• \(x^2 + y^2 - 1\) defines a circle, say, over \({\mathbb{R}}\)

• \(y^2 = x^3-x\) gives an elliptic curve:

• \(x^n+y^n-1\): does it even contain a \({\mathbb{Q}}{\hbox{-}}\)point? (Fermat’s Last Theorem)

• \(x^2 + 1\), which has no \({\mathbb{R}}{\hbox{-}}\)points.

• \(x^2 + y^2 + 1/{\mathbb{R}}\) vanishes nowhere, so its ring of functions is not \({\mathbb{R}}[x, y] / \left\langle{x^2 + y^2 + 1}\right\rangle\). The problem: \({\mathbb{R}}\) is not algebraically closed.

• \(x^2 - y^2 = 0\) over \({\mathbb{C}}\) is not a manifold (no chart at the origin):

• \(x+y+1/{\mathbb{F}}_3\), which has 3 points over \({\mathbb{F}}_3^2\), but \(f(x, y) = (x^3 - x)(y^3-y)\) vanishes at every point

  • Not possible when algebraically closed. For example, is there a nonzero polynomial that vanishes on every point in \({\mathbb{C}}\)?

  • \(V(f) = {\mathbb{F}}_3^2\), so the coordinate ring is zero instead of \({\mathbb{F}}_3[x, y]/\left\langle{f}\right\rangle\) This is addressed by scheme theory.

Take the curve \begin{align*} X = \left\{{(x, y, z) = (t^3, t^4, t^5) \in {\mathbb{C}}^3 {~\mathrel{\Big|}~}t\in {\mathbb{C}}}\right\} .\end{align*}

Then \(X\) is cut out by three equations:

• \(y^2 = xz\)
• \(x^2 = yz\)
• \(z^2 = x^2 y\)

Given the Riemann surface \begin{align*} y^2 = (x-1)(x-2)\cdots(x-2n) ,\end{align*} how does one visualize its solution set?

Let \begin{align*} I &= (x_1, x_2^2) \subset {\mathbb{C}}[x_1, x_2] \\ &= \left\{{ f_1 x_1 + f_2 x_2 {~\mathrel{\Big|}~}f_1, f_2 \in {\mathbb{C}}[x_1, x_2]}\right\} \end{align*}

Then \(\sqrt{I} = (x_1, x_2)\), since \(x_2^2 \in I \implies x_2 \in \sqrt{I}\).

\(\deg(x_1 + x_2^2 + x_1 x_2^3) = 4\)

• Let \(f(x) = 0\), then \({\mathbb{A}}^n = V\qty{\left\{{f}\right\}}\) is an affine variety.
• Any point \((a_1, \cdots, a_n)\in {\mathbb{A}}^n\) is an affine variety, uniquely determined by \begin{align*} V(x_1 - a_1, \cdots, x_n - a_n) = \left\{{a_1, \cdots, a_n}\right\} \end{align*}
• For any finite set \(r_1, \cdots, r_k \in {\mathbb{A}}^1\), there exists a polynomial \(f\in k[x_1]\) whose roots are \(r_i\).

Let \(X\) be the union of the \(x_1\) and \(x_2\) axes in \({\mathbb{A}}^2\), then \begin{align*} I(X) = \left\langle{x_1 x_2}\right\rangle = \left\{{g x_1 x_2 {~\mathrel{\Big|}~}g\in k[x_1, x_2]}\right\} .\end{align*}

Consider \(k[x]\). If \(J\subset k[x]\) is an ideal, it is principal, so \(J = \left\langle{f}\right\rangle\). We can factor \(f(x) = \prod_{i=1}^k (x-a_i)^{n_i}\) and \(V(f) = \left\{{a_1, \cdots, a_k}\right\}\). Then \begin{align*} I(V(f)) = \left\langle{(x-a_1)(x-a_2)\cdots(x-a_k)}\right\rangle = \sqrt{J} \subsetneq J ,\end{align*} so this loses information.

Let \(J = \left\langle{x-a_1, \cdots, x-a_n}\right\rangle\), then \(I(V(J)) = \sqrt{J} = J\) with \(J\) maximal. Thus there is a correspondence \begin{align*} \left\{{\substack{\text{Points of } {\mathbb{A}}^n}}\right\} \iff \left\{{\substack{\text{Maximal ideals of }k[x_1, \cdots, x_n]}}\right\} .\end{align*}

Example of property (b):

Take \(X_1 = V(y-x^2)\) and \(X_2 = V(y)\), a parabola and the \(x{\hbox{-}}\)axis.

Then \(X_1 \cap X_2 = \left\{{(0, 0)}\right\}\), and \(I(X_1) + I(X_2) = \left\langle{y-x^2, y}\right\rangle = \left\langle{x^2, y}\right\rangle\), but \begin{align*} I(X_1 \cap X_2) = \left\langle{x, y}\right\rangle = \sqrt{\left\langle{x^2, y}\right\rangle} \end{align*}

For \(X\subset Y \subset {\mathbb{A}}^n\), we have \(I(X) \supset I(Y) \supset I({\mathbb{A}}^n)\), so we have maps

Let \(Y = V(y-x^2) \subset {\mathbb{A}}^2/{\mathbb{C}}\) and \(X = \left\{{(1, 1)}\right\} = V(x-1, y-1)\subset {\mathbb{A}}^2/{\mathbb{C}}\).

Then there is an inclusion \(\left\langle{y-x^2}\right\rangle \subset \left\langle{x-1, y-1}\right\rangle\), e.g. by Taylor expanding about the point \((1, 1)\). and thus there is a map

Compare the classical topology on \({\mathbb{A}}^1_{/{\mathbb{C}}}\) to the Zariski topology. Consider the set \begin{align*} A\mathrel{\vcenter{:}}=\left\{{x\in {\mathbb{A}}^1_{/{\mathbb{C}}} {~\mathrel{\Big|}~}{\left\lVert {x} \right\rVert} \leq 1}\right\} \end{align*} which is closed in the classical topology. However, \(A\) is not closed in the Zariski topology, since the closed subsets are finite sets or the whole space. In fact, the topology here is the cofinite topology.

Let \(f: {\mathbb{A}}^1_{/k} \to {\mathbb{A}}^1_{/k}\) be any injective map. Then \(f\) is necessarily continuous wrt the Zariski topology. Thus the notion of continuity is too weak in this situation.

Consider \({\mathbb{A}}^2_{/{\mathbb{C}}}\), so the closed sets are curves and points. Observation: \(V(x_1 x_2 ) \subset {\mathbb{A}}^2_{/{\mathbb{C}}}\) decomposed into the union of the coordinate axes \(X_1 \mathrel{\vcenter{:}}= V(x_1)\) and \(X_2 \mathrel{\vcenter{:}}= V(x_2)\). The Zariski topology can detect these decompositions.

\(V(x_1 x_2)\) is reducible but connected.

