Proposition

Draw pictures of $\Spec(\mathbb{Z}), \Spec(\mathbb{R}), \Spec(\mathbb{C}[x]), \Spec(\mathbb{R}[x]), \Spec(\mathbb{Z}[x])$.

Solution

I am not sure how to draw pictures of these spaces, so I will just describe them in words.

$\Spec(\mathbb{Z})$

$\Spec(\mathbb{Z}) = \{ (0) \} \cup \{ (p) \mid p \text{ prime} \}$.

Let $E \subset \mathbb{Z}$. If $E$ is empty or $E = \{ 0 \}$, $V(E) = \Spec(\mathbb{Z})$.

Let $p$ be a prime number.

• If there exists $x \in E$ such that $p \nmid x$, then $E \not\subset (p)$.
• If $\forall x \in E, p \mid x$, then $E \subset (p)$.

Thus this is a necessary and sufficient condition for $(p)$ to be in $V(E)$. Since $E$ contains a nonzero element, $V(E)$ must be finite.

On the other hand, given primes $p_1, \cdots, p_n$, $V(\{ p_1p_2 \cdots p_n \}) = \{ (p_1), \cdots, (p_n) \}$.

Therefore, all closed sets in $\Spec(\mathbb{Z})$ are:

• any finite set of the form $\{ (p_1), \cdots, (p_n) \}$,
• $\{ (0) \}$,
• $\Spec(\mathbb{Z})$.

$\Spec(\mathbb{R})$

The only ideals of $\mathbb{R}$ are $(0)$ and $\mathbb{R}$ because $\mathbb{R}$ is a field. Thus the only prime ideal is $(0)$.

$\Spec(\mathbb{C}[x])$.

From algebra, we know that $\mathbb{C}[x]$ is a PID. (For instance, Corollary 4 [P.300, Dummit and Foote].)

If an ideal is generated by a polynomial of degree 2 or above, it cannot be a prime ideal because every polynomial in $\mathbb{C}[x]$ splits into linear factors. If an ideal is generated by a polynomial of degree 1, then it must be a maximal ideal because any ideal containing it must be generated by a nonzero, degree-0 polynomial, which is a unit in $\mathbb{C}[x]$. Therefore, $\Spec(\mathbb{C}[x])$ is the set of ideals generated by a polynomial of degree 1.

Let $x - a \in \mathbb{C}[x]$ be a polynomial of degree 1. Let $E \subset \mathbb{C}[x]$. Suppose $E \ne \emptyset$ and $E \ne (0)$. $(x - a) \in V(E)$ if and only if $\forall p(x) \in E, p(a) = 0$.

$E$ contains a nonzero polynomial that has a finite number of roots. Therefore, $V(E)$ must contain a finite number of prime ideals. On the other hand, for any finite number of prime ideals $(x - a_1), \cdots, (x - a_k) \in \Spec(\mathbb{C}[x])$, $E = (x - a_1) \cdots (x - a_k)$ gives us $V(E) = \{ (x - a_1), \cdots, (x - a_k) \}$.

Therefore, all closed sets in $\Spec(\mathbb{C}[x])$ are:

• any finite set of the form $\{ (x - a_1), \cdots, (x - a_n) \}$,
• $\{ (0) \}$,
• $\Spec(\mathbb{C}[x])$.

$\Spec(\mathbb{R}[x])$.

This case is similar to $\Spec(\mathbb{C}[x])$. $\mathbb{R}[x]$ is a PID.

An ideal $(p(x))$ is prime if and only if $p(x)$ is irreducible.

Let $E \subset \mathbb{R}[x]$. Suppose $E \ne \emptyset$ and $E \ne (0)$. Then given an irreducible polynomial $p(x) \in \mathbb{R}[x]$, $(p(x)) \in V(E)$ if and only if $p(x)$ divides all polynomials in $E$. If $E$ contains a nonzero polynomial, it has finitely many factors (up to associate).

Therefore, all closed sets in $\Spec(\mathbb{R}[x])$ are:

• any finite set of the form $\{ (p_1(x)), \cdots, (p_n(x)) \}$ where each $p_i(x)$ is irreducible.
• $\{ (0) \}$,
• $\Spec(\mathbb{R}[x])$.

$\Spec(\mathbb{Z}[x])$

I can’t figure out how to solve this, but [The red book of varieties and schemes, 2nd ed., p. 75] has the answer.