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If $A_1$ and $A_2$ are algebras over a field $F$, and if they are isomorphic as vector space over $F$, can we say that these algebras are isomorphic?

(one may assume that algebras are finite dimensional, if necessary; I don't know about it. But I am just wondering about vector space isomorphism implies algebra isomorphism. )

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Yes, we can say that they are isomorphic as vector spaces. But how about $\mathbb{C}$ and $\mathbb{R} \times \mathbb{R}$ as $\mathbb{R}$-algebras? They certainly aren't isomorphic as algebras. – t.b. Jan 10 '12 at 8:11
@t.b.: your comment is correct; but in a question on Mackey's criteria on Mathoverflow, the first answer does a similar kind of work i.e, there, showing isomorphism of vector spaces will imply isomorphisms of group algebras in the first answer to question. (Link: – Radk Jan 10 '12 at 8:16
It does not seem to be correct: it is correct. – Mariano Suárez-Alvarez Jan 10 '12 at 8:18
up vote 1 down vote accepted

No. As soon as $n$ is large, there are many, many non-isomorphic algebras of dimension $n$.

For example, the algebras $F\times F$ and $F[x]/(x^2)$ are isomorphic as vector spaces (they both have dimension $2$) but they are not isomorphic as algebras.

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This is the kind of question which you yourself could have answered by looking for examples... – Mariano Suárez-Alvarez Jan 10 '12 at 8:13
Yes. But I didn't think abou it because of some arguments in the first answer to a question ("…; ) – Radk Jan 10 '12 at 8:18
In that context, Vladimir has a morphism of finite dimensional algebras $A\to B$. If you prive that it is an isomorphism of vector spaces, then it follows that it is an isomorphism of algebras. But you have to start with a morphism of algebras. (This is a classic example where explaining the motivation for your question would have allowed more useful answers...) – Mariano Suárez-Alvarez Jan 10 '12 at 8:21
how this vector space isomorphism is sufficient for isomorphism of algebras? (I confused about it). – Radk Jan 10 '12 at 8:22
Frobenius reciprocity has absolutely nothing to do with it, really. – Mariano Suárez-Alvarez Jan 10 '12 at 8:29

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