# Tensor Product of *A*-Algebras

Proposition 1.Let B, C be A-algebras. Their tensor product has a natural structure of an A-algebra which satisfies

.

**Proof**

Fix . The map is *A*-bilinear so it induces an *A*-linear map

Now varying (*b*, *c*) gives an *A*-bilinear map , so it induces an *A*-linear

i.e. an *A*-bilinear map . 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 with A-algebra homomorphisms

If E is an A-algebra and , are A-algebra homomorphisms, then there is a unique A-algebra homomorphism such that and .

**Proof**

The *A*-bilinear map induces the *A*-linear map

.

To check that it is a ring homomorphism, it suffices to take the pure tensors, and show that .

Clearly for , and conversely, any such *f* must satisfy which uniquely determines *f*. ♦

# Examples

### 1. Induced Algebra

In the above suppose we fix *B* and take the functor

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

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

Lemma 1.If is an object, and is the coproduct of B and C, then for each in we have a natural isomorphism

where is the canonical map and 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 for any *A*-algebra *B*. There is more than one way to show it.

Proof 1: use the fact that is free over *A* with basis for , so is free over *B* with the same basis. Show that product in this ring still satisfies so it is isomorphic to .

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

as sets functorially in *C*. Since we also have , the result follows by Yoneda lemma.

### 3. Localization

Let *B* be an *A*-algebra with given homomorphism . If is a multiplicative subset, then 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 where .*

What is ? This is the set of primes such that

In other words under the continuous map .

### 4. Quotient

Let *B* be an *A*-algebra with given homomorphism . If is an ideal, then as *A*-modules. Again it is in fact an isomorphism of *A*-algebras. Now is the set of primes such that

.

Thus .

**Exercise B**

For ideals , describe .

Let where is a collection of polynomials. Describe the functor .

# Geometric Applications

Throughout this section, *k* denotes an algebraically closed field.

## Fibre of a Morphism

Let be a ring homomorphism and be a prime ideal. Substitute in example 4 and in example 3. Thus the set of such that is given by:

where is the residue field of *A* at .

Definition.Let be a morphism of affine k-schemes, which corresponds to a ring homomorphism . For a given point , its

fibreis the affine k-scheme with coordinate ring:

**Example**

Let and with morphism , . Algebraically, this is:

.

For the point , we have so that

The point , on the other hand, gives .

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

## Product of Varieties

Suppose , are closed subsets. Then is also closed, being cut out by the same equations which define *V* and *W*, but with disjoint sets of variables.

The topology for is generally much finer than the product topology of *V* and *W*. For example, prove this for as an easy exercise.

Lemma 2..

**Proof**

The projection maps induce *k*-algebra homomorphisms and hence ; *h* is surjective by our concrete description of above.

For injectivity, let *S* be a basis of . Suppose for some and . Thus

.

If we fix *w*, we get a linear relation which holds in . Since the ‘s are linearly independent, for all so each and . ♦

Corollary 1.If k is an algebraically closed field and A, B are finitely generated reduced k-algebras, then so is .

**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 be closed subvarieties of a variety *V*. This induces surjective *k*-algebra homomorphisms and . Write down the coordinate ring of the scheme intersection 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. ]

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

Indeed! Thanks again. 🙂