\({\mathbb{A}}^1_{/{\mathbb{C}}}\) is not irreducible, since we can write \begin{align*} {\mathbb{A}}^1_{/{\mathbb{C}}} = \left\{{{\left\lVert {x} \right\rVert} \leq 1}\right\} \cup\left\{{{\left\lVert {x} \right\rVert} \geq 1}\right\} \end{align*}

Let \(X = \left\{{\mathbf{p}^1, \cdots, \mathbf{p}^d}\right\}\) be a finite set in \({\mathbb{A}}^n\). The Zariski topology on \(X\) is the discrete topology, and \(X = {\coprod}_{i=1}^d \left\{{\mathbf{p}^i}\right\}\). So \begin{align*} A(X) = A\qty{\displaystyle\coprod_{i=1}^d \left\{{\mathbf{p}^i}\right\}} = \prod_{i=1}^d A\qty{\left\{{\mathbf{p}^i}\right\}} = \prod_{i=1}^d \frac{ k[x_1, \cdots, x_n]} {\left\langle{x_1 - p^i_1, \cdots, x_n - p^i_n}\right\rangle } \end{align*} where \(p_j^i\) is the \(j\)th component of \(\mathbf{p}^i\).

Set \(V(x_1 x_2) = X\), then \(A(X) = k[x_1, x_2]/ \left\langle{x_1 x_2}\right\rangle\). This not being a domain (since \(x_1 x_2 = 0\)) corresponds to \(X = V(x_1) \cup V(x_2)\) not being irreducible.

Let \(X_1\) be the \(xy{\hbox{-}}\)plane and \(X_2\) be the line parallel to the \(y{\hbox{-}}\)axis through \({\left[ {0,0,1} \right]}\), and let \(X= X_1 {\coprod}X_2\):

Then \(X_1 = V(z)\) and \(X_2 = V(x, z-1)\), and \begin{align*} I(X) = \left\langle{z}\right\rangle \cdot \left\langle{x, z-1}\right\rangle= \left\langle{xz, z^2 - z}\right\rangle \end{align*} The coordinate ring is then given by \begin{align*} A(X) = { {\mathbb{C}}[x, y, z] \over \left\langle{xz, z^2 - z}\right\rangle } = { {\mathbb{C}}[x, y, z] \over \left\langle{z}\right\rangle } \oplus { {\mathbb{C}}[x, y,z] \over \left\langle{x, z-1}\right\rangle } \end{align*}

Consider \({\mathbb{A}}^2/{\mathbb{C}}\) and some subvarieties \(C_i\):

Then irreducible subvarieties correspond to prime ideals in \({\mathbb{C}}[x, y]\). Here \(C_1, C_3\) correspond to \(V(f), V(g)\) for \(f,g\) irreducible polynomials, whereas \(C_2\) corresponds to a maximal ideal, i.e. \(V(x_1 - a_1, x_2 - a_2)\). Note that \(I(C_1 \cup C_2 \cup C_3)\) is not a prime ideal, since the variety is reducible as the union of 3 closed subsets.

A finite set is irreducible iff it contains only one point.

Any irreducible topological space is connected, since irreducible requires a union but connectedness requires a disjoint union.

\({\mathbb{A}}^n/k\) is irreducible: by prop 2.8, its irreducible iff the coordinate ring is a domain. However \(A({\mathbb{A}}^n) = k[x_1, \cdots, x_n]\), which is a domain.

\(V(x_1 x_2)\) is not irreducible, since it’s equal to \(V(x_1) \cup V(x_2)\).

Consider \({\mathbb{A}}^1/k\), what are the closed subsets? The finite sets, the empty set, and the entire space.

What are the irreducible closed subsets? Every point is a closed subset, so sets with more than one point are reducible. So the only irreducible closed subsets are \(\left\{{a}\right\}, {\mathbb{A}}^1/k\), since an affine variety is irreducible iff its coordinate ring is a domain and \(A({\mathbb{A}}^1/k) = k[x]\). We can check

which is of length \(1\), since there is one nontrivial containment \(Y_1 \subsetneq Y_2\), and so \(\dim({\mathbb{A}}^1/k) = 1\).

Consider \(V(x_1 x_2) \subset {\mathbb{A}}^2/k\), the union of the \(x_i\) axes. Then the closed subsets are \(V(x_1), V(x_2)\), along with finite sets and their unions. What is the longest chain of irreducible closed subsets?

Note that \(k[x_1, x_2] / \left\langle{x_1}\right\rangle \cong k[x_2]\) is a domain, so \(V(x_i)\) are irreducible. So we can have a chain \begin{align*} \emptyset \subsetneq \left\{{a}\right\} \subsetneq V(x_1) \subset X ,\end{align*} where \(a\) is any point on the \(x_2{\hbox{-}}\)axis, so \(\dim(X) = 1\).

The only closed sets containing \(V(x_1)\) are \(V(x_1)\cup S\) for \(S\) some finite set, which can not be irreducible.

Take \(X = {\mathbb{N}}\) and define a topology by setting closed subsets be the sets \(\left\{{0, \cdots, n}\right\}\) as \(n\) ranges over \({\mathbb{N}}\), along with \({\mathbb{N}}\) itself. Is \(X\) Noetherian? Check descending chains of closed sets:

\begin{align*} {\mathbb{N}}\supsetneq \left\{{0, \cdots, N}\right\} \supsetneq \left\{{0, \cdots, N-1}\right\} \cdots ,\end{align*}

which has length at most \(N\), so it terminates and \(X\) is Noetherian. But note that all of these closed subsets \(X_N \mathrel{\vcenter{:}}=\left\{{0, \cdots, N}\right\}\) are irreducible. Why? If \(X_n = X_i \cup X_j\) then one of \(i, j\) is equal to \(N\), i.e \(X_i, X_j = X_N\). So for every \(N\), there exists a chain of irreducible closed subsets of length \(N\), implying that \(\dim({\mathbb{N}}) = \infty\).

Using the fact above, let’s compute \(\dim {\mathbb{A}}^n/k\). We can take the following chain of prime ideals in \(k[x_1, \cdots, x_{n}]\): \begin{align*} 0 \subsetneq \left\langle{x_1}\right\rangle \subsetneq \left\langle{x_1, x_2}\right\rangle \cdots \subsetneq \left\langle{x_1, \cdots, x_n}\right\rangle .\end{align*}

By applying \(V({\,\cdot\,})\) we obtain \begin{align*} {\mathbb{A}}^n/k \supsetneq {\mathbb{A}}^{n-1}/k \cdots \supsetneq {\mathbb{A}}^0/k = \left\{{0}\right\} \supsetneq \emptyset ,\end{align*} where we know each is irreducible and closed, and it’s easy to check that these are maximal:

If there were an ideal \(\left\langle{x_1, x_2}\right\rangle \subset P \subset \left\langle{x_1, x_2, x_3}\right\rangle\), then take \(P\cap k[x_1, x_2, x_3] / \left\langle{x_1, x_2}\right\rangle\) which would yield a polynomial ring in \(k[x_1]\). But we know the only irreducible sets in \({\mathbb{A}}^1/k\) are a point and the entire space.

So this is a chain of maximal length, implying \(\dim {\mathbb{A}}^n/k = n\).

Take \(f=x_1x_2 + x_3^2 x_4\) and map \(x_3 \mapsto x_3 + x_4\).

Take \(V(x^2+y^2) \subset {\mathbb{A}}^2_{/{\mathbb{R}}}\).

Apply this to \(R=A(X)\), we find that there is a bijection \begin{align*} \operatorname{codim}1 \text{ prime ideals} \iff \operatorname{codim}1 \text{ closed irreducible subsets }Y\subset X,\text{ i.e. hypersurfaces} .\end{align*}

Taking \(A(X) = {\mathbb{C}}[x,y,z]/\left\langle{x^2+y^2-z^2}\right\rangle\), whose real points form a cone:

Note that \(x^2 + y^2 = (x-iy)(x+iy) = z^2\) in this quotient, so this is not a UFD.

