I am going through the book The Geometry of Schemes currently, just want to make sure I am not messing something up here, also see (d) as I am unsure how to really show this. The question is find Spec(R) with R:

(a) R = $\mathbb{Z}$, I have that $Spec(\mathbb{Z}) = \{(p)| p \ prime\} \cup \{0\}$

This is because $\mathbb{Z}$ is an integral domain so that $(0)$ is prime and we know that every prime also gives a prime ideal.

(b) R = $\mathbb{Z} / 3\mathbb{Z}$, I have Spec(R) = $\{0\}$

This is because R is an integral domain so that $(0)$ is prime, but we have that $\overline{1}, \overline{2}$ generated the whole ring so that their ideals are not prime as a prime ideal is a proper subset.

(c) R = $\mathbb{Z} / 6\mathbb{Z}$, I have Spec(R) = $\{(2),(3),(5)\}$, here i mean the class of 2,3,5.

This is because R is not an integral domain so we cannot include 0. We also have $\overline{1}$ generates the whole ring so its ideal is not a proper subset. We notice that $\overline{2}, \overline{3}, \overline{5}$ generate prime ideals as when we mod out by them we get a field. Finally we don't include 4 as when we mod out by $(\overline{4})$ it is not an integral domain.

(d) R = $\mathbb{Z}_{(3)}$, I have Spec(R) = $\{0\} \cup \{$reduced fractions where 3 divides the numerator, but not the denominator$\}$

We have that $R = \{\frac{a}{b}|a,b \in \mathbb{Z}, 3 \ doesn't \ divide \ b\}$. Now R is an integral domain so we include 0. Now we must have that three divides $a$ for it to generate a prime ideal otherwise it would generate the whole ring as it has an inverse. Here i am guessing a bit but then I believe the prime ideals would be represented by reduced fractions in $\mathbb{Q}$ where the 3 divides the numerator, but does not divide the denominator. Any hints as to why this is wrong or how to show this is appreciated.

(e) R = $\mathbb{C}[X]$, I have Spec(R) = $\{0\} \cup \{(x-c)| c \in \mathbb{C}\}$

We include 0 as this is an integral domain. Now the prime ideals are of the form $(x-c)$ such that $c \in \mathbb{C}$.

(f) R = $\mathbb{C}[X] \over (x^2)$, I have Spec(R) = $\{(x-c)| c \in \mathbb{C}\}$

We do not include 0 as this is not an integral domain. But we have the same polynomials in (e) remain irreducible so that they are prime as R is a UFD.


a) You also need to know in $\mathbf Z$ (more generally in any P.I.D.), non-zero prime ideals are maximal (and generated by irreducible elements).

b) More simply, $\mathbf Z/3\mathbf Z$ is a field. A field has only one proper ideal, $(0)$.

c) $5$ is a unit in $\mathbf Z/6\mathbf Z$ and does not generate a prime ideal. Actually, in a quatient ring $A/I$, prime ideals correspond bijectively to the prime ideals of $A$ which contain $I$. In the case of $\mathbf Z$, this translates as ‘correspond to the primes which divide $6$’, i.e. $(\bar 2), (\bar 3)$.

d) It's a little different: the prime ideals of $\mathbf Z_{(3)}$ correspond bijectively to the primes of $\mathbf Z$ which do not divide $3$, i.e. the primes $p\neq 3$.

e) is correct, since the prime ideals in a polynomial ring over a field, which is principal, are genetaed by irreducible polynomials, and conversely.

f) is false; the prime ideals of $\mathbf C[x]/(x^2)$ correspond bijectively to the prime ideals of $\mathbf C[x]$ which contain x^2, i.e. to the irreducible divisors of $x^2$. There's only one of them. Guess which?

  • $\begingroup$ (c) Yes this makes sense didn't think to check if 5 was a unit, for (f) we would have to have it be (x) I'm guessing. It would help to know the precise statement for corresponds bijectively to prime ideals, ie when does it hold for integral domains or must this be a bit stronger? $\endgroup$ – Michael N Dec 19 '15 at 20:07
  • $\begingroup$ Sorry did not read carefully enough everything is cleared up, thank you very much for your insight! $\endgroup$ – Michael N Dec 19 '15 at 20:13

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