# I: Varieties

## I.1: Affine Varieties

### 1.1: Some plane curves #to-work

(a) Let $Y$ be the plane curve $y=x^2$ (i.e., $Y$ is the zero set of the polynomial $f=$ $y-x^2$ ). Show that $A(Y)$ is isomorphic to a polynomial ring in one variable over $k$.

(b) Let $Z$ be the plane curve $x y=1$. Show that $A(Z)$ is not isomorphic to a polynomial ring in one variable over $k$.

(c) \* Let $f$ be any irreducible quadratic polynomial in $k[x, y]$, and let $W$ be the conic defined by $f$. Show that $A(W)$ is isomorphic to $A(Y)$ or $A(Z)$. Which one is it when?

> [!solution]- Solution (Part a)
> 
> $A(Y) = k[x,y]/\gens{y-x^2} \cong k[t, t^2] \cong k[t]$.

> [!solution]- Solution  (Part b)
> 
> $A(Z) = k[x, y]/\gens{xy-1} \cong k[x^{\pm 1}]$.
> This is not isomorphic to any polynomial ring, since it contains an invertible element $x\inv$ which is not in $k$.

> [!solution]- Solution (Part c, incomplete)
> Idea: for $F$ a conic (degree 2 polynomial), an affine change of coordinates yields either $F(x, y) =xy-1$ or $F(x,y) = y-x^2$.
> One can write
> $$F(x,y) =\tv{x,y, 1}^t\left(\begin{array}{ccc}
A & B / 2 & D / 2 \\
B / 2 & C & E / 2 \\
D / 2 & E / 2 & F
\end{array}\right)\tv{x,y, 1}$$
> This is symmetric and can thus be diagonalized, correspond to an affine change of coordinates where $F(x,y) = \lambda_1 x^2 + \lambda_2 y^2 + \lambda_3 1$.


### 1.2 The Twisted Cubic Curve #completed
 
Let $Y \subseteq \mathbf{A}^3$ be the set $Y = \left\{{(t, t^2,t^3) {~\mathrel{\Big\vert}~}t\in k}\right\}$. 

- Show that $Y$ is an affine variety of dimension 1. 
- Find generators for the ideal $I\left(Y\right)$. 
- Show that $A(Y)$ is isomorphic to a polynomial ring in one variable over $k$. 

We say that $Y$ is given by the *parametric representation* $x=t . y=t^2, z=t^3$.

> Useful facts: $\sqrt{I} = \sqrt{\prod p_i^{a_i}} = \prod p_i$ in a UFD when $I$ is a principal ideal factored into irreducibles. An ideal is also radical iff the quotient is reduced, and $\gens{f}$ is radical when $f$ is irreducible.

> [!solution]- Solution
> 
> - For dimension in part 1, note $$A(Y) = k[x,y]/\gens{y-x^2, z-x^3} \cong k[t, t^2,t^3] \cong k[t].$$One can now use that $\krulldim k[t] = 1$ and Prop 1.7 ($\dim Y = \krulldim A(Y)$) to conclude $\dim Y = 1$. This also shows part (3)
> - To see that $Y$ is an affine variety, it STS it is an irreducible closed subset of $\AA^3$. It is closed since $Y = V(y-x^2, z-x^3)$. It is irreducible iff $I(Y) = \gens{y-x^2, z-x^3}$ is prime by Cor 1.4. It is prime, since $k[x,y]/I(Y) \cong k[t]$ is a domain and $I(Y)$ is radical since it is generated by irreducibles.

### 1.3: Factoring into irreducible components #completed

Let $Y$ be the algebraic set in $\mathbf{A}^3$ defined by the two polynomials $x^2-yz$ and $x z-x$. Show that $Y$ is a union of three irreducible components. Describe them and find their prime ideals.

> [!solution]- Solution
> $Y = V(x^2-yz)\intersect V(xz-x) = V(x^2-yz) \intersect (V(x) \union V(z-1))$.
> Any point in $Y$ is in either $V(x)$ or $V(z-1)$, so either $x=0$ or $z=1$.
> If a point is in $V(x)$, then $x=0$ and $0^2=yz$, so either 
> 	- $x=0$ and $y=0$, and $z$ is free, or
> 	- $x=0$ and $z=0$, and $y$ is free.
> In either case, this is an intersection of two orthogonal hyperplanes in $\AA^3$ and thus a line.
> So we obtain the two lines $V(x, y) \union V(x, z)$.
> If a point is in $V(z-1)$ then $z=1$ then $x^2 = y$ yielding a parabola $V(x^2-y)$.
> 
> So $Y =V(x, y) \union V(x, z) \union V(x^2-y)$ is a union of two lines and a parabola.