Then taking a line through its surface is a codimension 1 subvariety not cut out by a single polynomial. Such a line might be given by \(V(x + iy, z)\), which is 2 polynomials, so why not codimension 2?

Note that \(V(z)\) is the union of the lines

• \(z = 0, x + iy= 0\),
• \(z=0, x - iy = 0\).

Note that it suffices to show that this ring has an irreducible that is not prime. Supposing \(z = f_1 f_2\), some \(f_i\) is a unit, then \(z\) is not prime because \(z\divides xy\) but divides neither of \(x,y\).

Note that \(k[x_1, \cdots, x_{n}]\) is a UFD since \(k\) is a UFD. Applying the corollary, every hypersurface in \({\mathbb{A}}^n\) is cut out by a single irreducible polynomial.

• \(X\) a manifold or an open set in \({\mathbb{R}}^n\) has a ring of \(C^\infty\) functions.
• \(X \subset {\mathbb{C}}\) has a ring of holomorphic functions.
• \(X\subset {\mathbb{R}}\) has a ring of real analytic functions

These all share a common feature: it suffices to check if a function is a member on an arbitrary open set about a point, i.e. they are local.

For \(X\) an affine variety and \(f\in A(X)\), consider the open set \(U\mathrel{\vcenter{:}}= V(f)^c\). Then \({1\over f}\) is a regular function on \(U\), so for \(p\in U\) we can take \(U_p\) to be all of \(U\).

For \(X = {\mathbb{A}}^1\), take \(f=x-1\). Then \({x\over x-1}\) is a regular function on \({\mathbb{A}}^1 \setminus\left\{{1}\right\}\).

Let \(X + V(x_1 x_4 - x_2 x_3)\) and \begin{align*} U \mathrel{\vcenter{:}}= X\setminus V(x_2, x_4) = \left\{{{\left[ {x_1, x_2, x_3, x_4} \right]} {~\mathrel{\Big|}~}x_1 x_4 = x_2 x_3, x_2\neq 0 \text{ or } x_4\neq 0 }\right\} .\end{align*}

Define \begin{align*} \phi: U &\to K \\ {\left[ {x_1, x_2, x_3, x_4} \right]} &\mapsto \begin{cases} {x_1\over x_2} & \text{if } x_2 \neq 0 \\ {x_3\over x_4} & \text{if } x_4 \neq 0 \end{cases} .\end{align*}

This is well-defined on \(\left\{{x_2\neq 0}\right\} \cap\left\{{x_4 \neq 0}\right\}\), since \({x_1\over x_2} = {x_3 \over x_4}\). Note that this doesn’t define an element of \(k\) at \({\left[ {0,0,0,1} \right]}\in U\). So this is not globally a fraction.

\(X = V(x_1 x_4 - x_2 x_3)\) on \(U = V(x_2, x_4)^c\), the following function is regular: \begin{align*} \phi: U &\to k \\ x &\mapsto \begin{cases} {x_1\over x_2} & x_2 \neq 0 \\ \\ {x_3 \over x_4} & x_4 \neq 0 \end{cases} .\end{align*} Note that this is not globally a fraction.

Let \(f\in R\) and take \(S = \left\{{f^n {~\mathrel{\Big|}~}n\geq 1}\right\}\), then \(R_f \mathrel{\vcenter{:}}= R_S\).

The smooth functions on \({\mathbb{R}}\) with the standard topology, \(\mathcal{F} = C^\infty\) where \(C^\infty(U)\) is the set of smooth functions \(U\to {\mathbb{R}}\). It suffices to check the restriction condition, but the restriction of a smooth function is smooth: if \(f\) is smooth on \(U\), it is smooth at every point in \(U\), i.e. all derivatives exist at all points of \(U\). So if \(V\subset U\), all derivatives of \(f\) will exist at points \(x \in V\), so \(f\) will be smooth on \(V\).

Note that this also works with continuous functions.

• Smooth functions on \({\mathbb{R}}^n\)
• Holomorphic functions on \({\mathbb{C}}\)

Let \(X = \left\{{p, q}\right\}\) be a two-point space with the discrete topology, i.e. every subset is open. Then define a sheaf by \begin{align*} \emptyset &\mapsto 0 \\ \left\{{p}\right\} &\mapsto R \\ \left\{{q}\right\} &\mapsto S \\ \implies \left\{{p, q}\right\} &\mapsto R\times S ,\end{align*} where the sheaf condition forces the assignment of the whole space to be the product. Note that the first 3 assignments are automatically compatible, which means that we need a unique \(f\in \mathcal{F}(X)\) restricting to \(R\) and \(S\). In other words, \(\mathcal{F}(X)\) needs to be unique and have maps to \(R, S\), but this is exactly the universal property of the product.

Consider the presheaf on \(X\) given by \(\mathcal{F}(X) = R\times S \times T\). Taking \(T = {\mathbb{Z}}/2{\mathbb{Z}}\), we can force uniqueness to fail: by projecting to \(R, S\), there are two elements in the fiber, namely \((r,s,0)\mapsto r,s\) and \((r,s,1)\mapsto r,s\).

Let \(X = \left\{{a, b, c}\right\}\) and \(\tau = \left\{{\emptyset, \left\{{a}\right\}, \left\{{a, b}\right\}, \left\{{a, c}\right\}}\right\}\). Can check that it’s closed under finite intersections and arbitrary unions, so this forms a topology. Now make the assignments \begin{align*} \left\{{a}\right\} & \mapsto A \\ \left\{{b}\right\} & \mapsto B \\ \left\{{a, b}\right\} & \mapsto C \\ X & \mapsto ? .\end{align*}

We have a situation like the following:

Unique gluing says that given \(r\in B, s\in C\) such that \(\phi_B(r) = \phi_C(s)\), there should exist a unique \(t\in \mathcal{F}(X)\) such that \({ \left.{{t}} \right|_{{\left\{{a, b}\right\}}} } = r\) and \({ \left.{{t}} \right|_{{\left\{{a, c}\right\}}} } = s\). This recovers exactly the fiber product. \begin{align*} B \times_A C \mathrel{\vcenter{:}}=\left\{{(r, s) \in B\times C {~\mathrel{\Big|}~}\phi_B(r) = \phi_C(s) \in A}\right\} .\end{align*}

Let \(X\) be an affine variety with the Zariski topology and let \(\mathcal{F} \mathrel{\vcenter{:}}={\mathcal{O}}_X\) be the sheaf of regular functions: \begin{align*} {\mathcal{O}}_X(U) \mathrel{\vcenter{:}}=\left\{{f: U\to k {~\mathrel{\Big|}~}\forall p\in U,\, \exists U_p \ni p,\,\, { \left.{{f}} \right|_{{U_p}} } ={g_p \over h_p} }\right\} .\end{align*}

Is this a presheaf? We can check that there are restriction maps: \begin{align*} {\mathcal{O}}_X(U) &\to {\mathcal{O}}_X(V) \\ \left\{{f: U\to K}\right\} &\mapsto \left\{{{ \left.{{f}} \right|_{{V}} }(x) \mathrel{\vcenter{:}}= f(x) \text{ for } x \in V }\right\} .\end{align*} This makes sense because if \(V\subset U\), any \(x\in V\) is in the domain of \(f\). Given that \(f\) is locally a fraction, say \(\rho = g_p / h_p\) on \(U_p \ni p\), is \({ \left.{{\phi}} \right|_{{V}} }\) locally a fraction? Yes: for all \(p\in V\subset U\), \(\phi = g_p / f_p\) on \(U_p\) and this remains true on \(U_p \cap V\).

