Ideals in an algebra over a field as subspaces

Question

I was given a homework question about finite dimensional algebras over a field. In my algebra class, we define an $\mathbb{F}-$algebra (where $\mathbb{F}$ is a field) $A$, to be a ring $A$ Together with a ring homomorphism $\varphi:\mathbb{F}\to Z(A).$ This is the definition that Dummit and Foote uses so I suppose it is the usual definition (though I see many places define an algebra with a bilinear map). My question is the following: Are ideals in an $\mathbb{F}-$algebra by definition sub spaces of $A$ As an $\mathbb{F}$-vector space or is there something that requires a proof here? Moreover, when we have an $\mathbb{F}-$algebra, can questions about ideals be turned into questions about sub spaces?

Answer 1

If $B\subseteq A$ is any extension of rings, and $M$ is a right $A$ module, then it is automatically a right $B$ module just by restricting "scalars." All the axioms hold for $B$ because they hold for $A$. (Tack on appropriate caveats for identity and unitary conditions.)

In particular for an algebra defined as you have, $F\cong\phi(F)\subseteq A$ falls into this case. Any ideal $I\lhd A$ automatically gains an $F$-vector space structure by identification with the subring isomorphic to $F$.

Of course, there may be many $F$-subspaces that are not ideals at all. Subspaces do not have to be closed under multiplication by arbitrary ring elements. For example, in $M_2(\mathbb R)$ the right ideals all have even $\mathbb R$ dimension, because the simple submodules are all $2$-dimensional, and the other submodules are direct sums of those. But there are undoubtably odd-dimensional subspaces in there too, they just aren't left/right/two-sided ideals.

You may ask if left vs right multiplication matters, but the answer is "no" since the image of $\phi$ was required to be central in $A$.