### 1.4: Beware the product topology #completed

If we identify $\mathbf{A}^2$ with $\mathbf{A}^1 \times \mathbf{A}^1$ in the natural way, show that the Zariski topology on $\mathbf{A}^2$ is not the product topology of the Zariski topologies on the two copies of $\mathbf{A}^1$.

> [!hint]- Hints
> - Zariski-closed subsets of $\AA^2$ are finite unions of plane curves or the entire space. The Zariski topology on $\AA^1$ is the cofinite topology if $k$ is infinite, and is not Hausdorff.

> [!solution]- Solution
> If $k$ is infinite and algebraically closed, then $\AA^1\slice k$ is not Hausdorff: the Zariski topology is the cofinite topology, so every open set has finite complement, and every two distinct open sets intersect. Now use that $\Delta_{\AA^1\slice k}$ is closed **in the product topology** on $\AA^1\times \AA^1$ iff the base is Hausdorff, but $\Delta$ is closed in the Zariski topology since it can be written as $V(x-y)$.
> 
> Alternative: consider $V(xy-1) \subseteq \AA^2$. If this had the product topology, the projection maps $\pi_x, \pi_y: \AA^2\to \AA^1$ would be closed, but $\pi_x(V(xy-1)) = \AA^1\smz$ which is open and not closed since its complement $0 = V(x)$ is closed and not open.

### 1.5: Coordinate rings = reduced fg algebras #completed

Show that a $k$-algebra $B$ is isomorphic to the affine coordinate ring of some algebraic set in $\mathbf{A}^n$. for some $n$, if and only if $B$ is a finitely generated $k$-algebra with no nilpotent elements.

> [!hint]- Hint
> - $I$ is radical $\iff R/I$ is reduced.
> - $B$ is a finitely generated $k\dash$ algebra $\iff B\cong \kxn/I$ for some $n$ and some $I\normal \kxn$.
> - $A(Y)$ is always reduced since $I(Y)$ is always radical.

> [!solution]- Solution
> $\impliedby$: if $B\in \kalg^{\fg}$ then $I_B \injects \kxn \surjects B$ for some $n$ and $B\cong \kxn/I_B$. Note $A(Y)\da \kxn/I(Y)$ if $Y = V(I) \subseteq \AA^n\slice k$, so take $I \da I_B$ and $Y = V(I_B)$ to be the algebraic set to get $B\cong A(Y)$. Then $I(Y) = I(V(I_B)) = \sqrt{I_B} = I_B$, where $I_B$ is radical since $B = \kxn/I_B$ is reduced.
> 
> $\implies$:  if $B = A(Y) = \kxn/I(Y)$ is an affine coordinate ring, this presents $B$ is a finitely generated $k\dash$algebra. $I(Y)$ is always radical, so $A(Y)$ is always reduced, so $B$ is reduced.

### 1.6: Opens are dense in irreducible spaces  #completed

Any nonempty open subset of an irreducible topological space is dense and irreducible. If $Y$ is a subset of a topological space $X$, which is irreducible in its induced topology, then the closure $\mkern 1.5mu\overline{\mkern-1.5muY\mkern-1.5mu}\mkern 1.5mu$ is also irreducible.

> [!hint]- Hint
> - Connected: can not be decomposed into two proper closed *disjoint* sets.
> - Irreducible: can not be decomposed into two proper closed sets. So irreducible $\implies$ connected.
> 	- E.g. $V(xy)$ is reducible.
> - Irreducible iff $Y = A_1 \union A_2 \implies Y \subseteq A_i$ when $A_1, A_2$ are closed.

> [!solution]- Solution
> - $X$ is irreducible $\iff$ every open $U\subseteq X$ is dense: if not, let $U$ be open and nonempty and $X$ be irreducible. If $\cl_X(U) \neq X$, then write $X = (X\sm U) \union \cl_X(U)$ as a union (not necessarily disjoint) of proper closed sets to contradict irreducibility.
> - Suppose $Y$ is irreducible and write $\cl_X(Y) = A \union B$, with $A, B$ closed in $X$.  Write $Y = (A \intersect Y) \union (B\intersect Y)$, then wlog $Y \subseteq (A\intersect Y) \subseteq A$ forces $\cl_X(Y) \subseteq \cl_X(A\intersect Y)\subseteq \cl_X(A) = A$. Thus $\cl_X(Y)$ is irreducible.

### 1.7: Properties of Noetherian spaces #completed 

(a) Show that the following conditions are equivalent for a topological space $X$ :
    -   $X$ is noetherian;
    -   Every nonempty family of closed subsets has a minimal element;
    -   $X$ satisfies the ascending chain condition for open subsets;
    -   Every nonempty family of open subsets has a maximal element.
(b) A noetherian topological space is quasi-compact, i.e., every open cover has a finite subcover.
(c) Any subset of a noetherian topological space is noetherian in its induced topology.
(d) A noetherian space which is also Hausdorff must be a finite set with the discrete topology.

