What is difference between free $\mathbb{F}$-algebra $A$ and vector space $V$ over $\mathbb{F}$?

Since the algebra is free, it has basis and so it is a vector space over $\mathbb{F}$.

So we have to look what more property the algebra $A$ hold.

Both the vector space $V$ and the algebra $A$ hold scalar multiplication and addition.

Also $a,b \in A$ implies $ab \in A$ but $a, b \in V$ does not imply $ab \in V$.

So this is the difference.

Am I right?

Now if I omit the freeness then what will be the difference between the algebra and the vector space ?

help me


Yes. The fundamental difference between a vector space and an algebra is that elements of an algebra may be multiplied with one another, while elements of a vector space can't.

$a, b \in V$ does not imply $ab \in V$.

It's more fundamental than this, though. It's not that $ab$ might happen to lie outside of $V$ if $V$ is a vector space. It's that $ab$ doesn't really make sense at all if $V$ is just a vector space.

The definition of "free" also changes when you go from vector spaces to algebras, because you can multiply elements of an algebra together. "Free" still means that there is a generating set so that any element in the vector space / algebra can be written in exactly one way using the elements of the generating set, the base field, and the allowed operations. For vector spaces, that's just vector addition and scalar multiplication ("linear combinations"), but for algebras, that also includes multiplying the generators with one another.

So for instance, $\Bbb C$ is a free $\Bbb R$-vector space (with standard basis $\{1, i\}$), but not a free $\Bbb R$-algebra (using standard operations), because if we try a generating set like $\{i\}$, then $1\in \Bbb C$ may be written both as $-i^2$ and as $i^4$. And this kind of failure will happen no matter which generator we try.

  • $\begingroup$ thank you very much sir. It cleared my understandings $\endgroup$ – M. A. SARKAR Aug 7 '19 at 12:01

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