Commutative Algebra 10

Algebras Over a Ring

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


An Aalgebra 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.


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.


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. 🙂




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'.


(⇒) 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  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).



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', -).


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.


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).


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].


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).

warningIn 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.


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.


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.


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

5 Responses to Commutative Algebra 10

  1. Vanya says:

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

  2. Vanya says:

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

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