To check that \({\mathcal{O}}_X\) is a sheaf, given a set of regular functions \(\left\{{\phi_i: U_i \to k}\right\}\) agreeing on intersections, define \begin{align*} \phi: \cup U_i &\to k\\ \phi(x) &\mathrel{\vcenter{:}}=\phi_i(x) \text{ if }x\in U_i .\end{align*}

This is well-defined, since if \(x\in U_i \cap U_j\), \(\phi_i(x) = \phi_j(x)\) since both restrict to the same function on \(U_i \cap U_j\) by assumption.

Why is \(\phi\) locally a fraction? We need to check that for all \(p\in U \mathrel{\vcenter{:}}=\cup U_i\) there exists a \(U_p \ni p\) with \({ \left.{{\phi}} \right|_{{U_p}} } = g_p/h_p\). But any \(p\in \cup U_i\) implies \(p\in U_i\) for some \(i\). Then there exists an open set \(U_{i, p} \ni p\) in \(U_i\) such that \({ \left.{{\phi}} \right|_{{U_{i, p}}} } = g_p / h_p\) by definition of a regular function. So take \(U_p = U_{i, p}\) and use the fact that \({ \left.{{\phi}} \right|_{{U_i}} } = \phi_i\) along with compatibility of restriction.

What is the stalk of \(\operatorname{Hol}({\mathbb{C}})\) at \(p=0\)?

Examples of equivalent elements in this stalk:

In this case \begin{align*} \operatorname{Hol}({\mathbb{C}})_0 = \left\{{\phi = \sum_{i>0}c_i z^i {~\mathrel{\Big|}~}\phi \text{ has a positive radius of convergence}}\right\} .\end{align*}

If \(X\) is an affine variety with the Zariski topology, \(\mathcal{O}_X\) is a sheaf of regular functions, where we recall \(\mathcal{O}_X(U)\) are the functions \(\phi: U\to k\) that are locally a fraction.

Let \(X = \left\{{p, q}\right\}\) with the discrete topology with the sheaf \(\mathcal{F}\) given by \begin{align*} p &\mapsto R \\ q &\mapsto S \\ X &\mapsto R\times S .\end{align*} Then \(\mathcal{F}_p = R\), since if \(U\) is open and \(p\in\ U\) then either \(U= \left\{{p}\right\}\) or \(U = X\). We can check that for \((r, s)\) a section of \(\mathcal{F}\), we have an equivalence of germs \begin{align*} (X, (r, s)) \sim (\left\{{p}\right\}, r) \text{ since } \left\{{p}\right\} \subset X\cap\left\{{p}\right\} \end{align*} Here \(X\) plays the role of \(U\), \(\left\{{p}\right\}\) of \(U'\), and the last \(\left\{{p}\right\}\) the role of \(W \subset U\cap U'\).

\begin{align*} {\mathcal{O}}_{X, p} &\to A(X) \\ (\left\{{p}\right\}, r) &\mapsto r \\ \mathcal{F}_p &\cong R .\end{align*}

Let \(M\) be a manifold and consider the sheaf \(C^\infty\) of smooth functions on \(M\). Then the stalk \(C_p^\infty\) at \(p\) is defined as the set of smooth functions in a neighborhood of \(p\) modulo functions being equivalent if they agree on a small enough ball \(B_\varepsilon(p)\). This contains a maximal ideal \({\mathfrak{m}}_p\), the smooth functions vanishing at \(p\). Then \({\mathfrak{m}}_p^2\) is again an ideal, and as sets, \begin{align*} {\mathfrak{m}}^2 = \left\{{f {~\mathrel{\Big|}~}{\partial}_i {\partial}_j f\mathrel{\Big|}_p = 0,\, \forall i,j}\right\} .\end{align*} Thus \({\mathfrak{m}}_p/{\mathfrak{m}}_p^2 \cong \left\{{{\partial}_v}\right\}^\vee\), the dual of the set of directional derivatives.

For smooth manifolds, we similarly require that there is a pullback that preserves smooth functions: \begin{align*} f^*: C^\infty(U) \to C^\infty(f^{-1}(U)) .\end{align*}

1. \(X\) an affine variety and \({\mathcal{O}}_X\) its ring of regular functions.

2. \(X\) a manifold over \({\mathbb{R}}^n\) with \({\mathcal{O}}_X\) a ring of smooth or continuous functions on \(X\).

3. \(X = \left\{{p, q}\right\}\) with the discrete topology and \({\mathcal{O}}_X\) given by \(p\mapsto R, q\mapsto S\).

4. Let \(U\subset X\) an open subset of \(X\) an affine variety. Then declare \({\mathcal{O}}_U\) to be \({ \left.{{{\mathcal{O}}_X}} \right|_{{U}} }\).

Let \(X\) be a topological space and \(p\in X\) a point. The skyscraper sheaf at \(p\) is defined by \begin{align*} k_p(U) \mathrel{\vcenter{:}}= \begin{cases} k & p\in U \\ 0 & p\not\in U \end{cases} .\end{align*}

If \((X, {\mathcal{O}}_X)\) is a ringed space associated to an affine variety, then we assume \({\mathcal{O}}_X(U)\) are literally functions on \(U\). Morphisms of open subsets is again defined by morphisms of ringed spaces.

Let \(X = {\mathbb{A}}^1/k\) and \(U \mathrel{\vcenter{:}}= D(x)\), then \(D(f) = {\mathbb{A}}^1\setminus\left\{{0}\right\}\). Then \(\iota: U\hookrightarrow X\) is continuous. Given an arbitrary distinguished open set \(D(f) \subset {\mathbb{A}}^1\), we know from previous results that \begin{align*} {\mathcal{O}}_X(D(f)) \mathrel{\vcenter{:}}={\mathcal{O}}_{{\mathbb{A}}^1}(D(f)) = A({\mathbb{A}}^1)_{\left\langle{f}\right\rangle} = k[x]_{\left\langle{f}\right\rangle} \mathrel{\vcenter{:}}=\left\{{g/f^n {~\mathrel{\Big|}~}g\in k[x]}\right\} .\end{align*} We want to show that \(\iota: (U, {\mathcal{O}}_U) \hookrightarrow(X, {\mathcal{O}}_X)\) is a morphism of ringed spaces where \({\mathcal{O}}_U(V) \mathrel{\vcenter{:}}={\mathcal{O}}_X(V)\). Does \(\iota^*\) pull back regular functions to regular functions? Yes, since \begin{align*} \iota^{-1} (D(f)) = D(f) \cup D(x) = D(xf) \end{align*} and thus \begin{align*} g/f^n \in {\mathcal{O}}_U(\iota^{-1}(D(f))) = {\mathcal{O}}_U(D(xf)) \end{align*} where we’ve used that \(f^n \neq 0 \implies xf\neq 0\).

A non-example: take \begin{align*} h: {\mathbb{A}}^1 &\to {\mathbb{A}}^1 \\ x & \mapsto \begin{cases} x & x \neq \pm 1 \\ -x & x= \pm 1 \end{cases} .\end{align*} This is continuous because the Zariski topology on \({\mathbb{A}}^1\) is the cofinite topology (since the closed sets are finite), so any injective map is continuous since inverse images of cofinite sets are again cofinite.

