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Here is the question I want to answer:

Let $F$ be a field and $D$ an integral domain which is a finite dimensional vector space over $F.$ Prove that $D$ is also a field.

Here are my thoughts:

Since $F$ is a field, then every nonzero element in $F$ has a multiplicative inverse. Since $D$ is an integral domain, then it has no zero divisors i.e., if $ab = 0$ then either $a=0$ of $b =0$ for every $a,b \in D.$ I do not really understand why we need the assumption that $D$ is a finite dimensional vector space (in case there is a proof simpler than the one given in the question mentioned below). Do I have to define a map as given in the second answer here Let $R$ be an integral ring and let $F$ be a field contained in $R$. $R$ is a vector space over $F$. Show that $R$ is a field. or is there a very simple way of proving this question?

Brain
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  • $D$ being finite dimensional implies it is integral over $F$. Now use https://math.stackexchange.com/questions/450020/integral-extensions-of-rings-when-one-of-the-rings-is-a-field. – Pedro May 12 '22 at 13:24
  • There is indeed a very simple solution. Saying that $D$ is a field is equivalent to saying that all the $F$-linear maps $x\mapsto ax$ for $a\in D$ are bijections. Can you use the two hypotheses to prove that? – Captain Lama May 12 '22 at 13:27
  • This is probably one of the most duplicated questions in the ring-theory tag. – rschwieb May 12 '22 at 13:30
  • @CaptainLama so there is no way to solve it without defining this linear map? – Brain May 12 '22 at 14:29
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    @Brain Of course there are other ways. You can just note that if ${b_1,\ldots, b_n}$ is a basis then so is ${ab_1,\ldots, ab_n}$, and therefore there is a linear combination $\sum \lambda_iab_i=1$, whence you have an inverse for $a$, namely $\sum \lambda_ib_i$. This all assumes $a$ is nonzero, naturally. This used to be at one of the duplicates but I think possibly the associated post was deleted. – rschwieb May 12 '22 at 14:54
  • I want the easiest solution ..... is this only the solution? @rschwieb – Brain May 12 '22 at 18:17
  • @Brain Uh... I just told you a second solution so obviously there's no "only" solution. Both solutions are pretty dead simple. I'm not sure what more you're asking for at this point. – rschwieb May 12 '22 at 19:18
  • No I meant ..... is your second solution is a whole solution or there are more details I should fill in in it?@rschwieb – Brain May 13 '22 at 02:28

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