How compatible with the ring of scalars does an algebra over a ring need to be?

Question

What are the minimum requirements for compatibility with scalar multiplication that are needed to be an algebra? Do the conventions differ from author to author?

According to Wikipedia, compatibility with scalars is defined as $(ax)\cdot(by) = (ab)(x\cdot y)$, but my gut tells me that this isn't stringent enough. I'm reluctant to just dismiss the Wikipedia definition though because outright mistakes in Wikipedia are rare and odds are good I'm missing something obvious.

The wikipedia article in question is about algebras over fields, but I am interested in algebras over rings in general. Let rings be commutative and unital.

Let $k$ be the field $\mathbb{Z}/2\mathbb{Z}$, consider the algebra with elements taken from $\mathbb{Z}$. Call the algebra $A$.

By convention, I write scalar literals with an overline, i.e. $\overline{0}$ and $\overline{1}$.

Define multiplication as $\overline{0}(a) = 0$ and $\overline{1}(a) = a$.

I can think of $4$ reasonable properties that we might want an algebra over a ring to have. $A$ has three out of four.

• $r(a+b) = r(a) + r(b)$ is satisfied.
• $r(s(a)) = (rs)(a)$ is satisfied.
• $r(ab) = r(a)b = ar(b)$ is satisfied.
• $(r+s)(a) = r(a) + s(a)$ is not satisfied, e.g. $(\overline{1} + \overline{1})(2) = 0$, but $\overline{1}(2) + \overline{1}(2) = 4$

The failure of the last property makes $A$ an odd duck, but is it an algebra or not?

Answer 1

Just to have my answer from the comments also here so people can find it faster:

$\mathbb{Z}$ is NOT a $\mathbb{F_2}$ vectorspace, so it cannot be an $\mathbb{F}_2$ algebra (by Wikipedia).

I personally prefer the definition (less axioms):

An (unital) $R$-algebra for a commutative ring $R$ is a ring $A$ together with a ring morphism $R \to \mathrm{Z}\left(A\right)$ (,where $\mathrm{Z}$ denotes the center).

This also makes clear that you want your scalar elements to be represented by actual elements in your algebra (for matrices: diagonal matrices $\lambda \mathrm{id}$).

Answer 2

Here's another way of defining an $R$-algebra in such a way that we naturally get all four of the conditions listed above.

Let $R$ be a commutative, unital ring.

$A$ is an $R$-algebra if and only if $A$ is a perhaps-associative ring and an $R$-module (with the notion of addition shared between the two), subject to the following compatibility condition:

$$ r(ab) = r(a)b = ar(b) $$

The other conditions are implied by $A$ being an $R$-module.

It is very natural to insist that the $*$-free reduct of any $R$-algebra is an $R$-module, so we might as well use it when defining what an $R$-algebra is.