2

Let $A$ be an integral domain that is an algebra over a field $K$. Show that if $A$ is finite-dimensional as a $K$-vector space, it is a field. Is the converse true?

Obviously the converse isn't true in general, just set $A:=\mathbb C$ and $K:=\mathbb Q$. For the other direction, set $r:=\operatorname {dim}_KA$ and assume that $r\ge 2$ (if $r=1$, then $A\cong K$ and we'd be done). So $A\cong \bigoplus_{i=1}^r(K)_i$, and in particular $A$ contains $e_1:=(1,0,\dots , 0)$ and $e_2:=(0,1,\dots , 0)$. These two non-zero elements are such that $e_1e_2=0$, contradicting the hypothesis that $A$ is an integral domain, and so $r$ must be $1$.

It makes me suspicious is that $A$ ends up to be $K$ and not a field in general, but it should be correct, do you confirm it?

Dr. Scotti
  • 2,493

2 Answers2

4

To provide an alternative (faster) proof than the method suggested by tomasz, you may note that $A$ being integral means that for any $a\neq 0$, the application $A\to A$ defined $x\mapsto ax$ has a trivial kernel. Then you can think about what that implies when $A$ is finite-dimensional.

Captain Lama
  • 25,743
  • Another exercise in the same section (1.4, Bosch) says that any surjective endomorphism of finite $K$-modules is also injective; since the $K$-modules are an abelian category, can I take for granted that the dual holds, and so that an injective endomorphism (of finite $K$-modules) is also surjective? In that case for any $a\neq 0$ we have a $x\in A$ such that $ax=1$. – Dr. Scotti Jan 15 '22 at 15:05
  • 1
    This exercise is probably formulated for modules over commutative rings. For vector spaces, it's much easier, there is no need to go with deep notions such as abelian categories, you just need the rank formula (or however it is called where you're from). – Captain Lama Jan 15 '22 at 17:58
  • I think that in a case like this, where it has been already well established that this is a frequently duplicated question with many existing answers including this one, one should not duplicate solutions from duplicate questions. One would just write as a comment "the fast way to prove it is here: link" – rschwieb Jan 21 '22 at 19:53
2

It's not correct. $A$ is only isomorphic to $\bigoplus_{i=1}^r(K)_i$ as a $K$-vector space, not as a ring. It is also easy to find counterexamples: take $A=\mathbf Q[\sqrt 2]$, $K=\mathbf Q$.

To prove the result, notice that for any basis $\alpha_1,\alpha_2,\ldots,\alpha_n$ of $A$ over $K$, you have $A=K[\alpha_1,\alpha_2,\ldots,\alpha_n]$, so (by induction) it is enough to consider the case when $A=K[\alpha]$ for some $\alpha\in A$. Then show that $K[\alpha]$ is a field if it's finite dimensional.

tomasz
  • 35,474