Exercise 1.16

Solution. We use that $\{𝔭\}$ is closed if and only if $𝔭$ is maximal from Exercise 18.

$\mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(\mathbb{Z})$ is visualized as a discrete line whose closed points are the prime ideals of the form $(p)$ for $p$ prime numbers, together with the point $(0)$ whose closure is the whole line. Note our visualizations of $\mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(\mathbb{Z}),\mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(\mathbb{Z}[x])$ are from [Fig. 1.6]Rei95:

$\mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(ℝ)$ consists of one point $(0)$ since $(0)$ is the unique prime ideal in a field:

$(0)$

The prime ideals of $k[x]$ for $k$ a field are the ideals generated by irreducible polynomials and $(0)$ since $k[x]$ is a PID. Thus, $\mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(ℂ[x])$ consists of prime ideals generated by linear polynomials $x-\alpha$ where $\alpha \in ℂ$ together with the point $(0)$ whose closure is all of $\mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(ℂ[x])$, which we can visualize as a continuous line:

$\mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(ℝ[x])$ consists of prime ideals generated by linear polynomials $x-a$ for $a\in ℝ$ and quadratic polynomials $(x-\beta )(x-\bar{\beta })$ for $\beta \in ℂ$, which we can visualize as a continuous line with two groups of closed points:

Note that $\mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(ℂ[x])$ as a set is $ℂ\cup \{(0)\}$, and that $\mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(ℝ[x])$ as a set is $ℍ\cup \{(0)\}$ where $ℍ$ is the upper half-plane $\{z\in ℂ\mid \mathrm{Im}<!--\; FUNCTION\; APPLICATION\; -->z\ge 0\}$, but we chose to visualize $\mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(ℂ[x])$ and $\mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(ℝ[x])$ as above since they are one dimensional as rings.

$\mathrm{Spec}<!--\; FUNCTION\; APPLICATION\; -->(\mathbb{Z}[x])$ is visualized as a plane whose closed points are the maximal ideals of the form $(p,f)$ for $p$ a prime number and $f$ a polynomial irreducible mod $p$, with non-maximal prime ideals $(f)$ where $f$ is an irreducible polynomial, whose closure consist of all primes $(p,f)$, together with the point $(0)$ whose closure is everything:

For completeness, we prove that these are all the prime ideals in $\mathbb{Z}[x]$. Let $𝔭\subseteq \mathbb{Z}[x]$ be a prime ideal. If $𝔭$ is $(0)$ or principal it is $(0)$ or $(f)$ for $f$ irreducible, so suppose not. We can then find $f_1,f_2\in 𝔭$ with no common factor in $\mathbb{Z}[x]$.

We claim $f_1,f_2$ do not share a factor in $\mathbb{Q}[x]$. For, if $f_1=h{g}_1$ and $f_2=h{g}_2$ with $h,g_1,g_2\in \mathbb{Q}[x]$ and $\mathrm{deg}<!--\; FUNCTION\; APPLICATION\; -->h\ge 1$. Write $h=a{h}_0$, $g_1=b_1{\gamma }_1$, and $g_2=b_2{\gamma }_2$ with $a,b_1,b_2\in \mathbb{Q}$, and $h_0,{\gamma }_1,{\gamma }_2$ primitive elements of $\mathbb{Z}[x]$. By Exercise $\text{??}\mathit{iv})$, $h_0{\gamma }_1$ and $h_0{\gamma }_2$ are primitive, hence $f_1=h{g}_1=(a{b}_1)(h_0{\gamma }_1)\in \mathbb{Z}[x]$ implies that $a{b}_1\in \mathbb{Z}$, and similarly $a{b}_2\in \mathbb{Z}$. Therefore $h_0\mid f_1,f_2$, a contradiction.

Now we claim $(f_1,f_2)\cap \mathbb{Z}\ne (0)$. $\mathbb{Q}[x]$ is a PID and $\gcd <!--\; FUNCTION\; APPLICATION\; -->(f_1,f_2)=1$, and so there exist $a,b\in \mathbb{Q}[x]$ such that $a{f}_1+b{f}_2=1$. If $c\in \mathbb{Z}$ is a common denominator of all the coefficients of $a$ and $b$ then $\mathbb{Z}\ni (\mathit{ca})f_1+(\mathit{cb})f_2=c$. Now since $\mathbb{Z}$ is a PID, and $(f_1,f_2\cap \mathbb{Z}$ is a nonzero prime ideal, it is generated by $p$ prime in $\mathbb{Z}$. □