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If $F$ is a field an "algebra over $F$" is roughly "a vector space over $F$ that's also a ring", plus a few compatibililty axioms.

Now say $R$ is a ring. One might say an $R$-module is "a vector space over $R$, except you can't call it a vector space because $R$ is not a field". What's the right word for "algebra over $R$" (informally, an $R$-module which is also a ring)?

(It came up the other day in regard to polynomials: If $p\in R[x]$ then the objects $X$ such that $p(X)$ makes sense are precisely the elements of "$R$-algebras"; given that it seems there must be a word for such things.)

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  • $\begingroup$ people.virginia.edu/~mve2x/7752_Spring2010/lecture5.pdf $\endgroup$ – G Tony Jacobs Jan 1 '18 at 15:24
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    $\begingroup$ R-algebras are common objects of study; as far as I know, that's what everyone calls them. $\endgroup$ – G Tony Jacobs Jan 1 '18 at 15:24
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    $\begingroup$ @ziggurism I wasn't saying it was necessary, I was just assuming there was a separate term. If you read R-vector space you'd be scandalized, right? $\endgroup$ – David C. Ullrich Jan 1 '18 at 15:42
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    $\begingroup$ @DavidC.Ullrich Yes! Extremely scandalized! I just threw up in my mouth a little reading your comment. (Might be the NYE hangover though). $\endgroup$ – ziggurism Jan 1 '18 at 15:43
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    $\begingroup$ "Ring" is a slightly misleading word here. It's a bit subtle to give the correct definition of an $R$-algebra if $R$ isn't commutative. Anyway, "algebra" is used in an extremely general way in mathematics compared to most other words - think of e.g. Lie algebras, Boolean algebras, $\sigma$-algebras, relational algebras, algebras over functors, algebras over monads, algebras over operads... $\endgroup$ – Qiaochu Yuan Jan 1 '18 at 20:16
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There are a few equivalent ways of defining $R$-algebra, all of which generalize your intuition from the case where $R=k$ is a field. I'll assume that "ring" means "commutative ring with unit" for simplicity.

  1. An $R$-algebra is an $R$-module M equipped with an $R$-bilinear map $M\times M\to M$ which satisfies all the usual axioms for ring multiplication.

  2. An $R$-algebra is a ring $S$ equipped with an $R$-module structure compatible with multiplication in the sense that $r.(st) = (r.s)t =s(r.t)$ for any $r\in R$, and any $s,t\in S$

  3. An $R$-algebra is a ring $S$ equipped with a (unit preserving) homomorphism $R\to S$, called the structure map of $S$.

It's fairly clear that an $R$-algebra in the sense of (1) gives an $R$-algebra in the sense of (2). Also, given an $R$-algebra in the sense of (2), we can give a homomorphism as in (3) by sending $r\in R$ to $r.1_S\in S$. This is in fact the unique $R$-linear ring homorphism from $R$ to $S$. Finally, given an $R$-algebra in the sense of (3), i.e., given a ring homomorphism $f:R\to S$, we can turn $S$ into an $R$-algebra in the sense of (1) by using the $R$-module structure $r.s := f(r)s$. The $R$-bilinear map $S\times S \to S$ is given by the ring multiplication of $S$, sending $(s,t)\to st$.

These different perspectives have various advantages. For example, thinking in terms of (1), it's fairly clear that a ring is essentially by definition the same thing as a $\mathbb{Z}$-algebra. Moreover, replacing our $\mathbb{Z}$-bilinear map $M\times M\to M$ with a $\mathbb{Z}$-linear map $M \otimes_{\mathbb{Z} } M \to M$, we obtain a notion of algebra which generalizes very nicely: giving a ring is the same thing as giving a map $M \otimes_{\mathbb{Z} } M \to M$ in the category of abelian groups (i.e., $\mathbb{Z}$-modules) satisfying a few properties that can be described abstractly using diagrams of $\mathbb{Z}$-algebra homomorphisms. For $R$ a fixed ring which is not necessarily $\mathbb{Z}$, giving an $R$-algebra is the same thing as giving a map $M \otimes_{R} M \to M$ in the category of $R$-modules which satisfies several conditions expressed purely in terms of arrows between $R$-modules.This idea, called a commutative algebra object, makes sense in any category which behaves suitably like the category of abelian groups (i.e., an abelian category) which has a sufficiently nice symmetric tensor product (i.e., a symmetric monoidal structure).

For example, you can talk about algebras in the category of representations of some finite group, in the category of sheaves of abelian groups on some topological space, and so on. The objects you get with this method are usually the 'correct' definition of "algebra" in whatever context you're working in, and the result isn't always what you would naively guess - e.g., what are called twisted commutative algebras, the algebra objects in a certain category consisting of sequences $(V_n)$ of $S_n$-representations for $n=1,2,3,...$ are fairly intricate objects if you want to write out a full, concrete definition. This categorical perspective is very flexible, and generalizes nicely.

Definition (3), in terms of homomorphisms, gives rise to the relative point of view in commutative algebra or, more generally, in algebraic geometry. This is the idea that rarely do you want to think of a ring (or variety, or scheme...) as an object sitting in a vacuum. You prefer to think of it as an algebra over a base ring, which can always be taken as $\mathbb{Z}$ if you're in doubt. A homomorphism of rings over $R$ is just a homomorphism $S\to T$ which forms a commutative triangle with respect to the structure maps $R\to S$ and $R\to T$.

The advantage you get from doing this is simplicity. You are throwing out homomorphisms that you almost certainly don't care about. For example, $\mathbb{C}$ has a large number of very complicated $\mathbb{Z}$-automorphisms. But probably, if you're studying algebra over $\mathbb{C}$, you only really care about a handful. The $\mathbb{R}$-algebra given by the inclusion map $\mathbb{R}\to \mathbb{C}$ has only two automorphisms, given by the identity and by complex conjugation. The $\mathbb{C}$-algebra given by the identity map $\mathbb{C}\to\mathbb{C}$ has no nontrivial automorphisms. After 'stabilizing' your base ring $R$ in this sense, it often becomes much easier to understand how maps between rings over $R$ work. For $S$ any $R$-algebra, consider that a $\mathbb{Z}$-algebra map $R[x]\to S$ may be extraordinarily pathological in general. But an $R$-algebra map $R[x]\to S$ is determined completely by a choice of element $s\in S$, onto which you map the indeterminate $x$.

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  • $\begingroup$ Thanks. But all I really wanted to know was the correct terminology for "$R$-algebra" - given the existence of the word "module" I assumed it was something other than "$R$-algebra"... $\endgroup$ – David C. Ullrich Jan 1 '18 at 16:41

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