Algebras Over a Ring

Let A be any ring; we would like to look at A-modules with a compatible ring structure.

Definition.

An $A$ –algebra is an $A$ -module $B$ , together with a multiplication operator $\times : B \times B \to B$ such that

$(B, +, \times)$ becomes a commutative ring (with 1);

multiplication is compatible with scalar multiplication:

$a\in A, b, b'\in B \implies a\cdot(b\times b') = (a\cdot b)\times b'.$

In the second condition, we have used · for module multiplication and × for ring multiplication. That condition gives us

$a \in A, b, b'\in B \implies b\times (a\cdot b') = (a\cdot b') \times b = a\cdot (b'\times b) = a\cdot (b\times b')$ .

Hence, for any $a_1, \ldots, a_k \in A$ and $b_1, \ldots, b_k \in B$ we have

$(a_1 \cdot b_1) \times \ldots \times (a_k \cdot b_k) = (\overbrace{a_1 \ldots a_k}^{\times \text{ in } A})\cdot (\overbrace{b_1 \times\ldots \times b_k} ^{\times \text{ in } B}).$

An alternative way of describing A-algebras is as follows.

Lemma 1.

If B is an A-algebra, the map $\phi : A\to B$ , $\phi(a) = a\cdot 1_B$ is a ring homomorphism.

Conversely, for any ring homomorphism $\phi: A\to B$ , B takes the structure of an A-algebra, where the A-module map is:

$A\times B\to B, \quad a\cdot b := \phi(a) \times b.$

Proof

First claim: clearly $\phi$ is additive and takes 1 to 1. Also

$\phi(a a') = aa'\cdot (1_B \times 1_B) = (a\cdot 1_B) \times (a'\cdot 1_B) = \phi(a) \times \phi(a').$

Second claim: verify the axioms as follows.

$\begin{aligned}\phi(1_A) \times b &= 1_B\times b = b\\ \phi(a+a')\times b &= (\phi(a) + \phi(a'))\times b = (\phi(a) \times b) + (\phi(a') \times b),\\ \phi(a)\times (b+b') &= (\phi(a) \times b) + (\phi(a) \times b'), \\ \phi(aa')\times b = \phi(a) \times \phi(a') \times b &= \phi(a) \times (\phi(a') \times b), \\ \phi(a)\times (b\times b') &= (\phi(a) \times b)\times b'.\end{aligned}$

This completes the proof. ♦

Exercise A

Prove that the constructions in the lemma are mutually inverse.

Definition.

If B is an A-algebra, an A-subalgebra is a subset of B which is an A-submodule as well as a subring.

Exercise B

If the A-algebra B corresponds to $\phi : A\to B$ , prove that an A-subalgebra of B is precisely a subring of B containing $\phi(A)$ .

P. S. Please do the above exercises. Don’t be lazy. 🙂

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Homomorphisms

Definition.

Let B and B’ be A-algebras. A homomorphism of A-algebras from B to B’ is an A-linear map $f:B \to B'$ which is also a ring homomorphism.

If we consider the alternative way of defining A-algebras, we get:

Proposition 1.

Suppose B and B’ are A-algebras corresponding to ring homomorphisms $\phi : A\to B$ and $\phi' : A\to B'$ . A homomorphism of A-algebras is precisely a ring homomorphism $f:B\to B'$ such that $f\circ \phi = \phi'$ .

Proof

(⇒) Suppose $f:B\to B'$ is a homomorphism of A-algebras. For $a\in A$ ,

$f(\phi(a)) = f(a\cdot 1_B) = a\cdot f(1_B) = a\cdot 1_{B'} = \phi'(a).$

(⇐) Suppose f is a ring homomorphism such that $f\circ \phi = \phi'$ , so for all $a\in A$ , $f(a\cdot 1_B) = a\cdot 1_{B'}$ . Hence for $a\in A$ , $b\in B$ ,

$f(a\cdot b) = f((a\cdot 1_B) \times b) = f(a\cdot 1_B) \times f(b) = (a\cdot 1_{B'}) \times f(b) = a\cdot f(b).$ ♦

The following result follows from the corresponding results for A-modules and rings.

First Isomorphism Theorem.

If $f:B\to B'$ is a homomorphism of A-algebras, the image of f is an A-subalgebra of B’ and we get an isomorphism of A-algebras.

$\overline f : A/\mathrm{ker} f \longrightarrow \mathrm{im} f, \quad a + \mathrm{ker} f \mapsto f(a).$

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Representability

For A-algebras B and C, let $\mathrm{Hom}_{A-\text{alg}}(B, C)$ be the set of A-algebra homomorphisms $f : B\to C$ . Unlike the case of modules, this set has no canonical additive structure, since if $f_1, f_2 : B\to C$ are homomorphisms of A-algebras, $f_1 + f_2$ usually is not (e.g. it does not take 1 to 1).

However, $\mathrm{Hom}_{A-\text{alg}}(B, C)$ is useful for classifying certain constructions from A-algebras.

As an example fix $B = A[X, Y]/(XY - 1)$ . Any A-algebra homomorphism $f:B\to C$ corresponds to a pair of elements $(\alpha, \beta) \in C \times C$ such that $\alpha\beta = 1$ . Thus we have a bijection:

$\begin{aligned}\mathrm{Hom}_{A-\text{alg}}(B, C) &\stackrel \cong\longrightarrow U(C),\\ f &\mapsto f(X), \end{aligned}$

where U(C) is the set of all units of C. Furthermore, any A-algebra homomorphism $g : C\to C'$ induces a map

$g_* : \mathrm{Hom}_{A-\text{alg}}(B, C) \longrightarrow \mathrm{Hom}_{A-\text{alg}}(B, C'), \quad f \mapsto g\circ f,$

which translates to the map g restricted to $U(C) \to U(C')$ .

