one define $\mathbb C[T]$? Is it $\mathbb C[X] / \operatorname{charpoly}(T)$? If so, I don't understand why the prime spectra will be the eigenvalues of $T$. Is this a general theorem that the prime spectra of the quotient polynomial ring $\mathbb C[X]/(p)$ is going to have $(p - r_i)$ as the prime spectra, where $r_i$ are the roots of $p$? – Siddharth Bhat Jul 12 '20 at 7:50
• The result you seek is just a consequence of the fact $\Bbb C$ is algebraically closed and thus every polynomial in one variable over $\Bbb C$ factors. (I would characterize this as probably too basic to be a "general theorem", but I guess it depends on how far along in your mathematical life you are.) – KReiser Jul 12 '20 at 8:56
• @KReiser I am unable to see the connection between this fact and the prime ideals of $\mathbb C[X] / (p)$. For example, if $p = (x - 1)$, then $\mathbb C[X]/(x - 1) \simeq \mathbb C$. So I don't see how $(p - 1)$ is the prime spectra of $\mathbb C[X]/(x - 1)$. Is the idea that $( p - 1)$ goes to $\{ 0 \}$ in $\mathbb C[X]/(x - 1)$, and $\{ 0 \}$ is the only ideal in the prime spectrum of $\mathbb C$? The same situation happens in general? – Siddharth Bhat Jul 12 '20 at 9:03
• A polynomial in $\Bbb C[t]$ determines a closed subscheme of $\Bbb A^1_\Bbb C$, and the (underlying) points of this closed subscheme are just the roots. In your example, the (underlying) points of $V(x-1)$ is just the point $(x-1)$, which we usually call $1\in\Bbb A^1_\Bbb C$. – KReiser Jul 12 '20 at 9:12