Morphisms in Algebraic Geometry

Next we study the “nice” functions between closed subspaces of $\mathbb A^n$.

Definition.

Suppose $V\subseteq \mathbb A^n$ and $W\subseteq A^m$ are closed subsets. A morphism $f:V\to W$ is a function which can be expressed as:

$f(v_1, \ldots, v_n) = (f_1(v_1, \ldots, v_n), f_2(v_1, \ldots, v_n), \ldots, f_m(v_1, \ldots, v_n))$

for some polynomials $f_1, f_2, \ldots, f_m \in k[X_1, \ldots, X_n]$. We also say f is a regular map.

Example

1. Let $V \subseteq \mathbb A^n$ be any closed subset. Regular maps of the form $V \to \mathbb A^1$ are given by polynomials $f \in k[X_1, \ldots, X_n]$.
2. Take $V = \mathbb A^1$ and $W = \{(x, y, z) \in \mathbb A^3 : x^2 + y^2 = z^2\}$. Define $f:V\to W$ by $t \mapsto (t^2 - 1, 2t, t^2 + 1)$. We write this as $x = t^2 - 1, y = 2t, z = t^2 + 1$.
3. Take $V = \mathbb A^1$ and $W = \{(x, y) \in \mathbb A^2 : y^2 = x^3\}$. Define $f:V\to W$ by $t\mapsto (t^2, t^3)$. We write this as $x = t^2, y = t^3$.

[Example 3: image edited from GeoGebra plot.]

The first example, although basic, is of huge importance.

Definition.

regular function for a closed subset $V\subseteq \mathbb A^n$ is a morphism of the form $V\to \mathbb A^1$.

The set of such functions is called the coordinate ring of V and is denoted by $k[V]$.

Note

Each regular function is given by $f\in k[X_1, \ldots, X_n]$ and two polynomials f, g induce the same function on V if and only if

$(\forall P \in V, f(P) = g(P)) \iff (\forall P \in V, (f-g)(P) = 0) \iff f-g \in I(V).$

Thus we have $k[V] \cong k[X_1, \ldots, X_n]/I(V)$, which gives k[V] its ring structure. To describe this ring structure in terms of regular functions, we have:

$\text{morphisms } f, g : V \to \mathbb A^1 \implies \begin{cases} (f+g) : V\to \mathbb A^1, \ &P \mapsto f(P) + g(P) \in k, \\ (f\cdot g) : V\to \mathbb A^1, \ &P \mapsto f(P)g(P)\in k. \end{cases}$

Morphisms in Algebra

General morphisms $f:V\to W$ can now be described in the language of their coordinate rings.

Proposition 1.

Let $V\subseteq \mathbb A^n$ and $W\subseteq \mathbb A^m$ be closed. There is a bijection between:

• morphisms $f:V\to W$;
• ring homomorphisms $f^* : k[W] \to k[V]$ which are linear over k.

Note

The second condition can be rephrased as: $k[W] \to k[V]$ is a homomorphism of k-algebras. In a later article, we will cover algebras over a ring in greater detail.

Proof

Given a morphism $f:V\to W$, upon composing with a regular function $g : W\to \mathbb A^1$, we obtain a regular function $g\circ f : V\to \mathbb A^1$ of V. This gives a ring homomorphism $k[W] \to k[V]$ which is clearly linear over k.

Conversely, suppose $\phi : k[W] \to k[V]$ is a ring homomorphism which is linear over k. Let $\overline X_1, \ldots, \overline X_m$ be the images of  $X_1, \ldots, X_m$ in $k[W] = k[X_1, \ldots, X_m]/I(W)$. Let $f_i = \phi(\overline X_i) \in k[V]$ which is a regular function $V\to \mathbb A^1$. We claim that the function

$f : V \to \mathbb A^m, \quad f(P) := (f_1(P), \ldots, f_m(P))$

has image in W. Indeed for any $g\in I(W)\subseteq k[X_1, \ldots, X_m]$ we have

$g\circ f = g(f_1, \ldots, f_m) = g(\phi(\overline X_1), \ldots, \phi(\overline X_m)) = \phi(g(\overline X_1, \ldots, \overline X_m)) = \phi(\overline{g(X_1, \ldots, X_m)})$

which is 0 since $\overline g =0$ in k[W]. This creates a bijection. ♦

Note

In order to swap $g(\phi(\ldots)) = \phi(g(\ldots))$ we require $\phi$ to be a ring homomorphism linear over k. E.g. if $g = \alpha X_1^2 +\beta X_2$ with $\alpha, \beta\in k$ then

$g(\phi(\overline X_1), \phi(\overline X_2)) = \alpha\phi(\overline X_1)^2 + \beta \phi(\overline X_2) = \phi(\alpha \overline X_1^2 + \beta \overline X_2) = \phi(g(\overline X_1, \overline X_2)).$

The following properties are obvious:

Lemma.

