4
$\begingroup$

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 $\operatorname{Spec}{k[x]} = \{0\} \cup \{u(x-\alpha)|u,\alpha \in k\}$?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Thomas Andrews May 9 '13 at 16:23
7
$\begingroup$

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 $k[x](x-a)$.

$\endgroup$
8
$\begingroup$

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 $\alpha$.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Nullstellensatz is overkill for the one-variable case. $\endgroup$ – Martin Brandenburg May 9 '13 at 17:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.