# A module over an algebra. Is it a vector space?

Let $A$ be an algebra over a field $k$. I would like to know if my understanding of the following correct or not.

What I want to clarify is the definition of a module $M$ over $A$. I know the definition of a module over a ring.

1. Is the definition of a module over an algebra $A$ the same as the ring theoretic definition replacing a ring by an algebra?

2. Or, a module is a $k$-module, plus $A$ action?

3. Or, are they the same?

If it is the definition 2 above, then $M$ is a vector space over $k$. Is $M$ a vector space in the case of 1?

• A small caveat. Let $M$ be a $k$-module. Suppose moreover that $M$ is an $A$-module. Then this last $A$-action induces another $k$-action that need not coincide with the original one. This reminds me of this: math.stackexchange.com/questions/889130/… Commented Jan 2, 2016 at 16:27
• Another thought: one could think that one could make a "relative" version of an $A$-module, as in that post: an $A$-module relative to $k$ should be a $k$-module $M$ with an action of $A$ such that the action of $k$ it induces coincides with the original one. But it's a boring notion, unlike the bimodule case, where it makes more sense: in that scenario, you have two induced actions of $k$, so it makes sense to require that they're equal (and moreover, equal to a given one). Commented Jan 2, 2016 at 16:37

1. Yes. A module $M$ over $A$ is just a module $M$ over the ring $A$; the additional structure of $A$ as a $k$-algebra plays no role.
2. (and 3.) That amounts to the same: since $A$ is a $k$-algebra, you already have a map $k \to A$ which turns $M$ into a $k$-module as well.
$M$, ultimately being a $k$-module as well, is a $k$-vector space. I'm not sure what you mean by "Is $M$ a vector space in the case of 1?"
• I meant that if $M$ is defined in the 1st way, I wanted to know if $M$ is a vector space. You answered this question too. Thank you!
• @Snow Just to spell out what the second point is describing here: an $R$ module $N$ amounts to a ring homomorphism of $R$ into the ring of abelian group endomorphisms $End(N)$. Since $A$ is a $k$ algebra, there is a ring homomorphism of $k$ into $A$ (actually into the center of $A$). So when you make $M$ into an $A$ module, you've got a ring homomorphism $A\to End(M)$, and composing this with $k\to A$ you can see that $M$ is both an $A$ and a $k$ module. Commented Mar 12, 2014 at 13:00
• What would the ring homomorphism of $k$ into $A$ be? I can imagine the ring homomorphism $A\to\operatorname{End}(M)$ to be $a\mapsto \gamma_a$, where $\gamma_a(x)=ax$, for $x\in M$. However, not sure about the case for a $k$ algebra. Commented Nov 23, 2019 at 13:06