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 (b, c) 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. ♦

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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 $B\otimes_A -$ .

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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.

warning The 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 A, B 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. ]

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2 Responses to Commutative Algebra 30

Vanya says:

April 25, 2020 at 7:56 am

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

- limsup says:
  
  April 25, 2020 at 4:04 pm
  
  Indeed! Thanks again. 🙂

Commutative Algebra 30

Tensor Product of A-Algebras