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?
Ideals in an algebra over a field as subspaces
abstract-algebraalgebraslinear algebraring-theory
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It only gets more ambiguous as you go along. Somewhere down the rabbit hole you learn that "algebras" might also mean general systems like monoid, loops, and quasi-groups. The point is that calling something "an algebra" is a context dependent definition. Authors should be responsible to explain what assumption they intend for their meaning of "an algebra".
Even so there are some top-level popular choices.
- When in doubt, default to associative. This is going to upset those who think of Lie algebras the most, but it is clear that to get to Lie theory you first would have been around associative algebras. So on that measure I think the correct default is to say products are associative and if they are not then you should, as an author, point that out very early on. For example, the AMS subject classifications list "nonassociative algebra" to emphasize that the algebras are possibly not associative. (Not non-associative means "not required to be associative" so it is larger class, not a complementary class of algebra).
When dealing with modules. If you have module, especially if everything is a vector space, you have some ring $K$ of scalars for you modules. So often we want everything in sight to remain linear, or bilinear, or multilinear, with respect to $K$. In the context it is most common to encounter $K$-algebra as a $K$-module $A$ equipped with a distributive product $*:A\times A\to A$ that is $K$-bilinear. That way the natural things you do, say left multiply $L_a(x)=ax$, right multiply $R_a(x)=xa$, remain $K$-linear. Now as for the associative assumption that depends on the modules you are studying. Example, if you modules are associative modules then sure, assume associative $*$. But if you have Lie modules then you would not want $*$ to be associative.
When dealing with finite or combinatorial forms of algebra. Here algebra probably means something more like generic names for products of any sort, say monoids, loops, etc. The name comes from some correspondences that arrose early on which showed combinatorial methods explained the working of the algebra (e.g. Magnus' treatment of free groups he called "Combinatorial Group Theory")
Some observations to make about the $K$ in a $K$-algebra (both associative and non-associative).
Every distributive product is $\mathbb{Z}$-bilinear and so every ring is a $\mathbb{Z}$-algebra.
Every bilinear product has a unique maximal ring $K$ of scalars over which it is bilinear (unique upto natural isomorphism). This is known as the centroid and can be computed by solving a system of linear equations. So long as your product has no degeneracy ($A*A=A$ and $a*A=0$ implies $a=0$, e.g. if $A$ has a $1$ or is simple) then the centroid is commutative. So it is common to assume $K$ is commutative to start with.
With those two points it means that insisting that something is a $K$-algebra instead of ring is mostly for convenience. You simply cannot make a distributive product that is not an algebra. You are just deciding to make that point evidently clear.
You can either define a $R$-algebra $A$ to be a ring with $R$-module structure, such that the multiplication is an $R$-linear if you fix a component. Or in case that $A$ has $1$ you can require $A \otimes_R A \to A$ to be $R$-linear, since then $A$ has a natural $R$-bimodule structure.
I think this definition has some disadvantages. First it depends on the definition of ring. Is it associative, commutative, has it a unit? Rings often are assumed to have a unit and are mostly associative. In case of an algebra you don't want this always.
An other point is that in most cases you have modules which you equip with a multiplication. For example in the case of polynomials, they have a quite natural module structure and you then define a multiplication.
So I prefer the definition a $R$-algebra is a $R$-module $A$ equipped with a $R$-bilinear map $A\times A \to A$. And then you can require this map to be associative, unital, commutative or whatever you need.
Now to answer you question. Mostly I guess you have a modules and equip it with an algebra structure. Then you usually want the natural structure of the module. And in the definition of algebra is the module structure included, so if you have an algebra that data is fixed.
You could define an other algebra from that by defining an other scalar multiplication, but this then is an other algebra.
Of cause you can also take a ring and then define the module structure, but then you should't think about the natural module structure, which this algebra may also has. Just the one you choose wich is included in the data of the algebra.
Best Answer
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 $A$.