Consider \begin{align*} \iota: ({\mathbb{R}}^2, C^\infty) \hookrightarrow({\mathbb{R}}^3, C^\infty) \end{align*} is the inclusion of a coordinate hyperplane. To say that this is a morphism of ringed spaces, we need that for all \(U\subset {\mathbb{R}}^3\) open and \(f:U\to {\mathbb{R}}\) a smooth function, we want \(i^* f\in C^\infty (\iota^{-1}(U))\). But this is the same as \(f\circ \iota \in C^\infty({\mathbb{R}}^2\cap U)\), which is true.

Let \(Y \mathrel{\vcenter{:}}= V(xy-1) \subset{\mathbb{A}}^2\) and \(U\mathrel{\vcenter{:}}= D(x) = {\mathbb{A}}^1\setminus\left\{{0}\right\}\subset {\mathbb{A}}^1\). Note that \begin{align*} A(Y) &= {k[x, y] \over \left\langle{xy-1}\right\rangle} \\ A({\mathbb{A}}^1) &= k[t] ,\end{align*} and if \(f_1 \mathrel{\vcenter{:}}= t, f_2\mathrel{\vcenter{:}}= t^{-1}\), then \(f_1, f_2 \in {\mathcal{O}}_U(U)\). So we can define a map \begin{align*} {\left[ {f_1, f_2} \right]}: U &\to Y \\ p &\mapsto {\left[ {p, {1\over p} } \right]} \end{align*} whose image lies in \(Y\). Conversely, there is a map \begin{align*} V(xy - 1) &\to U = D(0) \subset {\mathbb{A}}^1 \\ {\left[ {x, y} \right]} &\mapsto x .\end{align*} This a morphism from \(V(xy - 1)\) to \({\mathbb{A}}^1\), since the coordinates are regular functions. Since the image is contained in \(U\), the definitions imply that this is in fact a morphism of ringed spaces. We thus have mutually inverse maps \begin{align*} U &\mathrel{\operatorname*{\rightleftharpoons}_{t\mapsto {\left[ {t, t^{-1} } \right]}}^{x\mapsfrom {\left[ {x, y} \right]}}} V(xy-1) \\ ,\end{align*} so \(U\cong V(xy-1)\) as ringed spaces.

\({\mathbb{A}}^2\) vs \(V(x) \subset {\mathbb{A}}^n\). In fact, the map \begin{align*} f: {\mathbb{A}}^2 &\to {\mathbb{A}}^3 (y,z) &\mapsto (0, y, z) .\end{align*} This is continuous and the pullback of regular functions are again regular.

\({\mathbb{A}}^2 \setminus\left\{{0}\right\}\) is not an affine variety. Note that this is also not a distinguished open. We showed on a HW problem that the regular functions on \({\mathbb{A}}^2\setminus\left\{{0}\right\}\) are \(k[x, y]\), which are also the regular functions on \({\mathbb{A}}^2\). So there is a map inducing a pullback \begin{align*} \iota: {\mathbb{A}}^2\setminus\left\{{0}\right\}&\to {\mathbb{A}}^2 \\ \\ \iota^*: k[x, y] &\xrightarrow{\sim} k[x, y] .\end{align*} Note that \(\iota^*\) is an isomorphism on the space of regular functions, but \(\iota\) itself is not an isomorphism of topological spaces. Why? \(i^{-1}\) is not defined at zero.

The prototypical example is \({\mathbb{P}}^1_{/k}\) constructed from two copies of \({\mathbb{A}}^1_{/k}\). Set \(X_1 = {\mathbb{A}}^1, X_2 = {\mathbb{A}}^2\), with \(U_{12} \mathrel{\vcenter{:}}= D(x) \subset X_1\) and \(U_{21} \mathrel{\vcenter{:}}= D(y) \subset X_2\). Then let \begin{align*} f: U_{12} &\to U_{21} \\ x & \mapsto {1\over x} .\end{align*} This defines a regular function on \(U_{12}\) so defines a morphism \(U_{12} \xrightarrow{\sim} {\mathbb{A}}^1\).

Over \({\mathbb{C}}\), topologically this yields a sphere

Given a ringed space \(X = X_1\cup X_2\) with a structure sheaf \({\mathcal{O}}_X\), what is \({\mathcal{O}}_X(X)\)? By definition, it’s \begin{align*} {\mathcal{O}}_X(X) \mathrel{\vcenter{:}}=\left\{{\phi: X\to k {~\mathrel{\Big|}~}{ \left.{{\phi}} \right|_{{X_1}} }, { \left.{{\phi}} \right|_{{X_2}} } \text{ are regular} }\right\} .\end{align*}

Then if \({ \left.{{\phi}} \right|_{{X_1}} } = f(x)\) and \({ \left.{{\phi}} \right|_{{X_2}} } = g(y)\), we have \(y=1/x\) on the overlap and so \begin{align*} { \left.{{f(x)}} \right|_{{D(x)}} } = { \left.{{g(1/x)}} \right|_{{D(x)}} } \end{align*} Since \(f, g\) are rational functions agreeing on an infinite set, \(f(x) = g(1/x)\) both being polynomial forces \(f = g = c\) for some constant \(c \in k\). Thus \({\mathcal{O}}_X(X) = k\).

What about \({\mathcal{O}}_X(X_1)\)? This is just \(k[x]\), and similarly \({\mathcal{O}}_X(X_2) = k[y]\). We can also consider \({\mathcal{O}}_X(X_1\cap X_2) = D(x) \subset X\), so this yields \(k[x, 1/x]\). We thus have a diagram

\({\mathbb{P}}^1_{/k} = X_1 \cup X_2\) where \(X_1 \cong X_2 \cong {\mathbb{A}}^1_{/k}\). Take \(U_{12} \mathrel{\vcenter{:}}= D(x)\) and \(U_{21} \mathrel{\vcenter{:}}= D(y)\) with \begin{align*} f: U_{12} &\to U_{21} \\ x &\mapsto y \mathrel{\vcenter{:}}={1\over x} .\end{align*}

Let \(X_i = {\mathbb{A}}^1\) and \(U_{12} = D(x), U_{21} = D(y)\) with \begin{align*} f: U_{12} &\to U_{21} \\ x &\mapsto y\mathrel{\vcenter{:}}= x .\end{align*}

Define \(X\mathrel{\vcenter{:}}= X_1 {\coprod}_f X_2\), then \({\mathcal{O}}_X = \left\{{\phi: X\to k {~\mathrel{\Big|}~}{ \left.{{\phi}} \right|_{{X_i}} } \text{ is regular}}\right\} \cong k[x]\).

Let \(X_1 = X_2 = X_3 \mathrel{\vcenter{:}}={\mathbb{A}}^2/k\). Glue by the following instructions:

where \(f\) is a map with whatever formula is required to make the diagram commute. Here

• \(D(x_1 x_2) \subseteq X_1\)
• \(U_{12} = D(x_1)\)
• \(U_{21} = D(x_2)\).

Here \(X_1 = [1: y/x: z/x]\), \(X_2 = [x/y: 1: z/y]\).

