# Coordinate Rings as k-algebras

Let k be an algebraically closed field. Recall that a closed subset $V \subseteq \mathbb A^n_k$ is identified by its coordinate ring k[V], which is a finitely generated k-algebra since

$k[V] = k[X_1, \ldots, X_n] / I(V).$

Definition.

An affine k-variety is a finitely generated k-algebra A which is a reduced ring. Formally, we write V for the variety and $A = k[V]$ for the algebra instead.

We say V is irreducible if A is an integral domain. A morphism of affine k-varieties $f : V\to W$ is a homomorphism of the corresponding k-algebras $f^* : k[W] \to k[V].$

Thus, each closed set V gives an affine variety; isomorphic closed sets give isomorphic affine varieties. Conversely, we have:

Lemma.

Any affine k-variety A is the coordinate ring of some closed $V\subseteq \mathbb A^n_k$.

Proof

Indeed since A is finitely generated as a k-algebra, there is a surjective homomorphism (of k-algebras) $k[X_1, \ldots, X_n] \to A$; its kernel $\mathfrak a$ is a radical ideal because A is a reduced ring. Hence $A \cong k[V(\mathfrak a)]$. ♦

Note that taking the set:

$B \mapsto \mathrm{Hom}_{k\text{-alg}}(B, k),$

recovers the set of points for Bk[V]. This gives another application of the “Hom” construction.

Now, the reader might question the utility of this point of view, since we are just back to dealing with closed subsets $V\subseteq \mathbb A^n$. The advantage here is that now we can expand the class of objects of interest.

# Affine k-Schemes

Recall that to study multiplicity of intersection, we should really be looking at general ideals of k[V] and not just radical ones. Hence we define the following.

Definition.

An affine k-scheme is a finitely generated k-algebra A. Again we write V for the scheme and $A = k[V]$.

An affine k-scheme V is a k-variety if and only if $k[V]$ is reduced.

From our understanding of closed sets, we may now define the following for any affine kschemes V and W. Note that now we have the capacity to define more general constructions (see the last few rows).

 What we say What we actually mean P is a point on V. $\mathfrak m_P$ is a maximal ideal of k[V]; equivalently we have a k-algebra homomorphism $k[V] \to k$. $\phi : V \to W$ is a morphism of affine k-schemes. $\phi^* : k[W] \to k[V]$ is a homomorphism of k-algebras. W is a closed subvariety of V. $k[W] = k[V] /\mathfrak a$ for some radical ideal $\mathfrak a$ of $k[V]$. W is an irreducible closed subvariety of V. $k[W] = k[V]/\mathfrak p$ for some prime ideal $\mathfrak p$ of $k[V]$. $W = \cap W_i$ is a set-theoretic intersection in V. For $k[W_i] = k[V] / \mathfrak a_i$, we have $k[W] = k[V] / r(\sum_i \mathfrak a_i)$. $W = W_1 \cup W_2$ is a union in V. For $k[W_i] = k[V]/\mathfrak a_i$, we have $k[W] = k[V]/(\mathfrak a_1 \cap \mathfrak a_2)$. W is a closed subscheme of V. $k[W] = k[V]/\mathfrak a$ for some ideal $\mathfrak a$ of $k[V]$. $W = \cap W_i$ is a scheme intersection in V. For $k[W_i] = k[V]/\mathfrak a_i$, we have $k[W] = k[V] / (\sum_i \mathfrak a_i)$.

Exercise A

Let A and B be any rings. Prove the following.

• An ideal of $A\times B$ must be of the form $\mathfrak a \times \mathfrak b$, where $\mathfrak a$ (resp. $\mathfrak b$) is an ideal of A (resp. B).
• prime ideal of $A\times B$ must be of the form $\mathfrak p \times B$ or $A \times \mathfrak q$, where $\mathfrak p$ (resp. $\mathfrak q$) is a prime ideal of A (resp. B).
• maximal ideal of $A\times B$ must be of the form $\mathfrak m \times B$ or $A \times \mathfrak n$, where $\mathfrak m$ (resp. $\mathfrak n$) is a maximal ideal of A (resp. B).

Define the corresponding construction for disjoint union $V\amalg W$ in the above table.

# Tangent Spaces

The affine k-variety for a singleton point $*$ is just $k[*] = k$. A point on an affine k-scheme V is thus a morphism $* \to V$.

Now we take the affine k-scheme # such that $k[\#] := k[X]/(X^2)$. We will write this as $k[\#] = k[\epsilon]$ with $\epsilon^2 = 0$. Geometrically, $\epsilon$ corresponds to a “small perturbation” such that $\epsilon^2 = 0$. Note that # has exactly one point, so there is a unique map $* \to \#$; the corresponding homomorphism $k[\epsilon] \to k$ takes $\epsilon \mapsto 0$.

We imagine # as a point with an arrow sticking out. Thus $\# \to V$ takes the point to a point P on V and the arrow to a tangent vector at P.

Definition.

A tangent vector on the k-scheme V is a morphism of k-schemes $\phi : \# \to V$.

The base point of the tangent vector $\phi$ is given by composing $* \to \# \stackrel{\phi}\to V$. The tangent space of a point $P : * \to V$ is the set of all tangent vectors with base point P, denoted by $T_P V$.

Note

In terms of k-algebras, a tangent vector is a k-algebra homomorphism

$\phi^* : k[V] \to k[\epsilon] = k[X]/(X^2)$.

