# Monday April 20th

Exercise
: If $f: R\to S$ is a ring morphism, then $f^*: \spec S \to \sec R$ is a continuous map of topological spaces.

Exercise
: Defining $R^{\text{red}} \definedas R/\mfm R$, show that $q^*: \spec R^{\text{red}} \to \spec R$ is a homeomorphism.

Exercise
: The bijection $\spec\qty{\prod R_i} = \coprod \spec R_i$ is a canonical homeomorphism for finite (co)products.

Proposition (Spec Is Quasicompact)
: For any ring, $\spec R$ is quasi-compact, i.e. every open cover has a finite subcover.

> Here, "compact" will denote quasi-compact and Hausdorff.

Proof (Sketch, Important to Check)
:   The open sets $U(f) \definedas \spec R \setminus V(f)$ are a base for the Zariski topology.
    It suffices to check that every open cover by basic open sets has a finite subcover.
    Can check that that the unit ideal is generated by finitely many elements.

Theorem (Idempotents Corresponds to Clopen Subsets of Spec)
:   For any ring $R$, there is a bijection
    \begin{align*}
    \correspond{\text{Idempotents of }R} &\iff \correspond{\text{clopen subsets of }\spec R} \\
    e &\mapsto U(e)
    \end{align*}

    where on the RHS we have $e(1-e) = 0 \implies V(e) = V(1-e)$.

Note that a space is connected iff the only clopen sets are trivial, thus $\spec R$ is connected iff the only idempotents of $R$ are $0, 1$.
Thus a ring is connected, i.e. it's not a nontrivial product of rings, iff its spectrum is a connected space.

Theorem (Topological Properties of Zero Dimensional Rings)
:   For a ring $R$, TFAE

    - $\dim R = 0$, so every prime ideal is maximal.
    - $\spec R$ is separated, i.e. $T1$, points are closed, i.e. for disjoint points, disjoint neighborhoods exist (note that $\maxspec R$ is always separated and quasicompact)
    - $\spec R$ is Hausdorff
    - $\spec R$ is a Boolean space (compact and totally disconnected)

Theorem (13.10: Characterization of Gelfand Rings)
:   A ring $R$ is *Gelfand* iff every prime ideal is contained in a maximal ideal (e.g. a local ring).
    TFAE: 

    1. $R/\mcj(R)$ is Gelfand.
    2. $\maxspec R$ is compact.
    3. $\maxspec R$ is Hausdorff.

  Theorem (5.9,11: Maximal Ideals in Rings of Continuous Real-Valued Functions)
:   For a topological (Hausdorff?) space $X$, define $C(X)$ to be the ring of $\RR\dash$valued functions $f$ on $X$. 
    If $X$ is compact, then there is a bijection
    \begin{align*}
    X &\mapsvia{\cong} \maxspec C(X) \\
    c &\mapsto \mfm_c = \theset{f\in C(X) \suchthat f(c) = 0 }
    .\end{align*}

Recall Kolmogorov spaces are defined as those for which distinct points never have precisely the same set of neighborhoods.

Theorem (Hocharter: When Spaces Are the Spec of a Ring)
:   Let $X$ be a topological space.

    a. $X \cong_{\text{Top}} \maxspec R$ for some $R \iff X$ is quasi-compact and separated.
    b. $X \cong_{\text{Top}} \spec R \iff X = \inverselim Y_i$ where $Y_i$ is a finite Kolmogorov space.

> For fun: find where this theorem is applied in the literature!

Definition (Irreducible)
: A topological space is *irreducible* if it is nonempty and not the union of two proper closed subspaces.

Exercise
: Show that $X$ is Hausdorff and irreducible iff $\# X = 1$.

Exercise
:   For a topological space $X$, TFAR:

    1. $X$ is irreducible.
    2. Finite intersections of nonempty open sets are nonempty.
    3. Every open subset is dense.
    4. Every open subset is connected.

Proposition (Properties of Irreducible Spaces)
:   For $X$ a nonempty topological space,

    a. If $X$ is irreducible, then every nonempty open subset is irreducible.
    b. If $Y \subset X$ is irreducible, then its closure $\bar Y$ is irreducible.
    c. If $X$ is is covered by $\theset{U_i}$ and for all $i, j$ we have $U_i \intersect U_j \neq \emptyset$, then $X$ is irreducible.
    d. The continuous image of an irreducible space is irreducible.

Recall that we have a correspondence
\begin{align*}
\correspond{\text{Radical ideals of }R} \stackrel[V]{I}{\tofrom} \correspond{\text{Closed subsets of }\spec R} \\
.\end{align*}

Question:
What is $V(\mfp)$ for $\mfp$ prime?

> Note that the radical of an ideal could be prime without the ideal itself being prime.

Proposition (Important: Variety is Irreducible iff Radical is Prime)
: For $I\normal R$, then $V(I)$ is irreducible iff $\rad I$ is prime.

Proof
:   WLOG, we can take $I = \rad I$.
    If $\mfp \in \spec R$ is prime and $V(\mfp)$ is reducible, then we will have a contradiction.
    In this case, there exist $I, J \normal R$ such that $V(I), V(J) \subsetneq V(\mfp)$ with $V(\mfp) = V(I) \union V(J) = V(IJ)$.
    So $\mfp = \rad IJ \supset IJ$, so $\mfp \supset I$ or $\mfp \supset J$.
    WLOG $\mfp \supset J$, then $V(\mfp) \subset V(I)$, which is a contradiction.
    
    \
    
    Conversely, suppose $V(I)$ is irreducible and I is radical but not empty?
    Then there exist $a, b\in R$ such that $ab\in I$ but $a, b\not\in I$.
    So $\rad a, \rad b \not\subset I$, so $V(I) \not\subset V(a)$ or $V(b)$.
    But then $V(a) \intersect V(I)$ and $V(b) \intersect V(I)$ are proper closed subsets with intersection (?) $V(I)$.
    Then $(ab) \subset I$ implies that $V(I) \subset V(b) = V(ab) = V(a) V(b)$ (?).


Under what hypotheses does $\spec R$ have only finitely many closed subsets?
Uses this correspondence, and the next proof will show that this is true when the ring is Noetherian.