From Gathmann 5.10, open and closed subprevarieties. Let \(X\) be a prevariety and suppose \(U\subset X\) is open. Then \((U, {\mathcal{O}}_U)\) is a prevariety where \({\mathcal{O}}_U = { \left.{{{\mathcal{O}}_X}} \right|_{{U}} }\). How can we write \(U\) as (locally) an affine variety? Since the \(U_i\) are covered by distinguished opens \(D_{ij}\) in \(X_i\) where \(X = \cup X_i\) with \(X_i\) affine varieties, we can write \(U = \bigcup_i U_i = \bigcup_{i, j} D_{ij}\).

Let \(Y\subset X\) be a closed subset of a prevariety \(X\). We need to define \({\mathcal{O}}_Y(U)\) for all \(U\subset Y\) open, so we set \begin{align*} {\mathcal{O}}_Y(U) = \left\{{\phi: U\to k {~\mathrel{\Big|}~}\forall p\in U, \, \exists V_p \text{ with } p\in V_p \mathrel{\stackunder[2pt]{\stackon[4pt]{$\subseteq$}{$\scriptscriptstyle$}}{ $\scriptscriptstyle\text{open}$}} X \text{ and } \psi\in {\mathcal{O}}_X(V_p) \text{ s.t. } { \left.{{\psi}} \right|_{{U\cap V}} } = \phi }\right\} .\end{align*}

What’s the picture?

Consider \({\mathbb{A}}^1\), whose polynomial functions are \(k[x]\). Consider now \(D(x) \subset {\mathbb{A}}^1\), which is isomorphic to the affine variety \(V(xy-1)\). Then the regular functions on \(D(x)\) are given by \begin{align*} A(D(x)) = \frac{ k[x, y] }{ \left\langle{xy-1}\right\rangle } \cong k[x, x^{-1} ] .\end{align*}

Consider \({\mathbb{A}}^1 \times {\mathbb{A}}^1\), then the product is (and should be) \({\mathbb{A}}^2\), but \({\mathbb{A}}^2\) does not have the product topology. For example, one problem is that the Zariski open set \(D(x-y)\) is not covered by products of open sets in \({\mathbb{A}}^1\).

Glue \(D(x) \subset {\mathbb{A}}^1\) to \(D(y) \subset {\mathbb{A}}^1\) by the isomorphism \begin{align*} D(x) & \xrightarrow{\sim} D(y) \\ x &\mapsto y .\end{align*} This yields an affine line with two origins:

Consider the product:

Since the diagonal is given by \(\Delta_X = \left\{{(x, x) {~\mathrel{\Big|}~}x\in X}\right\}\), we have the following situation in blue:

We claim that \(\Delta_X\) is not closed, and for example \((0, 0') \in \mkern 1.5mu\overline{\mkern-1.5mu\Delta\mkern-1.5mu}\mkern 1.5mu_X\). Consider \(U\times U' \subset X\times X\) where \(U, U'\) are the two copies of \({\mathbb{A}}^1\) in \(X\). This is an affine open set, since it’s isomorphic to \({\mathbb{A}}^1\times{\mathbb{A}}^1\). If \(\Delta_X\) were closed, then \(S \mathrel{\vcenter{:}}=\Delta_X \cap(U\times U') = \left\{{(x, x) {~\mathrel{\Big|}~}x\neq 0}\right\}\) would be closed in \(U\times U'\):

This is because any polynomial vanishing on \(S\) must vanish at \((0, 0)\), so \(S\) is an affine variety. But then \(V(I(S)) = \Delta_{{\mathbb{A}}^1}\).

Let \(X = Y={\mathbb{A}}^1\) and set \(Z \mathrel{\vcenter{:}}= V(xy-1)\subset X\times Y\). Then \(\pi_Y(Z) = D(y) \subset Y \subset {\mathbb{A}}^1\) is not closed.

Note that \({\mathbb{P}}^1\) contains a copy of \({\mathbb{A}}^1\) where \(x_0 \neq 0\) and a second copy where \(x_1 \neq 0\), yielding maps \begin{align*} f_1: {\mathbb{P}}^1 &\to {\mathbb{A}}^1 \\ {\left[ {x_0: x_1} \right]} &\mapsto {\left[ {{x_0 \over x_1}} \right]} \\ \\ f_2: {\mathbb{P}}^1 &\to {\mathbb{A}}^1 \\ {\left[ {x_0: x_1} \right]} &\mapsto {\left[ {{x_1 \over x_0}} \right]} ,\end{align*} since every point in \({\mathbb{P}}^1\) corresponds to some line in \({\mathbb{A}}^2\), and thus has either \(x_0\neq 0\) or \(x_1 \neq 0\). These two copies cover \({\mathbb{P}}^1\), and the “transition map” is inversion.

Consider \({\mathbb{P}}^1\), then \(D(x_0) \cong {\mathbb{A}}^1\) with which contains a copy of \({\mathbb{A}}^1\) with coordinate ring \(k[x_1 / x_0]\) and a subset \(D(x_1 / x_0)\) with coordinate ring \(k[y, y^{-1}]\), and similarly, \(D(x_1) \cong {\mathbb{A}}^1\) has coordinate ring \(k[x_0\over x_1]\) and contains \(D(x_0/ x_1)\) with coordinate ring \(k[z, z^{-1}]\). Consider their overlap \(D(x_0) \cap D(x_1)\).

When do \(y, z\) denote the same point in \({\mathbb{P}}^1\)? When \(y = z^{-1}\). We can conclude that the \(n+1\) copies \(D(x_i) \subset {\mathbb{P}}^n\) are affine varieties isomorphic as ringed spaces on the overlaps, so the gluing construction makes \({\mathbb{P}}^n\) a prevariety.

The polynomial \begin{align*} f(x_0, x_1, x_2) = x_0^3 + x_1 x_2^2 + x_0 x_1 x_2 .\end{align*} has homogeneous degree 3.

The dehomogenization is usually not homogeneous. Take \begin{align*} f &\mathrel{\vcenter{:}}= x_0^3 + x_0 x_1^2 + x_0 x_1 x_2 + x_0^2 + x_1 \\ \implies f^i &= 1 +x_1^2 + x_1 x_2 + x_1 ,\end{align*} which has terms of mixed degrees.

Set \begin{align*} f(x_1, x_2) &\mathrel{\vcenter{:}}= 1 + x_1^2 + x_1 x_2 + x_2^3 \\ \implies f^h(x_0, x_1, x_2) &= x_0^3 + x_0 x_1^2 + x_0 x_1 x_2 + x_2^3 ,\end{align*} which is a homogeneous polynomial of degree \(3\). Note that \((f^h)^i = f\).

It need not be the case that \((f^i)^h = f\). Take \(f = x_0^3 + x_0 x_1 x_2\), then \(f^i = 1 + x_1 x_2\) and \((f^i)^h = x_0^2 + x_1 x_2\). Note that the total degree dropped, since everything was divisible by \(x_0\).