If $\phi^*$ has base point P, then $\mathfrak m_P \mapsto 0$ in the composition $k[V] \stackrel {\phi^*}\to k[\epsilon] \to k$. This means $\phi^*(\mathfrak m_P) \subseteq k\cdot \epsilon$ and so $\phi^*(\mathfrak m_P^2) = 0$. Thus $\phi^*$ factors through the following:

$\phi^* : k[V] \longrightarrow k[V]/\mathfrak m_P^2 \stackrel f\longrightarrow k[\epsilon].$

So it suffices to consider all k-algebra homomorphisms $f : k[V]/\mathfrak m_P^2 \to k[\epsilon]$ which satisfies $f(\mathfrak m_P/\mathfrak m_P^2) \subseteq k\cdot \epsilon$. This gives a k-linear map $\mathfrak m_P / \mathfrak m_P^2 \to k$, i.e. the dual space of $\mathfrak m_P / \mathfrak m_P^2$ as a k-vector space.

Conversely, a k-linear map $g:\mathfrak m_P / \mathfrak m_P^2 \to k$ also gives a k-algebra homomorphism $k[V]/\mathfrak m_P^2 \to k[\epsilon]$ by mapping k to k. Hence we have shown:

Proposition.

$T_P V$ is parametrized by the set of all k-linear maps $\mathfrak m_P / \mathfrak m_P^2 \to k$. In particular, it forms a finite-dimensional vector space over k.

Exercise B

Explain why $\mathfrak m_P / \mathfrak m_P^2$ is finite-dimensional over k.

## Example 1

Consider a simple example: $V = \mathbb A^2$ with $k[V] = k[X, Y]$. A point $P = (\alpha, \beta)\in V$ corresponds to $\mathfrak m_P = (X - \alpha, Y - \beta) \subset k[X, Y]$. A tangent vector at P then corresponds to a k-linear map

$g : \mathfrak m_P / \mathfrak m_P^2 \longrightarrow k.$

But $\mathfrak m_P^2$ is generated by $(X-\alpha)^2, (X-\alpha)(Y-\beta), (Y-\beta)^2$ so g is uniquely determined by $c := g(X-\alpha) \in k$ and $d := g(Y-\beta) \in k$. Clearly $\dim_k \mathfrak m_P / \mathfrak m_P^2 = 2$.

To compute the resulting $\phi^* : k[X, Y] \to k[\epsilon]$ we take, for each $f(X, Y) \in k[X, Y]$,

$f(X, Y) = f(\alpha, \beta) + \overbrace{\left.\frac{\partial f}{\partial X}\right|_P}^{\in k}\cdot (X - \alpha) + \overbrace{ \left.\frac{\partial f}{\partial Y}\right|_P} ^{\in k}\cdot (Y - \beta) + \ldots$

ignoring the terms of degree 2 or more in $(X-\alpha), (Y-\beta)$. Hence

$\phi^*(f) = f(\alpha, \beta) + \left.\frac{\partial f}{\partial X}\right|_P\cdot c + \left.\frac{\partial f}{\partial Y}\right|_P \cdot d.$

## Example 2

Suppose $V \subset \mathbb A^2$ is cut out by $Y^2 = X^3 - X$, so $k[V] = k[X, Y]/(Y^2 - X^3 + X)$. Take the point P = (0, 0), which corresponds to $\mathfrak m_P = (X, Y)$. Let us compute the space of tangent vectors at P.

[Edited from GeoGebra plot.]

As before we have $\mathfrak m_P^2 = (X^2, XY, Y^2) \subset k[V]$. Instead of looking at $\mathfrak m_P$ and $\mathfrak m_P^2$, we look at their preimages $\mathfrak n_P = (X, Y), \mathfrak n_P' = (X^2, XY, Y^2, Y^2 - X^3 + X)$ as ideals of $k[X, Y]$. (Remember the correspondence of ideals between a ring and its quotient!) Thus

$\mathfrak n_P' = (X^2, XY, Y^2, Y^2 - X^3 + X) = (X, Y^2) \implies \dim_k \mathfrak n_P / \mathfrak n_P' = 1$

since Y is a basis for this space. Hence $\dim_k T_P V = 1$.

## Example 3

Now take $V\subset \mathbb A^2$ cut out by $Y^2 = X^3 + X^2$ so $k[V] = k[X, Y]/(Y^2 - X^2 - X^3)$. Take P = (0, 0) again.

[Edited from GeoGebra plot.]

As before $\mathfrak m_P = (X, Y)$ and $\mathfrak m_P^2 = (X^2, XY, Y^2)$. Again, we take their preimages $\mathfrak n_P = (X, Y)$ and $\mathfrak n_P' = (X^2, XY, Y^2, Y^2 - X^2 - X^3)$ as ideals of $k[X, Y]$. This time we get $\mathfrak n_P' = (X^2, XY, Y^2)$ and so

$\dim T_P V = \dim_k \mathfrak m_P /\mathfrak m_P^2 = \dim_k \mathfrak n_P / \mathfrak n_P' = 2$.

Summary.

Geometrically, the tangent space at P is larger than the dimension when there is a singularity at P, as the reader can see from the graphs above. Thus this gives an algebraic definition of singularity at various points. We will of course need an algebraic definition of dimension for this.

Exercise C

In each of the following varieties, compute the dimension of the tangent space at the origin. Intuitively, which varieties are singular at the origin?

• $V = \{ (x, y) \in \mathbb A^2 : y^2 = x^3\}$.
• $V = \{ (x, y, z) \in \mathbb A^3 : y + z = z^2 + xz + xyz\}$.
• $V = \{ (w, x, y, z) \in \mathbb A^4 : w^2z + x^2 = z + x + y^2, z + y^3(w+1) = x^3 - w^2\}$.

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