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Note: I have solved the question but there is this other line of attack that's giving me no hint on how to proceed. This is an attempt to solve this question in the following manner:

If we start with with the fact that an element $a \in R$ is not a unit element in the finite commutative ring, how can we prove that it has to be a zero-divisor?

I don't know why it has been voted down, because I seriously think that it does involve some imagination. I don't want to just solve the problem. My intent here was to start with the fact that we have a non-unit element $a$ and then prove that it is a zero-divisor. [moving from non-unit to proving a zero-divisor is what I am looking for]

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    Show how you solved the problem so that we can know what a new approach would be. And in the link you wrote there are several very, very nice answers. Do you want something different to them all ? – DonAntonio Sep 24 '16 at 10:58
  • I solved it by starting with the fact that if $a$ is not a zero divisor, then it is an unit element and that for some $k \in \mathbb{Z} , ; a = a^{k}$ (since it is finite ring) and that gives $a^{k-2} $ as its inverse element. Now, Proving the question I wrote above, is logically equivalent. Since, Here I have proven that If $a$ is not a zero-divisor then it is an unit element which is same as proving that if $a$ is not an unit element then it is a zero divisor but I don't know how to start with this fact and then prove the result. – MUH Sep 24 '16 at 11:04
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    Did you read the answers in the link you gave? I think all possible more or less reasonable approachs appear there. – DonAntonio Sep 24 '16 at 11:05
  • The mapping argument proves the this, but I am not sure that one should prove this fact by just showing it for one functional mapping. I guess we should prove this that it is valid for all mappings (why mappings even, function is a special kind of relation, so I am not even sure about this as we don't need the relation to be a function). That is why I am not convinced with the mapping $x \rightarrow ax$ to prove the result. – MUH Sep 24 '16 at 11:12
  • You lost me: I can't see how that, perhaps the most basic answer in that link, doesn't make it: the map $;x\to ax;$ is either surjective or not. if $;a;$ is not a zero divisor then it is surjective and thus also injective (finiteness) and etc., as written there. – DonAntonio Sep 24 '16 at 11:24
  • My doubt was that we have taken a very specific map i.e. $x \rightarrow ax$ and showed that for this mapping, the result hold true. See that the methods they used is based on this mapping. Now, this result is general and independent of any mapping. Hence, we must prove this result either for all mapping or find other method to proceed. \ Any way, How did you infer that if $a$ is not a zero-divisor then the map is surjective. – MUH Sep 24 '16 at 11:39
  • Because then $;ax=0\implies x=0\implies;$ the map is injective and thus surjective... – DonAntonio Sep 24 '16 at 11:41

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