Is it precise to say a ring is an algebra?

Question

For example, when we say that every ring $R$ is a $\mathbb{Z}$-algebra with the homomorphism $\varphi: \mathbb{Z} \to R$ sending $n \mapsto n \cdot 1_R$, what exactly does it mean for 'a ring to be an algebra'? A ring is a ring; but an algebra is a ring with a structure map. What is the benefit or insight gained from equating $R$ with a $\mathbb{Z}$-algebra? Are there other such examples?

Answer 1

Algebra is full of structures which are "things over things."

For groups, we talk about group actions, which are just "sets over a group." A group action can be described two equivalent ways. For a group $G$ acting on a set $X$, we sometimes call $X$ a $G$-set. This can be defined two equivalent ways:

1. There exists a map $G\times X\to X$, written $(g,x)\mapsto g\cdot x$, satisfying the properties $1\cdot x=x$, $(gh)\cdot x=g\cdot (h\cdot x)$

2. There exists a homomorphism $G\to \operatorname{Aut}(X)$, where $\operatorname{Aut}(X)$ is the group of permutations of $X$.

Next, we encounter modules, which are "abelian groups over a ring." An $R$-module $M$ can be defined as

1. An abelian group $M$ along with an action $R\times M\to M$ satisfying $1\cdot x=x$, $r\cdot(x+y)=r\cdot x+r\cdot y$, $(r+s)\cdot x=r\cdot x+s\cdot x$, and $(rs)\cdot x=r\cdot (s\cdot x)$
2. An abelian group $M$ along with a ring homomorphism $R\to \operatorname{End}(M)$, where $\operatorname{End}(M)$ is the group of maps $M\to M$.

Then we have algebras, which are like "rings over a ring." An $R$-algebra $A$ can be defined as

1. An $R$-module $A$, where $A$ is also a ring, and the action of $R$ is required to commute with the multiplication in $A$. That is, we have a map $R\times A\to A$ satisfying $(r\cdot a)(s\cdot b)=rs\cdot ab$.
2. A ring $A$ together with a ring homomorphism $R\to Z(A)$, where $Z(A)$ is the center of $A$.

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I'm mentioning all these things because I am procrastinating, but also because it helped me a lot to realize how analogous these things are.

In the category of rings, $\Bbb Z$ is an initial object. If you don't know categories, all this means is that, for any ring $R$, there is a unique map $\Bbb Z\to R$. Now, since $\Bbb Z$ is a commutative ring, we see that the image of this map must be in the center of $R$, and so this map makes any ring $R$ into a $\Bbb Z$-algebra. Conversely, given a $\Bbb Z$-algebra $R$, by definition, $R$ is a ring.

There is one important note here, though: the use of the word "algebra" is not standard. Some people assume "algebra" to mean "algebra over a field." In this case, every algebra is a ring, but not every ring is an algebra. Also, some authors talk about non-unital or non-associative algebras. Usually a ring is assumed to be unital and associative, so in this case, algebras and rings are incomparable.

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If you are confused about these statements, it is a good exercise to just prove them. For example, consider the two definitions I gave for (unital associative) $R$-algebras. Given such a map $R\times A\to A$, note that for each $r\in R$, we have a corresponding map $A\to A$. But this is determined just by where it sends $1$. This gives us a way to define a map $R\to A$. You can also show that the image is necessarily central. Conversely, start with a ring homomorphism $R\to Z(A)$ and try to define an $R$-algebra structure on $A$ with the first definition.

Answer 2

When we say that every ring "is" a $\mathbb{Z}$-algebra, what we mean precisely is that there is an equivalence of categories between the category of rings and the category of $\mathbb{Z}$-algebras (given by forgetting the structure map). The word "is" is doing a lot of work here, and really people typically mean not only that there is such an equivalence but that a particular map they have in mind produces it.

This equivalence can be thought of as coming from a more basic equivalence between abelian groups and $\mathbb{Z}$-modules (we get rings resp. $\mathbb{Z}$-algebras by taking monoid objects with respect to the tensor product), and one practical upshot is that it reminds you that extension of scalars functors are available: if $f : R \to S$ is any ring homomorphism, it induces an extension of scalars functor $(-) \otimes_R S$ from $R$-modules to $S$-modules, and if $R$ and $S$ are both commutative this functor also sends $R$-algebras to $S$-algebras.