This is not a ring morphism, since \((f+g)^h \neq f^h + g^h\) in general. Taking \(f = x_0^2 + x_1\) and \(g= -x_0^2 + x_2\), we have \begin{align*} f^h + g^h &= x_0 x_1 + x_0 x_2 \\ (f+g)^h &= x_1 + x_2 .\end{align*}

Consider \(f(x_1, x_2) = x_1^2 - x_2^2 - 1\) and consider \(V(f) \subset {\mathbb{A}}^2_{/{\mathbb{C}}}\):

Note that for real projective space, we can view this as a sphere with antipodal points identified. We can thus visualize this in the following way:

We can normalize the \(x_0\) coordinate to one, hence the plane. We can also project \(V(f)\) from the plane onto the sphere, mirroring to antipodal points:

This misses some points on the equator, since we aren’t including points where \(x_0 = 0\). Consider the homogenization \(V(f^h) \subset {\mathbb{P}}^2_{/{\mathbb{C}}}\): \begin{align*} V(f^h) = V(x_1^2 - x_2^2 - x_0^2) .\end{align*} Then \begin{align*} V(f^h) \cap V(x_0) = \left\{{{\left[ {0:x_1:x_2} \right]} {~\mathrel{\Big|}~}f^h(0, x_1, x_2) = 0 }\right\} = \left\{{{\left[ {0:1:1} \right]}, {\left[ {0:1:-1} \right]}}\right\} ,\end{align*} which can be seen in the picture as the points at infinity:

Note that the equator is \(V(x_0) = {\mathbb{P}}^2_{/{\mathbb{C}}}\setminus U_0 \cong {\mathbb{P}}^2\setminus{\mathbb{A}}^2\). So we get a circle of points at infinity, i.e. \(V(x_0) = {\mathbb{P}}^1 = \left\{{{\left[ {0:v_1:v_2} \right]}}\right\}\).

Consider \(V(f)\) where \(f\) is a line in \({\mathbb{A}}^2_{/{\mathbb{C}}}\), say \(f(x_1, x_2) = ax_1 + bx_2 + c\). This yields \(f^h = ax_1 + bx_2 + cx_0\) and we can consider \(V(f^h) \cong {\mathbb{P}}^2_{{\mathbb{C}}}\). We know \({\mathbb{P}}^1_{{\mathbb{C}}}\) is topologically a sphere and \({\mathbb{A}}^1_{/{\mathbb{C}}}\) is a point:

The points at infinity correspond to \begin{align*} V(f^h) = V(f^h) \cap V(x_0) = \left\{{{\left[ {0:-b:a} \right]}}\right\} ,\end{align*} which is a single point not depending on \(c\).

Consider \(X = V(x_0 x_2 - x_1^2) \subset {\mathbb{P}}^2\), which defines a conic, and the projection \({\mathbb{P}}^2 \setminus\left\{{[1:0:0]}\right\} \to {\mathbb{P}}^1\):

This morphism can be restricted to \(\phi: X\setminus\left\{{[1:0:0]}\right\} \to {\mathbb{P}}^2\), and the claim is that this morphism extends to all of \(X\). The secant lines approach a tangent line at \([1:0:0]\), which \(V(x_0)\) at a unique point. So we define \begin{align*} \mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu(x) \mathrel{\vcenter{:}}= \begin{cases} [x_1: x_2] & x \neq [1:0:0] \\ [x_0: x_1] & x \neq [0:0:1] \end{cases} .\end{align*}

This locally writes \(\phi\) as a morphism, so we only need to check that they agree on the overlap. Note that on \(X\), we have \([x_1: x_2] = [x_0 : x_1]\) wherever both are well-defined. In fact, \(\mkern 1.5mu\overline{\mkern-1.5mu\phi\mkern-1.5mu}\mkern 1.5mu\) is an isomorphism, since an inverse can be explicitly written. Thus \(X\cong {\mathbb{P}}^1\), and in fact all nondegenerate¹³ conics are isomorphic to \({\mathbb{P}}^1\) as well. Note that such a \(Q\) is a quadratic form, so \(Q(x) = B(x, x)\) for some bilinear form, and \(Q\) is nondegenerate iff \(\det B \neq 0\) where \(B_{ij} = B(e_i, e_j)\).

Take

\begin{align*} f: {\mathbb{P}}^1 \times{\mathbb{P}}^1 &\to {\mathbb{P}}^3 \\ \qty{ {\left[ {x_0: x_1} \right]}, {\left[ {y_0: y_1} \right]}} &\mapsto {\left[ {z_{00} : z_{01} : z_{10} : z_{11} } \right]} \mathrel{\vcenter{:}}={\left[ {x_0 y_0: x_0 y_1 : x_1 y_0 : x_1 y_1} \right]} .\end{align*} Restricting to \(\operatorname{im}(f)\) yields an isomorphism to \(X \subseteq {\mathbb{P}}^3\) given by the quadric¹⁴ \(X = V(z_{00} z_{11} - z_{10} z_{01} )\).

For \({\mathbb{P}}^1 \times{\mathbb{P}}^1 \to {\mathbb{P}}^3\), its image is \(X = V_p(xy - zw)\), which is a quadric (vanishing locus of a degree 4 polynomial).

The projection map has fibers, which induce a ruling¹⁵ which we can see from the real points:

Let \(n=1\), then we get the embedding \begin{align*} {\mathbb{P}}^1 &\hookrightarrow{\mathbb{P}}^d \\ [x_0: x_1] &\mapsto [x_0^d: x_0^{d-1}x_1 : \cdots : x_0 x_1^{d-1} : x_1^d] .\end{align*} Note that there are \(d+1\) such monomials, and not all can simultaneously vanish. The image of this \({\mathbb{P}}^1\) is called the twisted normal curve.

Take \begin{align*} {\mathbb{P}}^1 &\hookrightarrow{\mathbb{P}}^2 \\ [x_0 : x_1] &\mapsto [x_0^2 : x_0 x_1: x_1^2] .\end{align*} What homogeneous polynomials cut out \(\phi({\mathbb{P}}^1)\)? I.e., what is \(I(\phi({\mathbb{P}}^1)) \subset S({\mathbb{P}}^2)\)? Note that \(w_0 w_2 - w_1^2 { \left.{{}} \right|_{{\phi({\mathbb{P}}^1)}} }\), so this is an element. Is it a generator? I.e., given any \(p\in V(w_0 w_2 - w_1^2)\) is of the form \(p = [x_0^2 : x_0 x_1: x_1^2]\) for some \(x_), x_1 \in k\)? The answer is yes, by choosing signs of \(\sqrt{w_0}, \sqrt{w_2}\).

Take \begin{align*} \phi: {\mathbb{P}}^1 &\hookrightarrow{\mathbb{P}}^3 \\ [x_0: x_1] &\mapsto [x_0^3: x_0^2 x_1 : x_0 x_1^2: x_1^3] .\end{align*} What are some elements of this ideal?

• \(w_0 w_3 - w_1 w_2\)
• \(w_0 w_2 - w_1^2\)
• \(w_1 w_3 - w_2^2\)

Note that the first is not a \(k{\hbox{-}}\)linear combination of the other two. There is also a pattern: \(w_0/w_1 = w_1 / w_2 = w_2/w_3 = \cdots\). However, there will be issues when the denominators are zero.

In this case, \(\phi({\mathbb{P}}^1)\) is the twisted cubic. What is \(V(w_0 w_2 - w_1^2, w_1 w_3 - w_2^2) \setminus\phi({\mathbb{P}}^1)\)? Note that being in \(\phi({\mathbb{P}}^1)\) means \(w_1, w_2, w_3 \neq 0\), and similarly if \(w_0, w_1, w_2 \neq 0\). We can conclude that \(V(w_1, w_2) \subset V(w_0 w_2 - w_1^2, w_1 w_3 - w_2^2)\):

This variety has two components: the twisted cubic, and a line. This variety has degree 4, since any generic hyperplane intersects it at 4 points. Why? Pulling back a hyperplane yields a cubic, which generally vanishes at three points in affine space.