Thus, we obtain a natural isomorphism $U(-) \cong \mathrm{Hom}_{A-\text{alg}}(B, -)$ . To express this rigorously, one needs the language of category theory.

Exercise C

For any A-algebra C, let

$\begin{aligned} SL_2 C &= \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} : a, b, c, d \in C, ad - bc = 1\right\}, \\ GL_2 C &= \left\{ \begin{pmatrix} a & b \\ c & d \end{pmatrix} : a, b, c, d \in C, ad - bc \text{ unit}\right\}. \end{aligned}$

Find A-algebras B and B’ such that

$SL_2(-) \cong \mathrm{Hom}_{A-\text{alg}}(B, -), \quad GL_2(-) \cong \mathrm{Hom}_{A-\text{alg}}(B', -)$ .

Note

Finally observe that $U(C), SL_2 C, GL_2 C$ are not just sets, but have group structures as well. To specify these group structures rigorously, one requires B to have a Hopf algebra structure.

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Generating a Subalgebra

If B is an A-algebra and $C_i \subseteq B$ is a collection of subalgebras, their intersection $C = \cap_i C_i$ is clearly also a subalgebra of B. This follows from the fact that intersection of submodules (resp. subring) is a submodule (resp. subring).

Definition.

For subset $S \subseteq B$ , let $\Sigma$ be the collection of all A-subalgebras of B containing S. Their intersection $\cap \{C : C\in S\}$ is the A-subalgebra of B generated by S; it is denoted by $A[S].$

This generated subalgebra can be concretely described as follows. For a finite sequence $\alpha_1, \ldots, \alpha_k \in S$ , and polynomial $f(X_1, \ldots, X_k) \in A[X_1, \ldots, X_k]$ in k variables, we take the element $f(\alpha_1, \ldots, \alpha_k) \in B$ . Taking the set of all these elements over all k, all $\alpha_1, \ldots, \alpha_k\in S$ and all $f(X_1, \ldots, X_k)$ then gives us $A[S]$ .

Definition.

An A-algebra B is finitely generated (or of finite type) if $B = A[S]$ for some finite subset $S\subseteq B$ .

If $S = \{\alpha_1, \ldots, \alpha_n\}$ , it suffices to take only polynomials f in exactly n variables and consider all $f(\alpha_1, \ldots, \alpha_n)$ , in which case we have a surjective homomorphism of A-algebras

$A[X_1, \ldots, X_n] \longrightarrow B, \qquad f(X_1, \ldots, X_n) \mapsto f(\alpha_1, \ldots, \alpha_n).$

warning In future articles, if B is an A-algebra, we say B is a finite A-algebra if it is finitely generated as an A-module; we say it is of finite type if it is finitely generated as an A-algebra.

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Direct Product of Algebras

Let $(B_i)_{i\in I}$ be a collection of A-algebras. Let $B := \prod_{i\in I} B_i$ be the set-theoretic product; we saw earlier that B has a natural structure of an A-module. It also has a natural structure of a ring via component-wise multiplication: $(b_i)_i \times (b'_i)_i := (b_i b'_i)_i$ , which is compatible with its A-module structure. Thus:

Theorem (Universal Property of Products).

For each $j\in I$ define the projection map

$\pi_j : \prod_{i\in I} B_i \longrightarrow B_j, \quad (b_i)_i \mapsto b_j,$

clearly a homomorphism of A-algebras. This collection $(B, (\pi_i : B \to B_i))$ satisfies the following.

For any A-algebra C and collection $(C, (\phi_i : C \to B_i))$ where each $\phi_i$ is an A-algebra homomorphism, there is a unique A-algebra homomorphism $f : C \to B$ such that $\pi_i\circ f = \phi_i$ for each $i\in I$ .

Proof

Easy exercise. ♦

Thus $\prod_i B_i$ is the A-algebra analogue of the direct product. Is there an A-algebra analogue of the direct sum?

The natural inclination is to take the direct sum $B = \oplus_{i\in I} B_i$ as an A-module then define multiplication component-wise, but it would not work since $\oplus_i B_i$ does not contain $(1, 1, \ldots)$ . The correct answer is obtained via tensor products, which will be covered in a later article.

In the next article, we will revisit algebraic geometry with a more general framework.

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5 Responses to Commutative Algebra 10

Vanya says:

April 2, 2020 at 5:09 am

In the definition of algebra , should the ” compatible multiplication x: B x B -> B ” be ” A x B -> B” ?

- Vanya says:
  
  April 2, 2020 at 5:12 am
  
  Sorry, my mistake.
  
Vanya says:

April 5, 2020 at 3:20 pm

I wonder why $\text{Hom}_{A-alg}(B,C$ has no canonical additive structure (under the section representability). Could you please clarify?

- limsup says:
  
  April 5, 2020 at 3:35 pm
  
  Because if $f_1, f_2 : B\to C$ are homomorphisms of A-algebras, then $f_1 + f_2$ is not. For example, the map does not take 1 to 1.
  
- limsup says:
  
  April 5, 2020 at 3:41 pm
  
  Added a remark in the main text. Thanks!

Commutative Algebra 10

Algebras Over a Ring

Homomorphisms

Representability

Generating a Subalgebra

Direct Product of Algebras

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Commutative Algebra 10

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