Commutative Algebra 30

Tensor Product of A-Algebras

Proposition 1.

Let B, C be A-algebras. Their tensor product D = B\otimes_A C has a natural structure of an A-algebra which satisfies

(b \otimes c) \times (b'\otimes c') = (bb') \otimes (cc').

Proof

Fix (b, c) \in B \times C. The map B\times C \longrightarrow B\otimes_A C, (b',c') \mapsto bb' \otimes cc' is A-bilinear so it induces an A-linear map

m_{b,c} : B\otimes_A C \longrightarrow B\otimes_A C, \quad b'\otimes c' \mapsto bb' \otimes cc'.

Now varying (bc) gives an A-bilinear map B\times C \to \mathrm{Hom}_A (B\otimes_A C, B\otimes_A C), (b, c) \mapsto m_{b,c} so it induces an A-linear

B\otimes_A C \longrightarrow \mathrm{Hom}_A(B\otimes_A C, B\otimes_A C), \quad b\otimes c \mapsto (b' \otimes c' \mapsto bb' \otimes cc'),

i.e. an A-bilinear map D \times D \to D. Note that A-bilinearity means distributivity in either term. To prove associativity, show that it holds on the pure tensors (exercise). ♦

It turns out this is the coproduct of B and C in the category of A-algebras. This was briefly aluded to earlier.

Proposition 2.

Let D = B\otimes_A C with A-algebra homomorphisms

\begin{aligned} \phi_1 : B\longrightarrow B\otimes_A C, &\quad b \mapsto b\otimes 1_C, \\ \phi_2 : C \longrightarrow B\otimes_A C, &\quad c \mapsto 1_B \otimes c.\end{aligned}

If E is an A-algebra and \psi_1 : B \to E, \psi_2 : C\to E are A-algebra homomorphisms, then there is a unique A-algebra homomorphism f:D \to E such that f\circ \phi_1 = \psi_1 and f\circ \phi_2 = \psi_2.

Proof

The A-bilinear map B\times C\to E, (b,c) \mapsto \psi_1(b)\psi_2(c) induces the A-linear map

f : B\otimes_A C\longrightarrow E, \quad b\otimes c\mapsto \psi_1(b) \psi_2(c).

To check that it is a ring homomorphism, it suffices to take the pure tensors, and show that f((b\otimes c)(b'\otimes c')) = f(b\otimes c)f(b'\otimes c').

Clearly f\circ\phi_i = \psi_i for i =1,2, and conversely, any such f must satisfy f(b\otimes c) = \psi_1(b) \psi_2(c) which uniquely determines f. ♦

blue-lin

Examples

1. Induced Algebra

In the above suppose we fix B and take the functor

A\text{-}\mathbf{Mod} \longrightarrow B\text{-}\mathbf{Mod},\quad C \mapsto B\otimes_A C

This functor satisfies the universal property: for any B-algebra D we have

\mathrm{Hom}_{B\text{-alg}}(B\otimes_A C, D) \cong \mathrm{Hom}_{A\text{-alg}}(C, D).

In fact, this is a general property from category theory.

Lemma 1.

If B\in \mathcal C is an object, and B\amalg C is the coproduct of B and C, then for each f:B\to D in \mathcal C we have a natural isomorphism

\mathrm{hom}_{\mathcal C}(C, D) \stackrel\cong \longrightarrow \mathrm{hom}_{B \downarrow \mathcal C} (i_B, f),

where i_B : B\to B \amalg C is the canonical map and B\downarrow \mathcal C is the coslice category.

Exercise A

Do at least one of the following.

  • Prove the universal property directly.
  • Prove the lemma and use it to prove the universal property.

2. Polynomial Ring

We have B \otimes_A A[X] \cong B[X] for any A-algebra B. There is more than one way to show it.

Proof 1: use the fact that A[X] is free over A with basis X^n for n\ge 0, so B \otimes_A A[X] is free over B with the same basis. Show that product in this ring still satisfies X^m \times X^n = X^{m+n} so it is isomorphic to B[X].

Proof 2: for any B-algebra C, example 1 gives

\mathrm{Hom}_{B\text{-alg}}(B \otimes_A A[X], C) \cong \mathrm{Hom}_{A\text{-alg}}(A[X], C) \cong C

as sets functorially in C. Since we also have \mathrm{Hom}_{B\text{-alg}} (B[X], C)\cong C, the result follows by Yoneda lemma.

3. Localization

Let B be an A-algebra with given homomorphism f:A\to B. If S\subseteq A is a multiplicative subset, then S^{-1}A \otimes_A B \cong S^{-1}B as A-modules as we saw earlier. Clearly the isomorphism preserves the ring structure so we get an isomorphism of A-modules.

Technically you have to localise with respect to a multiplicative subset of the ring, so it would be more correct to write T^{-1}B where T = f(S).

What is \mathrm{Spec} S^{-1}B? This is the set of primes \mathfrak q \subset B such that

\mathfrak q \cap T = \mathfrak q \cap f(S) = \emptyset \iff f^{-1}(\mathfrak q) \cap S = \emptyset.

In other words \mathrm{Spec} (S^{-1}A \otimes_A B) = (f^*)^{-1} (\mathrm{Spec} S^{-1}A) under the continuous map f^* : \mathrm{Spec} B \to \mathrm{Spec} A.

