# $\spec A$ as a space (Monday, August 30)

## The prime spectrum $\spec A$

:::{.remark}
Let $R\in\CRing$ be a commutative unital ring in which $0\neq 1$ unless $R=0$.
The goal is to define a space $X$ such that $R$ is the ring of functions on $X$, imitating the correspondence between $X\in \Mfd$ and $R \da C^0(X; \RR)$.
Recall that an ideal $\mfp \in \Id(R)$ is **prime** iff $\mfp \subset A$ is a proper subset and $fg\in \mfp \implies f\in \mfp$ or $g\in \mfp$, or equivalently $R/\mfp$ is a field.
:::

:::{.slogan}
Ideals are "contagious" under multiplication, and *prime* ideals have "reverse contagion".
:::

:::{.definition title="Spectrum of a ring"}
For $A\in\CRing$ as above, 
\[
\spec A \da \ts{\mfp \normal A \st \mfp \text{ is a prime ideal}} \qquad \in \Set
.\]
We topologize $\spec A$ by defining a topology of closed sets as follows:
\[
\tau(A) \da \ts{V(I) \st I\normal A},\qquad V(I) \da \ts{ \mfp \in \spec(A) \st \mfp \contains I }
.\]
:::

:::{.exercise title="The topology is really a topology"}
Prove that $(\spec A, \tau(A))$ yields a well-defined topological space. 
:::

:::{.example title="Spec of a field"}
For $A$ a field, $\spec(A) = \ts{\gens{0}}$ is a point -- any other nonzero element $\mfp \in \spec A$ would contain a unit $u$, in which case $u\inv u = 1\in \mfp \implies \mfp = A$. 
:::

:::{.example title="Spec of a polynomial ring"}
For $k$ an algebraically closed field, 
\[
\spec k[t] = \ts{ \gens{0}, \gens{t-a} \st a\in k}
.\]
This is a PID, so every ideal is of the form $I = \gens{f}$, and one can check that
\[
V(\gens{f})
= 
\begin{cases}
\spec k[t] & f=0 
\\
\gens{x-a_1, \cdots, a-a_k} & f(x) = \displaystyle\prod_{i=1}^k (x-a_i)
\end{cases}
.\]

Note that this is **not** the cofinite topology on $\spec A$, since $f=0$ defines a generic point $\eta \da \gens{0}$.
:::