1

While there is a similar question here but that was marked as a duplicate to this question. The latter question, at the level that I am at doesn't give me much insight. I also thought that if I could see differences depending on the spectrum over a ring $\operatorname{Spec} \mathbb{Z}[x]$, a field $\operatorname{Spec} \mathbb{R}[x]$ and $\operatorname{Spec} \mathbb{C}[x,y]$, that might shed some light.

I know that all prime numbers $p$ would generate prime ideals in $\operatorname{Spec} \mathbb{Z}[x]$ How is this denoted - is it correct to say that all polynomials generated by $\langle (x-p)\rangle$ i.e $\{(x-p),(x-p)(x-p),... \}$ for each $p$ are prime prime ideals? hence $\langle (x-p)\rangle$ are all of them in in $\operatorname{Spec} \mathbb{Z}[x]$?

Understanding that better would be a start and perhaps someone could give me examples in each of the 3 cases $Spec\mathbb{Z}[x]$, $Spec\mathbb{R}[x]$, and $Spec\mathbb{C}[x,y]$? i.e. $\langle (3x-5) \rangle$ are prime ideals in $Spec\mathbb{Z}[x]$ as all generated polynomials have $(3x-5)$ in them. But, and I can't think of an example but suppose it would be of the form $(ax^2 +c); a,c \in \mathbb{Z}$ which would be another generator for a prime in $Spec\mathbb{Z}[x]$, but not in $Spec\mathbb{R}[x]$ as it could be reduced to some $(dx + e)(fx +g); d,e,f,g \in \mathbb{R}$.

I think that this would be very insightful as I have not been able to find any examples of such and such is a prime ideal in this ring because... . Most resources (at least online) springboard off of abstracted examples, and I understand why as things become categorical. But I think it very useful to have some working numerical examples - but am having difficulty finding them.

Thanks,

Brian

Community
  • 1
Relative0
  • 1,419
  • @Relativeo Hey, could you read your question through again, there are several places where it is not clear (to me!) what exactly you are asking? – user50948 Jun 18 '14 at 13:16
  • "all prime numbers $p$ in $X$" makes no sense. Are you reading a book on algebraic geometry? – KCd Jun 18 '14 at 13:32
  • 1
    You don't need $p$ to be prime. $(X-a)$ is a prime ideal for all $a\in Z$, – adrido Jun 18 '14 at 20:59
  • adrido - That is really helpful actually! I would have thought that not any number $a$ would work because as per the example here: http://mathworld.wolfram.com/Ideal.html there the generators $\langle 4,10 \rangle = \langle 2 \rangle$ thus are not unique generators - I was guessing that this would bring about only primes working in the generators $\langle (X-a) \rangle$. – Relative0 Jun 19 '14 at 13:29

1 Answers1

2

$Spec \mathbb{Z}$ which is the spectrum of the integers has a closed point for every prime number $p$. Furthermore every $p$ corresponds to the maximal ideal $(p) \subset \mathbb{Z}$ along with a non-closed generic point (whose closure is the whole space) corresponding to the zero ideal $(0)$. Thus the closed subsets of $Spec \mathbb{Z}$ are the finite unions of closed points and the whole space.

Now $Spec \mathbb{R}[x]$, which is the polynomial ring over the field $\mathbb{R}$ which is also denoted $\mathbb{A}^1$ is not algebraically closed then non-linear irreducible polynomials exist. An example is such that $Spec \mathbb{R}[x]$ consists of closed points $(x-a)$ for $a \in \mathbb{R}$ and $(x^2 + px + q); p,q \in \mathbb{R}$ such that the discriminant $p^2-4q < 0$ along with the generic point $(0)$.

As with $Spec(\mathbb{R}[x])$, $\mathbb{C}[x]$ is the spectrum of the polynomial ring over the field $\mathbb{C}$ which is also denoted as $\mathbb{A}^1$, the affine line. Now since $\mathbb{C}$ is algebraically closed a non-constant polynomial is irreducible if and only if it is linear of the form $x-a$ for some element $a \in \mathbb{C}$. That however is for the affine line.

This was found at the following wikipedia page on Closed Points

Relative0
  • 1,419