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Reading answers for one of my homework, I found a statement:

Let $R$ be a finite commutative ring, $a \in R$, and $F\colon R \to R$ is given by $F(b) = ab$. If a is not a unit, then $F$ is not surjective. So as $R$ is finite, $F$ is not injective.

How does the implication work here? I understand the definitions of injection and surjection but I find it difficult to apply them here. Why does the finiteness of a ring imply its injection? Hope some experts can help me.

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  • $\begingroup$ Are you assuming $\;R\;$ is a unitary ring? $\endgroup$ – DonAntonio Apr 27 '18 at 21:01
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    $\begingroup$ A function from a finite set to itself is injective if and only if it is surjective. $\endgroup$ – Antoine Giard Apr 27 '18 at 21:02
  • $\begingroup$ @JamieCarr That's what "unitary ring" means: a ring with a multiplicative unit/ $\endgroup$ – DonAntonio Apr 27 '18 at 21:04
  • $\begingroup$ @DonAntonio Yeah obviously, thank you! $\endgroup$ – Antoine Giard Apr 27 '18 at 21:05
  • $\begingroup$ @DonAntonio XD! $\endgroup$ – Jamie Carr Apr 27 '18 at 21:06
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If $\;f\;$ is surjective, then there is $\;b\in R\;$ s.t. $\;ab=1\;$ (assuming $\;R\;$ is unitary) , and thus $\;a\;$ is a unit...contradiction.

And now just observe that a function from any finite set fo itself is injective iff it is surjective.

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  • $\begingroup$ Thanks for your answer! I am just bad at these logics..... $\endgroup$ – Jamie Carr Apr 27 '18 at 21:15

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