Finding prime spectrum of given rings 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: 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?