• For any closed set V, we have $(1_V)^* = 1_{k[V]}$.
• For any morphisms $f:V \to V', g:V' \to V''$ of closed sets we have

$(g\circ f)^* =f^* \circ g^* : k[V''] \longrightarrow k[V].$

Proof

The first is clear. For the second, pick any $h\in k[V'']$, a regular map $V''\to \mathbb A^1$. Then

$f^*(g^*(h)) = f^*(h\circ g) = (h \circ g)\circ f = h\circ (g\circ f) = (g\circ f)^*(h)$. ♦

Examples

Let us interpret the earlier examples as homomorphisms $k[W] \to k[V]$.

Example 2. We have:

$f : \mathbb A^1 \to \{(x, y, z) \in \mathbb A^3 : x^2 + y^2 = z^2\}, \quad t \mapsto (t^2 - 1, 2t, t^2 + 1)$.

We wrote this as $x = t^2 - 1, y = 2t, z = t^2 + 1$ for a reason, for f corresponds to:

$f^* : k[X, Y, Z]/(X^2 + Y^2 - Z^2) \longrightarrow k[T], \quad X \mapsto T^2 - 1, Y \mapsto 2T, Z \mapsto T^2 + 1.$

Example 3. Similarly

$f : \mathbb A^1 \to \{(x,y) \in \mathbb A^2 : y^2 = x^3\}, \quad t\mapsto (t^2, t^3)$

was written as $x = t^2, y = t^3$. Algebraically,

$f^* : k[X,Y]/(Y^2 - X^3) \longrightarrow k[T], \quad X \mapsto T^2, Y \mapsto T^3.$

Properties

Since we defined a topology on closed sets, the morphisms should be continuous.

Proposition 2.

A morphism $f : (V\subseteq \mathbb A^n) \to (W\subseteq \mathbb A^m)$ is a continuous map, with respect to the subspace topology on both sets.

Proof

Let W’ be a closed subset of W, so it is also a closed subset of $\mathbb A^m$. We need to show $f^{-1}(W')$ is closed in V (equivalently, in $\mathbb A^n)$. Now W’ can be written as

$W' = \{Q \in W: \text{ for each } g \in S', g(Q) = 0\}$

for some subset $S' \subseteq k[X_1, \ldots, X_m]$. It follows that

$f^{-1}(W') = \{P \in V : \text{ for each } g \in S', g(f(P)) = 0\}$

is cut out from V by equations $\{g\circ f : g\in S'\}$. Hence $f^{-1}(W')$ is closed. ♦

Isomorphisms

Definition.

Closed subsets $V \subseteq \mathbb A^n$ and $W\subseteq \mathbb A^m$ are said to be isomorphic if there exist regular maps $f : V\to W$ and $g:W\to V$ such that $g\circ f = 1_V$ and $f\circ g = 1_W$.

This implies:

$f^* : k[W] \to k[V],\ g^* : k[V] \to k[W], \quad f^*\circ g^* = 1_{k[V]}, \ g^* \circ f^* = 1_{k[W]}.$

Hence isomorphism of the closed subsets corresponds to isomorphism of the underlying k-algebras!

Example

Let us take example 3 from above, where $f : \mathbb A^1 \to \{(x,y) \in \mathbb A^2 : y^2 = x^3\}$ is defined by $f(t) = (t^2, t^3)$. Note that f is bijective on the points, and it corresponds to:

$f^* : k[X,Y]/(Y^2 - X^3) \longrightarrow k[T], \quad X \mapsto T^2, Y \mapsto T^3.$

The image of $f^*$ is not surjective since it does not contain T. We have thus learnt:

There exist bijective regular maps $V\to W$ which are not isomorphisms.

More General Correspondence

Putting it together, we obtain the following bijective correspondences:

• The top correspondence was the original one.
• The left correspondence follows from point-set topology.
• The right correspondence follows from the correspondence between ideals of $A/\mathfrak a$ and ideals of $A$ containing $\mathfrak a$.
• The correspondence preserves radical ideals because $\mathfrak b$ is a radical ideal of A if and only if $A/\mathfrak b$ is a reduced ring; now apply $(A/\mathfrak a)\, /\, (\mathfrak b/\mathfrak a) \cong A/\mathfrak a$.

Summary.

In other words, we have a bijection between radical ideals of the coordinate ring k[V] and closed subsets of V. This enables us to look at V and its coordinate ring k[V], ignoring the ambient affine space it sits in.

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2 Responses to Commutative Algebra 5

1. Vanya says:

Could you please explain further the last sentence “This enables us to look at V and its coordinate ring k[V], ignoring the ambient affine space it sits in.”? Is it not the case that we need to consider some ambient affine space while we consider some closed subspace V it sits in?

• limsup says:

Yes, for now. In a few articles, you will see how this pan out, where we define an abstract variety as a k-algebra.