4. Quotient

Let B be an A-algebra with given homomorphism f:A\to B. If \mathfrak a \subseteq A is an ideal, then (A/\mathfrak a)\otimes_A B \cong B/\mathfrak a B as A-modules. Again it is in fact an isomorphism of A-algebras. Now \mathrm{Spec} B/\mathfrak a B is the set of primes \mathfrak q such that

\mathfrak q \supseteq \mathfrak a B \iff \mathfrak q \supseteq f(\mathfrak a) \iff f^{-1}(\mathfrak q) \supseteq \mathfrak a.

Thus \mathrm{Spec} (A/\mathfrak a \otimes_A B) = (f^*)^{-1}(\mathrm{Spec} A/\mathfrak a).

Exercise B

For ideals \mathfrak a, \mathfrak b, describe (A/\mathfrak a)\otimes_A (A/\mathfrak b).

Let B =A[X_1, \ldots X_n]/(f_i) where f_i\in A[X_1, \ldots, X_n] is a collection of polynomials. Describe the functor B\otimes_A -.

blue-lin

Geometric Applications

Throughout this section, k denotes an algebraically closed field.

Fibre of a Morphism

Let f:A\to B be a ring homomorphism and \mathfrak p \subset A be a prime ideal. Substitute \mathfrak a = \mathfrak p in example 4 and S = A - \mathfrak p in example 3. Thus the set of \mathfrak q \in \mathrm{Spec} B such that f^*(\mathfrak q) = f^{-1}(\mathfrak q) = \mathfrak p is given by:

S^{-1}A \otimes_A (A/\mathfrak p) \otimes_A B \cong k(\mathfrak p) \otimes_A B

where k(\mathfrak p) is the residue field of A at \mathfrak p.

Definition.

Let \phi : V\to W be a morphism of affine k-schemes, which corresponds to a ring homomorphism f :k[W] \to k[V]. For a given point P\in W, its fibre is the affine k-scheme \phi^{-1}(P) with coordinate ring:

k[\phi^{-1}(P)] = (k[W]/\mathfrak m_P) \otimes_{k[W]} k[V] \cong k[V]/f(\mathfrak m_P) k[V].

Example

Let V = \{(a,b,c,x,y) \in \mathbb A^5 : ax^2 + by^2 = c\} and W = \mathbb A^3 with morphism \phi:V \to W, (a,b,c,x,y) \mapsto (a,b,c). Algebraically, this is:

f : k[A, B, C] \hookrightarrow k[A, B, C, X, Y]/(AX^2 + BY^2 - C).

For the point (1, 1, 1) \in W, we have \mathfrak m = (A-1, B-1, C-1) so that

\begin{aligned} k[\phi^{-1}(P)] &= k[A, B, C, X, Y]/(AX^2 + BY^2 - C, A-1, B-1, C-1)\\ &\cong k[X, Y]/(X^2 + Y^2 - 1).\end{aligned}

The point Q = (0, 0, 0)\in W, on the other hand, gives k[\phi^{-1}(Q)] \cong k[X, Y].

Geometrically, V is a family of ellipses parametrized by the parameters a, b, c.

Product of Varieties

Suppose V\subseteq \mathbb A^n, W\subseteq \mathbb A^m are closed subsets. Then V\times W \subseteq \mathbb A^{m+n} is also closed, being cut out by the same equations which define V and W, but with disjoint sets of variables.

warningThe topology for V\times W is generally much finer than the product topology of V and W. For example, prove this for V = W = \mathbb A^1 as an easy exercise.

Lemma 2.

k[V\times W] \cong k[V] \otimes_k k[W].

Proof

The projection maps V\times W \to V, W induce k-algebra homomorphisms k[V], k[W] \to k[V\times W] and hence h : k[V]\otimes_k k[W] \to k[V\times W]; h is surjective by our concrete description of V\times W above.

For injectivity, let S be a basis of k[V]. Suppose h(\sum_{i=1}^n f_i \otimes g_i)= 0 for some f_1, \ldots, f_n \in S and g_1, \ldots, g_n \in k[W]. Thus

v\in V, w\in W \implies \sum_{i=1}^n f_i(v) g_i(w) = 0.

If we fix w, we get a linear relation \sum_{i=1}^n g_i(w) f_i = 0 which holds in k[V]. Since the f_i‘s are linearly independent, g_1(w) = \ldots = g_n(w) = 0 for all w\in W so each g_i = 0 and \sum_{i=1}^n f_i \otimes g_i = 0. ♦

Corollary 1.

If k is an algebraically closed field and A, B are finitely generated reduced k-algebras, then so is A\otimes_k B.

Proof

Indeed AB are isomorphic to k[V], k[W] for some k-varieties V and W. Apply the above. ♦

Exercise C

Find a counter-example when k is not algebraically closed. [ Hint: you need a base field of positive characteristic. ]

Intersection

Finally, we leave the following as an exercise.

Exercise D

Let W_1, W_2 be closed subvarieties of a variety V. This induces surjective k-algebra homomorphisms k[V] \to k[W_1] and k[V] \to k[W_2]. Write down the coordinate ring of the scheme intersection W_1 \cap W_2 using the language of tensor product.

Question to Ponder

Intuitively, there seems to be some commonality among the above three constructions. Explain this in terms of category theory. [ Hint: you might want to wait until the article on fibre products. ]

blue-lin

This entry was posted in Advanced Algebra and tagged , , , , , . Bookmark the permalink.

2 Responses to Commutative Algebra 30

  1. Vanya says:

    In the sentence before second proposition is it ” .. .in the category of A-Algebras” instead of $A-modules”?

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s