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Let $V = (V, +, \cdot, *)$ be an $n$-dimensional associative algebra over an algebraically closed field $F$. Let $\{v_1, \dots, v_n\}$ be a basis for $V$. I am asked to show that, for each $i = 1, \dots, n$, there are only two possible values that $v_i * v_i$ can take: either $0$ or $v_i$.

Actually the exercise I am trying to solve is only considering the case $n = 2$, but I guess the more general case is also true and not much more difficult.

Maybe related to this?

Tom
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    Actually it's pretty clearly false, consider $\mathbb{C}[x]/(x^2-2x)$. Or any example really, and if $v_i^2=v_i$ then change $v_i$ into $2v_i$. – Captain Lama Nov 21 '19 at 17:54
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    Yeah, I think you might mean that there exists a basis ${v_i}$ with this property. It is certainly not true for all bases. – Thomas Andrews Nov 21 '19 at 17:56
  • @CaptainLama you mean false even if n = 2? – Tom Nov 21 '19 at 17:56
  • @Tom Yes. But maybe Thomas in on to something. It might be existence that you are looking for. – Captain Lama Nov 21 '19 at 17:57
  • I'll copy paste the text of the exercise, verbatim, word by word – Tom Nov 21 '19 at 17:58
  • Let $F$ be an algebraically-closed field and let $(K,*)$ be an associative $F$-algebra having a basis ${v_1, v_2}$ as a vector space over $F$. Show that $v_2^2 = v_2$ or $v_2^2 = 0_K$. – Tom Nov 21 '19 at 18:00
  • @CaptainLama This is exercise 188 on page 82 of Golan - The Linear Algebra a Beginnign Graduate Student Ought to Know – Tom Nov 21 '19 at 18:02
  • @ThomasAndrews maybe I misinterpreted the text? – Tom Nov 21 '19 at 18:03
  • No, your interpretation is correct, the exercise is just false as it is written. It's very likely it actually should say "there is a basis such that", which is true. – Captain Lama Nov 21 '19 at 18:04
  • Hard for me to tell, since I don't have the original source. If you post the source question in English, it might help. – Thomas Andrews Nov 21 '19 at 18:05
  • @ThomasAndrews I just did – Tom Nov 21 '19 at 18:05

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Since you posted the exact wording of the problem in the comments, we can establish that as it is written, the exercise is just false. It's pretty easy to see: take $V=F\times F$, and take the basis $v_1=(2,0)$ and $v_2=(0,2)$. Then clearly $v_i^2=2v_i$.

On the other hand, what is true, and probably what the exercise is about, is that there exists a basis such that $v_i^2=0$ or $v_i$. This is another way of saying that there are exactly two $2$-dimensional algebras over $F$, up to isomorphism: $F\times F$ and $F[\varepsilon]=F[x]/(x^2)$.

To see that, note first that there is a natural embedding $F\subset V$ since $V$ has a unit, and we can take $v_1=1$ as our first basis element. Then take any $x\in V\setminus F$, and check that it satisfies a quadratic equation with coefficient of $F$. Then try to see that up to an affine transformation, you can assume that this equation is either $x^2=0$ or $x^2=x$.

Captain Lama
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  • Thanks a lot! I m actually struggling to understand. For example: why are you assuming that the algebra in question has a unit? – Tom Nov 21 '19 at 18:47
  • Oh yes sorry, I thought that was in the question, but actually it's not. I'll have to modify my answer then. – Captain Lama Nov 21 '19 at 18:55
  • ok I ll appreciate it! – Tom Nov 22 '19 at 16:49
  • I had been stuck on this same question for hours until I found this page! However this solution still seems incomplete. The assumption that there is a unit element has still not been justified. It. seems you need a so called "unit theorem" (which seems to require considerable amounts of further effort to prove) to justify it: https://pure.mpg.de/rest/items/item_3254490/component/file_3254491/content – Just_a_fool May 31 '22 at 13:13
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    @Just_a_fool I don't think this "unit theorem" says what you think it does. I think you have been misled by the unfortunate use in English of the word "unit" to mean "invertible element" and not "neutral element for the multiplication". You are right that in my proof I assume the algebra is with unit (it seemed obvious to me that it was part of the assumptions, but it is indeed not actually written in the question). But the theorem in your link also assumes that, and it then shows that certain subgroups contain invertible elements. – Captain Lama May 31 '22 at 13:58
  • Thanks for the reply! I see, so you are saying that in the linked paper, they use the term "unit" to refer to, simply, an invertible element (not necessarily an element $e$ s.t. \forall $v \in V, v \bullet e = e \bullet v = v$). However, your comment that all two dimensional algebras are isomorphic to one of two unital algebras implies that units can always be found. So how could we actually show this? Is there an elegant way? I have tried via brute force (first showing that our algebra must be commutative and then trying to get a handle on all $v_i \bullet v_j$ products, but is gets so messy. – Just_a_fool May 31 '22 at 14:25
  • In short, I've tried to show that there will always be some $\alpha, \beta \in F$, s.t. $\alpha v_{1} + \beta v_{2}$ acts as a unit, that is: $(\alpha v_{1} + \beta v_{2})\bullet v = v \bullet (\alpha v_{1} + \beta v_{2}) = v $, but the resulting equations I get blow up a bit (with no near solution in sight) and I wonder if there is a standard/canonical way of handling this question. – Just_a_fool May 31 '22 at 14:34
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    @Just_a_fool As I said in my previous comment, I do assume that the algebra is unital. It is not true that all $2$-dimensional algebras are unital: just take the algebra with the zero product. On the other hand it's not too difficult to show that even if the algebra is not unital there is still always a basis consisting in nilpotents or idempotent elements. Maybe I'll edit the answer if I have time. – Captain Lama May 31 '22 at 17:11