# Monday, October 04

:::{.remark}
We'll set up some properties for schemes.
A scheme can be:

- Irreducible 
- Smooth
- Reduced
- Noetherian

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:::{.remark}
Recall that $X\in \Top$ is **connected** iff whenever $X = X_1 \disjoint X_2$ with $X_i$ closed, one of $X_i = X$ and the other is empty.
$X$ is **irreducible** iff this holds in the weaker case when $X = X_1 \union X_2$ is not necessarily disjoint.
Note that irreducible implies connected.
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:::{.definition title="Connectedness and irreducibility for schemes"}
$(X, \OO_X) \in \Sch$ is **connected** (resp. **irreducible**) iff $\realize{X} \in \Top$ is connected (resp. irreducible).
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:::{.example title="Connected and irreducible"}
$X\da \AA^n\slice k$ is connected and irreducible.
Write $\AA^n = \spec \kxn$, whose closed sets are $V(I) \da \ts{p\contains I}$.
Suppose $X = V(I) \union V(J) = V(IJ)$, then all primes $p$ satisfy $IJ \contains p$, and this $IJ \contains \intersect_{p\in \spec R} p = \nilrad{R} = 0$, using a fact from commutative algebra.
Then $IJ = 0$ implies $I=0$ (wlog), so $V(I) = X$.
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:::{.example title="Connected but not irreducible"}
Let $X \da \spec k[x, y] / \gens{xy}$, for $k$ not necessarily algebraically closed.
This is roughly the coordinate axis in $k^2$, except there are generic points corresponding to irreducible closed subsets.
Points $\gens{x-a, y-b}$ are closed and contained in $X$ when $(a, b)$ is on the axis.
There are non-maximal primes $\gens{x}, \gens{y}$.

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Note $X \contains V(\gens{x}) = \ts{p\in x} = \ts{\gens{x}} \union \ts{\gens{x, y-b}}$, and similarly, $V(\gens{y}) = \ts{\gens{y}} \union \ts{\gens{x-a, y}}$.
However $V(x) \union V(y) = X$ but $V(x) \intersect V(y) = \ts{\gens{x, y}}$, making $X$ connected but not irreducible.
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:::{.example title="Not connected"}
Let $X\da \spec ( k[x]/\gens{x(x-1)} ) \cong \spec ( k[x]/\gens{x} ) \oplus \spec (k[x]/\gens{x-1}) \cong \spec (k \oplus k)$, using the Chinese remainder theorem.
Note that this has two prime ideals, $\ts{0 \oplus k, k \oplus 0}$, making it a discrete space and thus a disjoint union of its two closed points.
Note that $0 \oplus 0$ isn't prime, consider $(a, 0) \oplus (0, b)$.
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:::{.example title="Connected and irreducible"}
Consider $X \da \spec \ZZpadic = \ts{ \gens 0, \gens p}$.
Then $\cl_X\qty{ \ts{\gens{0}}} = X$, since $0$ is not a closed point.
This is connected and irreducible, and the picture is a point with a fuzzy point:

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:::{.exercise title="$\spec \ZZ$ is connected and irreducible"}
Show $\spec \ZZ$ is connected and irreducible.
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:::{.definition title="Reduced schemes"}
$(X, \OO_X) \in \Sch$ is **reduced** iff $\OO_X(U)$ is a reduced ring for all open $U \subseteq X$, i.e. contains no nilpotents.
Equivalently, $\OO_{X, p}$ is reduced for all $p\in X$.
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:::{.remark}
Note that this is a genuinely new concepts for schemes, since functions valued in a field always yields a reduced ring.
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:::{.example title="$\spec R$ is reduced for $R$ reduced"}
Let $R$ be reduced and take $X\da \spec R$, then for $U \subseteq X$ with $U = V(I)^c \supseteq D(f) \da V(f)$ for any $f\in I$.
In fact $U = \Union_{f\in I} D(f)$, and $\OO_X(U)$ by the sheaf property can be written as 
\[
\OO_X(U) 
&= \ts{\phi_f \in \OO_X(D(f)) \st \ro{\phi_f}{D(f) \intersect D(g) } = \ro{\phi_f}{D(f) \intersect D(g)} } \\
&\subseteq \prod_{f\in I} \OO_X(D(f)) \\
&= \prod_{f\in I} R\localize{f}
,\]
by the Dan-Changho theorem, and the claim is that $R\localize{f}$ is reduced for all $f$.
This follows since $(a/f^k)^d \sim 0/1 \implies f^d a^d = 0$, so $f^{dm}a^d = (f^ma)^d = 0$, so $f^ma=0$ since $R$ is reduced and this $a/f^k \sim 0$, so the localization has no nilpotents.
Then $\OO_X(U)$ is a subring of a reduced ring and thus reduced, and $\spec R$ is a reduced scheme.
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:::{.definition title="Integral schemes"}
$(X, \OO_X)\in \Sch$ is **integral** iff $\OO_X(U)$ is an integral domain for all $U$.
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:::{.example title="$\spec R$ is integral for $R$ integral"}
For $R$ an integral domain, $\spec R$ is integral.
Supposing $R$ is not an integral domain, so $xy=0$ with $x,y\neq 0$.
Then $0\in p$ for any prime, so $x$ or $y$ is in any prime ideal, so $V(x) \union V(y) = \spec R$.
Alternatively, one can use $\OO_{\spec R}(\spec R) = R$.
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:::{.proposition title="?"}
$(X, \OO_X)$ is integral $\iff$ it is irreducible and reduced.
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:::{.proof title="of proposition"}
**Integral $\implies$ reduced**:
Just use the domains have no nilpotents.

**Integral $\implies$ irreducible**:
For the sake of contradiction, suppose $X = X_1 \union X_2$ with $X_i$ closed.
Then there exist $U_i = X_i^c$ open and disjoint (using $X^c = X_1^c \intersect X_2^c$), so $\OO_(U_1 \disjoint U_2) = \OO(U_1) \times \OO(U_2)$ by the sheaf property for $\OO$.
However, this is not an integral domain.

**Irreducible and reduced $\implies$ integral**:
Next time!
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