# Why is this function neither surjective nor injective?

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.

• Are you assuming $\;R\;$ is a unitary ring? – DonAntonio Apr 27 '18 at 21:01
• A function from a finite set to itself is injective if and only if it is surjective. – Antoine Giard Apr 27 '18 at 21:02
• @JamieCarr That's what "unitary ring" means: a ring with a multiplicative unit/ – DonAntonio Apr 27 '18 at 21:04
• @DonAntonio Yeah obviously, thank you! – Antoine Giard Apr 27 '18 at 21:05
• @DonAntonio XD! – Jamie Carr Apr 27 '18 at 21:06

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.