Consider \(T_{(1, 1)} V(xy-1)\). First translate \((1, 1)\) to the origin, so \begin{align*} T_{(1, 1)} V(xy-1) = T_{(0, 0)} = V((x-1)(y-1) - 1) = T_{(0, 0)} V(xy-x-y) = V(-x-y) \end{align*} On the other hand, \(T_{(0, 0)} V(xy) = V(0) = {\mathbb{C}}^2\).

Consider \begin{align*} T_0 V(xy) = V(f_1 {~\mathrel{\Big|}~}f\in \left\langle{xy}\right\rangle) = V(0) = {\mathbb{A}}^2 ,\end{align*} since every polynomial in this ideal has degree at least 2. Letting \(X = V(xy)\), note that we could embed \(X\hookrightarrow{\mathbb{A}}^3\) as \(X\cong V(xy, z)\). In this case we have \begin{align*} T_0 X = V(f_1 {~\mathrel{\Big|}~}f\in \left\langle{xy, z}\right\rangle) = V(z) \cong {\mathbb{A}}^2 \end{align*} So we get a vector space of a different dimension from this different affine embedding, but \(\dim T_0 X\) is the same.

Let \(X = V_p(xy-z^2) \subset {\mathbb{P}}^2\), which is a projective curve. What is \(T_p X\) for \(p = [0:1:0]\)? Take an affine chart \(\left\{{y\neq 0}\right\} \cap X\), noting that \(\left\{{y\neq 0}\right\} \cong {\mathbb{A}}^2\). We could dehomogenize the ideal \({ \left.{{\left\langle{xy-z^2}\right\rangle}} \right|_{{y=1}} } = \left\langle{x-z^2}\right\rangle\). Thus \(X \cap D(y) = V(x-z^2) \subset {\mathbb{A}}^2\) and the point \([0:1:0] \in X\) gives \((0, 0)\) in this affine chart. Then \begin{align*} T_p X = V(f_1 {~\mathrel{\Big|}~}f\in \left\langle{x-z^2}\right\rangle) = V(x) \end{align*} Then \(f = (x-z^2)g\) implies that \(f_1 = (xg)_1 = g_0 x\), the constant term of \(g\) multiplied by \(x\), since \(z^2\) kills any degree 1 part of \(g\). So \(T_p X\) is a line.

Take \(X\) to be the union of the coordinate axes in \({\mathbb{A}}^3\):

Then \(I(X) = \left\langle{xy, yz, xz}\right\rangle\) and \begin{align*} T_0 X = V(f_1 {~\mathrel{\Big|}~}f\in I(X)) = V(0) = {\mathbb{A}}^3 \end{align*} since the minimal degree of any such polynomial is 2. Note that \(\dim X = 1\) but \(\dim T_0 X = 3\)

Take \(Y = V(xy(x-y)) \subset {\mathbb{A}}^2\):

Then \(T_0 X = V(0) = {\mathbb{A}}^2\).

First move \((1,0,0)\) to the origin, yielding \(T_{(0,0,0)} V((x+1)y, yz, (x+1) z)\). This is a different choice of affine embedding into \({\mathbb{A}}^3\) which sends \((1,0,0) \mapsto (0,0,0)\). Taking the vanishing locus of linear parts, it suffices to take the linear parts of the generators, which yields the \(x{\hbox{-}}\)axis \(V(y, z)\), making the dimension of the tangent space 1.

\({\mathbb{A}}^n\) is smooth since \(T_p {\mathbb{A}}^n = k^n\) for all points \(p\), which has dimension \(n\).

\({\mathbb{A}}^n {\coprod}{\mathbb{A}}^{n-1}\) is also smooth since each connected component is smooth.

Consider \(X\mathrel{\vcenter{:}}= V(y^2 - x^3) \subset {\mathbb{A}}^2\):

Note that \(\dim T_0 X = 2\) is easy to see since it’s equal to \(V\qty{f_1 {~\mathrel{\Big|}~}f\in \left\langle{y^2 - x^3}\right\rangle} = V(0) = k^2\). Thus \(p\neq 0\) are smooth points and \(p=0\) is the unique singular point. So \(X\) is not smooth, but \(X\setminus\left\{{0}\right\}\) is.

Is \(V(x^2 - y^2 - 1) \subset {\mathbb{A}}^2\) smooth? We have \(I(X) = \left\langle{f_1}\right\rangle \mathrel{\vcenter{:}}=\left\langle{x^2 - y^2 - 1}\right\rangle\), so let \(p\in X\). Then consider the matrix \begin{align*} \begin{bmatrix} J \mathrel{\vcenter{:}}= {\frac{\partial f}{\partial x}\,} & {\frac{\partial f}{\partial y}\,} \\ \end{bmatrix} = \begin{bmatrix} 2x & -2y \\ \end{bmatrix} .\end{align*} We want to show that at any \(p\in X\), we have \(\operatorname{rank}(J) = 1\). This is true for \(p\neq (0, 0)\), but this is not a point in \(X\).

Consider \(X \mathrel{\vcenter{:}}= V(y^2 - x^3 + x^2) \subset {\mathbb{A}}^2\):

Then \(I(X) = \left\langle{y^2 - x^3 + x^2}\right\rangle = \left\langle{f}\right\rangle\), and \begin{align*} J = \begin{bmatrix} 2y & -3x^2 + 2x \\ \end{bmatrix} .\end{align*} Then \(\operatorname{rank}(J) = 0\) at \(p = (0, 0)\), which is a point in \(X\), so \(X\) is not smooth.

Consider \(X\mathrel{\vcenter{:}}= V(x^2 + y^2, 1+z^3) \subset{\mathbb{A}}^3\), then \(I(X) = \left\langle{x^2 + y^2, 1 + z^3}\right\rangle \left\langle{f, g}\right\rangle\) which is clearly a radical ideal.

We then have \begin{align*} J = \begin{bmatrix} f_x & f_y & f_z \\ g_x & g_y & g_z \\ \end{bmatrix} = \begin{bmatrix} 2x & 2y & 0 \\ 0 & 0 & 3z^2 \end{bmatrix} ,\end{align*} and thus \begin{align*} \operatorname{rank}(J) = \begin{cases} 0 & x=y=z=0 \\ 1 & x=y=0 \text{ xor } z=0 \\ 2 & \text{else}. \end{cases} \end{align*}

We can check that \(\dim X = 1\) and \(\operatorname{codim}_{{\mathbb{A}}^3} X = 3-1 = 2\), so a point \((x,y,z) \in X\) is smooth iff \(\operatorname{rank}(J) = 2\). The singular locus is where \(x=y=0\) and \(z= \zeta_6\) is any generator of the 6th roots of unity, i.e. \(\zeta_6, \zeta_6^3, \zeta_6^5\), along with the point \(0\). Note that \(z=0\) is not a point on \(X\), since \(1+z^3\neq 0\) in this case.

Thus the singular locus is \(V(x^2 + y^2) = V((x+iy)(x-iy)) \cap V(1+z^3)\), which results in 3 singular points after intersecting:

Note that it doesn’t matter that \(V(1+z^3)\) was intersected here, as long as it’s anything that intersects the \(z{\hbox{-}}\)axis nontrivially we will still get something singular.