> [!hint]- Hint
> - Noetherian is the DCC on closed sets.

> [!solution]- Solution (Part a)
> - $1\implies 2$: Let $\ts{Y_j}_{j\in J}$ be a family of closed subsets. By AOC, choose a $Y_1$ -- if not minimal, there is some $Y_2 \subsetneq Y_1$, etc. This produces a decreasing chain, which stabilizes since $X$ is noetherian, so $Y_n = Y_{n+1} = \cdots$ for some $n$ and $Y_n$ is minimal.
> - $3\implies 4$: Similar proof as $1\implies 2$.
> - $1\iff 3$: take $U_1 \subseteq U_2\subseteq \cdots$ an ascending chain of opens; then $X\sm U_1 \supseteq X\sm U_2\supseteq \cdots$ is a descending chain of closed sets. Since $X$ is Noetherian, $X\sm U_n = X\sm U_{n+1} = \cdots$ so $U_n = U_{n+1} = \cdots$. A similar proof works for the other direction.
> - $2\implies 4$: take complements.
> - $4\implies 3$: if $U_1\subseteq U_2\subseteq \cdots$ is a chain of opens, $\ts{U_i}$ has a maximal element by assumption, so the chain stabilizes.

> [!solution]- Solution (Part b)
> Let $\mcu\covers X$ and build an ascending chain of opens: pick $U_1, U_2$ so that $U_1 \subseteq V_2 \da \union_{i\leq 2} U_i$, pick $U_3$ so that $U_2 \subseteq V_3\da \union_{i\leq 3} U_i$, etc. Then $V_1 \da U_1 \subseteq V_2 \subseteq \cdots \subseteq V_n = V_{n+1} = \cdots$ by Noetherianness. Thus $V_n = \union_{i\geq 0} U_i = \mcu$, so $V_n \covers X$ is a finite subcover.

> [!solution]- Solution (Part c)
> Let $U_1\subseteq \cdots$ be a chain of opens in $Y$, then by definition $U_i = V_i \intersect Y$ for some $V_i$ open in $X$, although $U_i$ need not be open in $X$. Then the family $\ts{V_i}$ has a maximal element $V_n$ which is open in $X$, so $U_n = V_n \intersect Y$ is a maximal open in $Y$, making $Y$ Noetherian.

> [!solution]- Solution (Part d)
> Take the decomposition into irreducible components $X=\union_{i\leq n} C_i$ for Noetherian spaces; it suffices to restrict to $C_0$ which is also Hausdorff. Then $C_0$ can only have one point, otherwise if $x\neq y\in C_0$ then there exist opens $U\ni x, V\ni Y$ open and disjoint by Hausdorfness, contradicting that $U$ must be dense as an open subset of an irreducible space. So $C_i =\ts{x_i}$ and $X = \ts{x_1,\cdots, x_n}$ is finite union of its closed singleton sets, making it discrete.


### 1.8: Irreducible components and minimal primes #completed 

Let $Y$ be an affine variety of dimension $r$ in $\mathbf{A}^n$. Let $H$ be a hypersurface in $\mathbf{A}^n$, and assume that $Y \nsubseteq H$. Then every irreducible component of $Y \cap H$ has dimension $r-1$.[^1]

> [!hint]- Hint
> - $V(I)$ irreducible $\implies I$ prime, so maximal irreducible subsets of $\spec R$ (i.e. irreducible components) correspond to minimal primes of $R$.
> - Irreducible components $Y$ of $X = V(\mfp)$ correspond to minimal primes $\mfq$ *containing* $\mfp$.
> - The Hauptidealsatz: if $A$ is Noetherian and $f\in A$ is not a zero divisor or a unit then every minimal prime containing $f$ has height 1.
> - If $p\in \spec A$ then $\dim A/p + \height(p) = \dim A$.

> [!solution]- Solution
> Idea: write $H = V(\gens f)$, then every irreducible component $C_i$ of $Y\intersect H \subseteq Y$ corresponds to aminimal prime $\mfp_i \in A(Y)$ containing $\gens{f}$. Every such $\mfp_i$ has height 1 by the Hauptidealsatz, so apply $\dim A(Y)/\mfp_i + \height(\mfp_i) = \dim A(Y) \implies \dim C_i + 1 = \dim Y$.
> #todo This is very sketchy, this elides some algebra details.
> 

### 1.9: Generators vs irreducible components #completed

Let $a \subseteq A=k\left[x_1, \ldots, x_n\right]$ be an ideal which can be generated by $r$ elements. Then every irreducible component of $Z(a)$ has dimension $\geqslant n-r$.

