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Quick proof that every finite integral domain is a field.

I'm not sure if I'm missing anything, but here is my attempt.

Let $R$ be a finite integral domain. So it is unital and commutative. Since $R$ is finite, it is of the form $R= \{a_1,a_2,...,a_n\}.$ Let $0\neq a\in R.$ Then if $aa_i = aa_j,$ we must have $a_i=a_j$ (because $R$ has no zero divisors). Since $R$ is closed under multiplication, $R= \{aa_1,aa_2,...,aa_n\}.$ Since $R$ is unital, there exists $a_i\in R$ such that $aa_i=1.$ Thus $R$ is a field.

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    This is indeed the idea. You can phrase in an other way as well: $a \mapsto ba$ is injective for every $b$. Since an injective map between two finite set of same cardinality (here, $R$) is bijective, this map is bijective so there exists $a$ such that $ba = 1$. –  Oct 30 '19 at 22:26
  • @RobinCarlier let me just confirm if I understood you right. $a\mapsto b\cdot a$ is the same as $R\mapsto R,$ right? And so this is bijective. $R$ is unital, so $b\cdot a$ can be $1.$ Since the map is bijective, $\exists a$ such that $b\cdot a=1.$ –  Oct 30 '19 at 22:37
  • The notation $a \mapsto b.a$ is just a shorthand for "the map R \to R that sends $a$ to $a.b$. And yep, that's it! –  Oct 31 '19 at 09:49
  • Your argument rehashes a handful of the solutions at the duplicate. – rschwieb Oct 31 '19 at 17:42

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