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We know $\text{an algebra over a field is a vector space}$.

So what is the basic difference between an algebra over a field and a vector space over a field ?

what is the difference between an algebra over a field and a group algebra ?

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MAS
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An algebra over a field is, even if you ignore the field, a ring. Therefore, it has both addition and multiplication. Additionally, of course, multiplication with scalars is also defined. So, for $a,b\in A$ and $\alpha \in K$, where $A$ is an algebra over a field $K$, the terms $a+b, ab, \alpha a$ are all defined, and all elements of $A$.

A vector space, on the other hand, is (ignoring the field) merely an (Abelian) group - it has only one operation defined on it, addition. $a+b$ and $\alpha a$ are defined, but $ab$ is not.

A group algebra is a particular case of an algebra over a field. That is, for a field $K$ and group $G$, $K(G)$ is an algebra over $K$. Not all algebras over $K$ need to be group algebras, however.

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  • so we can not consider an algebra as a vector space – MAS Jul 31 '19 at 08:09
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    @M.A.SARKAR Yes, every algebra over $K$ is a vector space over $K$, as even a short google search would tell you: http://mathworld.wolfram.com/Algebra.html – 5xum Jul 31 '19 at 08:11
  • ok, nice. Thanks – MAS Jul 31 '19 at 08:11
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    @M.A.SARKAR An algebra over a field is a vector space - but it is a vector space plus a bilinear product operation. – gandalf61 Jul 31 '19 at 08:21
  • @gandalf61, thank you – MAS Jul 31 '19 at 08:22
  • @gandalf61, so to show an algebra over a field, we have to at first satisfy vector space properties and then satisfy billinear product. Is it ? – MAS Jul 31 '19 at 08:31
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    @M.A.SARKAR You have to show that it is a vector space, and that it is a ring. I don't know what you mean by "bilinear product". – 5xum Jul 31 '19 at 08:33
  • @5xum, actually the comment by #gandalf61, says about billiear product above – MAS Jul 31 '19 at 08:41
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    @M.A.SARKAR See this Wikipedia article. A ring has two bilinear products - addition and multiplication. In an algebra bilinear addition is part of the vector space properties, and you also have to show that the multiplication operation is bilinear too. – gandalf61 Jul 31 '19 at 08:49
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It is not correct to say that an algebra over a field is a vector space. Rather, it is a vector space plus certain linear maps defined on this space that satisfy some requirements (depending on the definition, associativity, commutativity etc.).

So if you have an algebra, it gives you a well-defined vector space but a vector space does not give you a well-defined algebra, the data of the multiplication maps really matters. You can have two non-isomorphic algebras with isomorphic vector spaces, for example over $\mathbb{R}$ there is $\mathbb{R}[x]/(x^2+1)$ and $\mathbb{R}[x]/(x^2)$ (both are 2-dimensional as vector spaces but one has nilpotents and the other does not, so they are not isomorphic as rings). See here for some examples over an algebraically closed field.

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    It is correct to say that an algebra over a field is a vector space. Just like it is correct to say that a person from Germany is a person from Europe. – 5xum Jul 31 '19 at 08:09
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    @5xum there is "is" of predication and "is" of identity. An algebra over a field is a vector space, where "is" is "is" of predication. An algebra over a field is not a vector space, where "is" is "is" of identity. I think therefore that your comment does not contradict my answer, if that was the implication. –  Jul 31 '19 at 09:05
  • @5xum the analogy you suggest is not perfect, it seems, because two different people from Germany usually do not define the same person from Europe (whereas algebra vs vector space is about structure, German vs European is about property: https://ncatlab.org/nlab/show/stuff,+structure,+property). –  Jul 31 '19 at 11:34