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I am trying to prove that the set of zero divisors of the ring $\mathbb{Z}/_{b\mathbb{Z}} $ is equal to $\{[x] \in \mathbb{Z}/_{b\mathbb{Z}}: \gcd(x,b)>1\}$. Therefore I started with the set of zero divisors equals: $\{[x] \in \mathbb{Z}/_{b\mathbb{Z}}: cx = db \, \space \text{for some} \space c,d \in \mathbb{Z}\backslash\{0\}\}$. And now to my question: Do we have: $x$ divides $db \implies \gcd(x,b)>1$.

If this holds (why/why not?), I would be done.

Bill Dubuque
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Slyrack
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    Perhaps, if $x > 1$; since $x \mid db$ either $x\mid b$ or $x\mid d$ by Euclid's lemma, and so you may take $x\mid b$ and thus $\gcd(x,b) = x$. – Spectre Jun 23 '21 at 08:08
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    @Spectre How can i just take $x$ divides $b$, and ignore the case that $x$ could divide $d$ instead of b? – Slyrack Jun 23 '21 at 08:11
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    did you mean, $x \mid d$ in your comment? – Spectre Jun 23 '21 at 08:12
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    Yes, I will correct it – Slyrack Jun 23 '21 at 08:12
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    I only meant that you can take it like that. – Spectre Jun 23 '21 at 08:13
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    the 3 possibilities when $x \mid db$ are : a) $x \mid d$ , b) $x \mid b$ or c) $pf(x) \in (pf(b) \cup pf(d))$ (I used my own notation $pf(x)$ to represent the set of prime factors of $x$). – Spectre Jun 23 '21 at 08:15
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    That confuses me, though, so I can't give you an exact answer and so I'd say 'Wait till Euclid comes on your way' – Spectre Jun 23 '21 at 08:16
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    By definition, $x$ is a zero-divisor $!\bmod b$ if $,x\not\equiv 0,$ and for some $,c\not\equiv 0,$ we have $,cx\equiv 0,, $ i.e. iff $,b\mid cx,, b\nmid c,x,$ But if $,(b,x)=1,$ then by Euclid's Lemma $,b\mid cx\Rightarrow b\mid c,,$ contra hypothesis. Thus $(b,x)\neq 1,,$ so $(b,x)>1.,$ See the linked dupe for other proofs. – Bill Dubuque Jun 23 '21 at 08:59

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