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?

  • $\begingroup$ It is not an algebra, for me an algebra is a ring $A$ together with a morphism from your underlying commutative ring $R$ into your ring, I think this is the most standard definition as it allows to identify any algebra $A$ with an $R$-linear cat with one object. $\endgroup$
    – Enkidu
    Mar 30 at 6:01
  • $\begingroup$ @imtrying46, the Wikipedia article seems to be careful to distinguish multiplication ($x \cdot y$) from scalar multiplication $ax$ and multiplication of scalars $ab$. I don't think it follows from their definition that $A$ isn't a $k$-algebra. $\endgroup$ Mar 30 at 6:02
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    $\begingroup$ Wikipedia disallows your example since $\mathbb{Z}$ is NOT a $\mathbb{F}_2$-vectorspace $\endgroup$
    – Enkidu
    Mar 30 at 6:04
  • $\begingroup$ @Enkidu ... Ah, okay, that's probably where I'm messing up then. Also, wait, I think that an $R$-algebra is a pair of a ring $A$ and a morphism from $R$ to the endomorphism ring of $A$, right? Not from $R$ to $A$ itself. $\endgroup$ Mar 30 at 6:06
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    $\begingroup$ it is from $R\to A$ you literally want to interpret scalars as elements (for matrices these are identical diagonal matrices for example), what you are thinking is a representation. Then endormorphism ring of a representation is an "algebra" over your algebra/groupring (they are rarely commutative, and if they are, they usually are boring). $\endgroup$
    – Enkidu
    Mar 30 at 6:08

2 Answers 2


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 $\lambda \mathrm{id}$).


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.


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