I am trying to connect the definition of an algebra from baby rudin to a chapter in an abstract algebra text. It seems however, that an R-algebra isn’t the same as what rudin is talking about. What is the more precise name for the algebra given my Rudin?

Definition (Dummit and Foote) Let $R$ be a commutative ring with identity. An $R$ -algebra is a ring $A$ with identity together with a ring homomorphism $f: R \rightarrow A$ mapping $1_{R}$ to $1_{A}$ such that the subring $f(R)$ of $A$ is contained in the center of $A$.

7.28 Definition (Rudin PMA) A family $\mathscr{A}$ of complex functions defined on a set $E$ is said to be an algebra if (i) $f+g \in \mathscr{A}$, (ii) $f g \in \mathscr{A}$, and (iii) $c f \in \mathscr{A}$ for all $f \in \mathscr{A}, g \in \mathscr{A}$ and for all complex constants $c$, that is, if $\mathscr{A}$ is closed under addition, multiplication, and scalar multiplication. We shall also have to consider algebras of reai functions; in this case, (iii) is of course only required to hold for all real $c$. If $\mathscr{A}$ has the property that $f \in \mathscr{A}$ whenever $f_{n} \in \mathscr{A}(n=1,2,3, \ldots)$ and $f_{n} \rightarrow f$ uniformly on $E$, then $\mathscr{A}$ is said to be uniformly closed. Let $\mathscr{B}$ be the set of all functions which are limits of uniformly convergent sequences of members of $\mathscr{A}$. Then $\mathscr{B}$ is called the uniform closure of $\mathscr{A}$. (See Definition 7.14.)

  • $\begingroup$ Without (iii), this gives $\mathcal{A}$ the strucuture of a ring. But with (iii) included, it is also a $\mathbb{C}$-algebra (or whatever field your constants takes values in). Note that (i) and (iii) gives you a $\mathbb{C}$-vector space structure on $\mathcal{A}$. $\endgroup$
    – daruma
    Jun 9, 2021 at 18:59
  • $\begingroup$ @daruma When you say C-algebra is that in the sense of the definition by dummit and foote? $\endgroup$ Jun 9, 2021 at 19:10
  • $\begingroup$ For some motivation and insight see this post. Be sure to also read the linked posts (and their links ...). $\endgroup$ Jun 9, 2021 at 22:12
  • $\begingroup$ Btw, the answer you accepted does not specifically address your question. You will likely receive more specific (and more helpful) answers if you unaccept (many readers skip questions that already have accepted answers). Generally it is never a good idea to quickly accept an answer. $\endgroup$ Jun 9, 2021 at 23:40

3 Answers 3


For a possibly useful broader context:

What "$R$-algebra $A$" means is very context dependent, and there are many mutually incompatible conventions... so you just have to hope that either your source explains what they mean, or you can infer it from context.

In particular, it is essentially a waste of time to worry too much about comparison, much less reconciliation, of various versions.

For that matter, must a ring have a unit $1$? :) Certainly many attractive theorems use existence of $1$. But this can be weakened to "existence of sufficiently-many idempotents", meaning that for any finite (for example) subset $X$ of the ring, there is an idempotent $e$ (meaning that $e^2=e$, such that $ex=x$ for all $x\in X$. If the ring is not commutative, then we may want to specify left/right conditions.

Similarly, for a commutative ring $R$ with $1_R$, for a ring $A$ to be an "$R$-algebra", do we really need $R$ to inject to $A$? After all, $\mathbb Z/n$ is a pretty reasonable $\mathbb Z$-algebra.

Do we really need the image of $1_R$ in $A$ to be $1_A$? Or merely that $1_R\cdot a=a$ for all $a\in A$?

And so on.

I've come to think that there's no universally optimal definition of "algebra", but, rather, that there are many somewhat-different things that can be called "algebras", and needing some explanation for clarity. So it's not so much "definitions", but just "naming".


I think it can get confusing as there are different levels of abstraction in this definition.

(1). Algebras over fields: A ring $A$ that also happens to be a vector space over your field $F$. Even if your algebra $A$ is not commutative, it needs to commute with your scalars, $F$.

Very often your algebras are commutative as well and in that case, this is basically Rudin's definitions.

(2). Algebras over rings: Same definition as (1) except you replace vector space over $F$ with modules over $R$. This is the Dummit and Foote definition. When he is saying that $f(R)$ is contained in the center of $A$, that is saying you should think of your scalars $R$ as being embedded in your algebra $A$ and that they should really commute with any elements in $A$.

An example to have in mind might be polynomial rings like $R[x]$. This is a module over $R$ spanned by $\{1,x,x^2,...\}$. You do need your scalars $R$ to commute with other elements in $R[x]$.

  • $\begingroup$ Thank you for this answer. Now is Rudin's definition using the term "algebra" as a blanket term for both algebras over fields and algebras over rings? $\endgroup$ Jun 9, 2021 at 19:53
  • $\begingroup$ You never work with algebras over rings that aren't algebra over fields in real analysis because really all you care about are the reals and complex numbers. You also would never want your multiplication of functions to be noncommutative so in that essense, Rudin is (1) where everything commutes. $\endgroup$
    – daruma
    Jun 9, 2021 at 20:01

Here is an answer I found very satisfying from Answers to student's Math 104 questions

You ask about the relation between the sense given to the word "algebra" in Definition 7.28 (p.161) and other meanings of the term that you have seen.

These meanings are essentially the same -- one has a system with operations of addition, internal multiplication, and multiplication by some external "scalars". In Rudin, the "scalars" are real or complex numbers; in the concept of R-algebra that you have seen, they are members of a commutative ring (of which the real and complex numbers are special cases). Certain conditions (such as commutativity and associativity of addition, and distributivity for internal and scalar multiplication) are always assumed when one speaks of an algebra; others, such as commutativity and/or associativity of internal multiplication, may or may not be assumed.

You ask why one doesn't have more specific terms. One does. One can speak of R-algebra (R a commutative ring) or k-algebras (k a field); one can speak of commutative algebras, not-necessarily-commutative associative algebras, and nonassociative algebras such as Lie algebra. In a context where only one sort of algebra is being considered, an author may conveniently use "algebra" to refer to that concept; so Rudin here uses "algebra" to mean "subalgebra of the commutative C-algebra of all complex functions on a set E, under pointwise operations".

There is also a much more general sense of "algebra" than these: In the area of math called "universal algebra" (or "general algebra"), all of the sorts of objects considered in the area of algebra -- groups, rings, lattices, vector spaces, etc. -- and in general, all structures consisting of a set with a family of specified operations on it -- are called "algebras". The ambiguity between this sense of "algebra" and the more specific sense of the preceding paragraph is a real problem; but we don't have an alternative term for either concept, so it looks as though we will live with it for a long time to come.


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