# Spectrum of polynomial ring

In M. Reid's Undergraduate Commutative Algebra, the author states that if $k$ is an algebraically closed field then $\operatorname{Spec}{k[x]} = \{0\} \cup k$ (page 21). Is this correct? Instead, shouldn't he have written $\operatorname{Spec}{k[x]} = \{0\} \cup \{u(x-\alpha)|u,\alpha \in k\}$?

• It is slightly confusing claim since $0\in k$, so it would seem to mean $\text{Spec} k[x]=k$, but the author probably means disjoint union and that it is an isomorphism, not exact equality. – Thomas Andrews May 9 '13 at 16:23

## 3 Answers

I think the author is likely identifying each element of $k$ with the prime ideal that is the kernel of the evaluation at that element. In other words, $a \in k$ is identified with the ideal $k[x](x-a)$.

The prime ideals of $k[x]$ are $0$ and $(x-\alpha)$ for $\alpha \in k$. If you multiply a generator of an ideal by a unit, the ideal doesn't change. The author abbreviates/identifies $(x-\alpha)$ with $\alpha$.

These answers are correct, but just for clarity:

By the Nullstellensatz, k[x] has maximal (and therefore prime) ideals generated by x-a for each a in k. These are all the maximal ideals, but there is also one prime, but non-maximal ideal, namely (0), (which is different from (x-0) = (x)!). The (0) ideal is the generic point of Spec k[x], while the other points are closed in Spec k[x].

One usually compares this to Spec Z, which has similar properties: there's one closed point for each prime p, since each prime generates a maximal ideal, and one non-closed point generated by the zero ideal.

• Nullstellensatz is overkill for the one-variable case. – Martin Brandenburg May 9 '13 at 17:02