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This question already has an answer here:

If a nonzero element $a$ of a ring $\mathbb{D}$ does not have a multiplicative inverse, then $a$ must be a zero divisor.

If $\mathbb{D}$ has finitely many elements, then the statement is true. However, I don't see why, as for the infinite case, 7 isn't a zero divisor only because every element $a > 0$ when multiplied by a $b > 0$ gives you $ab > 0$ which has nothing to do with infinity or finite amounts of elements.

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marked as duplicate by Juniven, Daniel Fischer Mar 20 '17 at 15:35

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    $\begingroup$ You're right, in $\mathbb{Z}$ zero is the only zero-divisor, $1, -1$ are the only invertible elements, and everything else is neither/nor. Perhaps your ring has some extra conditions? $\endgroup$ – Andreas Caranti Sep 14 '13 at 21:18
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If $\mathbb D$ is finite, with $n$ elements, and $a\in \mathbb D$ does not have a multiplicative inverse, we consider the elements $$ad_1, ad_2 \dots ad_n$$where the $d_i$ are the $n$ distinct elements of $\mathbb D$. None of these elements is $1$, so the $n$ elements can have at most $n-1$ different values - so two of them must be equal.$$ad_i=ad_j \implies a(d_i-d_j)=0$$Since $d_i\neq d_j$, $a$ is a zero-divisor.

Note that this does use the finiteness assumption in an essential way. $2\in \mathbb Z$ does not have a multiplicative inverse, and is not a zero divisor. There are a number of proofs like this (eg a finite integral domain is a field, not true in the infinite case) which depend on being able to count the elements of a finite object.

Note that in the ring of integers mod $14$, $7$ does not have a multiplicative inverse, and $2\times 7=0$. In the ring of integers mod $15$ we have $7\times 13 =1$ and $7$ is not a zero divisor.

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  • $\begingroup$ Excellent @Mark Bennet. Just great. $\endgroup$ – Don Larynx Sep 17 '13 at 17:06
  • $\begingroup$ Let $ad_1 = 2, ad_2 = 4, ad_3 = 5$. Then three elements have three distinct values, so two of them are not equal. $\endgroup$ – Don Larynx Sep 17 '13 at 17:23
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    $\begingroup$ @Jossie The point is that there are $n$ products, and there are $n$ elements in the ring. But none of the products can be $1$, so there are only $n-1$ possible values. So two of the products have to be equal. You only get three products if there are only three elements in the ring. $\endgroup$ – Mark Bennet Sep 17 '13 at 17:54
  • $\begingroup$ I just showed that there are three possible distinct values in a set of three elements...I remain confused. $\endgroup$ – Don Larynx Sep 17 '13 at 20:53
  • $\begingroup$ @Jossie But when you go to $n$ elements there are not enough for all of them to be the same. $\endgroup$ – Mark Bennet Sep 17 '13 at 21:05
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Hint: Consider the map sending $r \to ra$. Use the fact that $a$ does not have a multiplicative inverse, so it's not onto; hence, in the finite case, it's not injective. Find a non-trivial kernel element and see what happens.


Note that the word finite is important here. The proof I've mentioned breaks down since the map $r \to 7r$ is injective, even though it's not surjective. More generally, in infinite rings, elements can be non-units and non-zero-divisors.

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  • $\begingroup$ This is a good answer, but I don't know the terminology (kernel)? @T. Bongers $\endgroup$ – Don Larynx Sep 17 '13 at 17:07
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Consider the applications $l_a:\mathbb{D}\rightarrow\mathbb{D}$ and $r_a:\mathbb{D}\rightarrow\mathbb{D}$ given by $l_a(x)=a\cdot x$ and $r_a(x)=x\cdot a$. Both are injective if $a$ is not a zero divisor. But, because $\mathbb{D}$ is finite both are surjective too. Then you have left and right inverses for $a$ and you are done.

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