> [!solution]- Solution
> Induct: $Y_1 = V(f_1), Y_2 = V(f_1, f_2), \cdots, Y_r = V(f_1,\cdots, f_r)$ and $Y_k = Y_{k-1}\intersect H_k$ where $H_k = V(f_k)$ is a hypersurface. If $\ts{f_i}$ form a minimal generating set, then every irreducible component of $Y_2$ has dimension $\dim Y_1 - 1 = (n-1) - 1 = n-2$, and similarly $\dim Y_k$ has components of dimension $\dim Y_{k-1} - 1 = (n-(k-1) ) - 1 = n - k$, applying the previous exercise at each stage. If $\ts{f_i}$ is not a minimal set of generators, then instead $\dim Y_k \geq n-k$ and one gets an inequality.

### 1.10 #completed 

(a) If $Y$ is any subset of a topological space $X$, then $\operatorname{dim} Y \leqslant \operatorname{dim} X$.
(b) If $X$ is a topological space which is covered by a family of open subsets $\ts{U_i}$, then $\operatorname{dim} X=\sup \operatorname{dim} U_i$.
(c) Give an example of a topological space $X$ and a dense open subset $U$ with $\operatorname{dim} U <\operatorname{dim} X$.
(d) If $Y$ is a closed subset of an irreducible finite-dimensional topological space $X$, and if $\operatorname{dim} Y=\operatorname{dim} X$, then $Y=X$.
(e) Give an example of a noetherian topological space of infinite dimension.

> [!solution]- Solution (Part a)
> Let $(U_i, \leq)_{0\leq i\leq  n}$ be a strict chain of opens in $Y$ witnessing that $\dim Y = n$, then $U_i = V_i \intersect Y$ for opens $V_i\subseteq X$ using the subspace topology. Issue: the $V_i$ may not form a chain. Fix this with the "partial sum" trick: let $\tilde V_n \da \Union_{i\leq n} V_i$, then $(\tilde V_n, \leq)$ is a strict chain of opens in $X$, meaning $\dim X \geq n$.

> [!solution]- Solution (Part b)
> By part (a), since $\dim U_i \leq \dim X$ for all $i$ and this is preserved by $\sup$, it STS $\dim U_i \geq \dim X$ for every $i$.. Let $(C_i, \subsetneq)_{0\leq i\leq n}$ be a strict chain witnessing $\dim X = n$, then for any $i_0$,  $(C_j \intersect U_{i_0}, \leq)_{0\leq j\leq n}$ is a chain in $U_{i_0}$. Its length is $n$, *unless* $U_{i_0} \intersect C_0$ is empty, so choose $i_0$ so it's nonempty. This yields a strict chain of length $n$, so $\dim U_{i_0} \geq n \implies \sup_i \dim U_i \geq n$.

> [!solution]- Solution (Part c)
> Any point has dimension 0, so take $\AA^1 \slice \CC \da \spec \CC[x]$ which has dimension 1 and take a generic point corresponding to the ideal $\gens 0$.

> [!solution]- Solution (Part d)
> If $\dim Y = n$, pick a witness chain $(C_i, \subsetneq)_{0\leq i\leq n}$. Then all the $C_i$ lie in some irreducible component $Z$, so $\dim Z = \dim Y$ and $Y = C_n$ wlog by picking a chain that includes $Y$. But then if $Y$ is properly contained in $X$, then so is $Z$ and thus $C_1 \subsetneq C_2 \subsetneq \cdots \subsetneq C_n = Z \subsetneq X$ which makes $\dim X \geq n+1$, a contradiction. So $X=Y$.

> [!solution]- Solution (Part e)
> Take $X = \ZZ_{\geq 0}$ with closed sets $U_i \da \ts{1,\cdots, i}$. This is infinite dimension since $U_1 \subset U_2 \subset \cdots$ is an infinite strict chain. This is Noetherian since it satisfies the DCC on closed sets: any chain of closed sets $V_1 \supset V_2\cdots$ is of the form $V_1 = U_n$ for some $n$, which is finite.


### 1.11 \* #to-work

Let $Y \subseteq \mathbf{A}^3$ be the curve given parametrically by $x=t^3, y=t^4, z=t^5$. Show that $I(Y)$ is a prime ideal of height 2 in $k[x, y ;-]$ which cannot be generated by 2 elements. We say $Y$ is not a local complete intersection-cf. (Ex. 2.17).

> [!solution]- Solution

### 1.12 #to-work

Give an example of an irreducible polynomial $f \in \mathbf{R}[x, y]$. whose zero set $Z(f)$ in $\mathbf{A}_{\mathbf{R}}^2$ is not irreducible (cf. 1.4.2).

[^1]: (See (7.1) for a generalization.)

> [!solution]- Solution