# constructing an algebra

Given generators and relations, would the following process construct an F-algebra?

1. Create a group-ring for each generator, with the field F as the ring and the group generated by the generator
2. Take the tensor product of these vector spaces
3. Mod out by the ideal generated by the given relations

1. Duquec

I don’t see any problems with your construction . . . in fact, if all you want to do is construct an $\mathbb{F}$-algebra, you could stop at 2.

However, on Step 2, I suspect that what you actually mean is “Take the tensor product of these rings.”

2. briansdumb

My goal was to construct an algebra satisfying the conditions given.
While it’s true that they are rings, I think they are also vector spaces over the field F.
The reason I suggest modding out by the ideal (step 3) is to encode into the algebra the relations originally given.

3. Brian Cruz

If I’m picturing what you’re saying properly, then each group ring would have all linear combinations involving all powers of its associated generator. Creating a tensor product of these would have to be done with care, as elements in the group could, and in general do, contain more than one instance (non-contiguously) of any given generator. We would have to include tensor products of all degrees considering all different orders of the vector spaces to take this into account. This is what we have to deal with if the group isn’t abelian.

I think we can start by generating a group of all the generators over multiplication at once (all finite products using any generators in any order), and then created a group ring using F as the group to take into account all the linear combinations. This seems like it’s the same as what is described in the comment for problem number 6: to consider the ring generated by the the generators as well as (the elements of) F.

If the relations only concern the group objects (elements of the group and multiplications) the modding could probably happen at the group level (like in Dummit Foote Sec. 5.4). Otherwise, the relations must generate an ideal to mod the algebra itself (like DF 11.5)

4. Brian Cruz

Responding to my own first paragraph: If the point is to use modulos to get commutativity later, then at the beginning we would want to assume we’re working with a non-Abelian group anyway.

5. briansdumb

My goal was to extend the method we used to construct tensor products from direct products of vector spaces by modding out by an ideal generated by some relations. In the case of the tensor product, we were given relations in the lecture notes, and he shows the construction of the ideal.
Whatever structure you want to impose on your algebra, I think you should be able to generate an ideal and mod out to create it from the generators.

6. Brian,
I interpret your question to mean this:
Suppose we have generators G and relations R and want to use them to present an algebra over a field K. What K-algebra are we talking about?

Take the free vector space V=KG with G as a basis, and consider the tensor algebra TV. Notice that TV is the “free K-algebra” generated by G. A basis is given by the finite words in the “alphabet” given by the generators G. We are not assuming that the generators commute.

Now let’s introduce the relations. They are given as non-commutative polynomial equations in G (such as g^2=1, x^2=0 or xg=-gx) and they will define a two-sided ideal I (in this case the ideal generated by g^2-1, x^2, xg+gx).

Then the algebra with generators G and relations R is the quotient TV/I.

• Brian: This construction is close to what you tried, but there are a couple of issues:
1. If you want to take the “group ring” of a generator g, you have to make g into a group. The natural thing would be to take the free group generated by g, but for that you would need to introduce the inverse of g, which we don’t want to do here.
2. You could instead try to take the polynomial ring K[g] with a generator g. But then when you tensor two of these, say K[g] and K[h], you will get the polynomial ring in two variables K[g,h] instead of the tensor algebra generated by g and h. Thus you would have no way of distinguishing gh from hg, which we want to do.

• Briansdumb

1) I wonder if, in your particular example, modding out by $g^2+1$ would force $g^{-1}$ to equal $g$? which is as we want?