Show $ab=0\Leftrightarrow (a)+(b)=R.$ for two zero divisors $a$ and $b$ in $R=\mathbb Z_{pq}$

Question

Let $a,b\in\mathbb Z_{pq}\backslash\{0\}$ where $p$ and $q$ are two distinct primes and $a$ and $b$ are zero divisors. Show that $$ab=0\Leftrightarrow (a)+(b)=R$$ where $R=\mathbb Z_{pq}$. Any hints (please no finished solutions)? Unfortunately I can't prove either direction. For $\Rightarrow$ we have $pq\mid ab$. What does that mean for the product $ab$? How can I derive that $a$ and $b$ are coprime? For the other direction I don't know how to grasp $(a)+(b)$. I can just say that there are $x,y\in\mathbb Z_{pq}$ such that I can write $ab=x\cdot a+y\cdot b$.

Answer 1

If $pq{\,\mid\,}ab$, then $p$ divides at least one of $a,b$, say $a$.

But then, since $p{\,\mid\,}a$, we can't also have $q{\,\mid\,}a$ (why?).

But $q$ must divide at least one of $a,b$.

Thus, $a$ is a multiple of $p$, but not a multiple of $q$, and $b$ is a multiple of $q$, but not a multiple of $p$.

If $(a,b)\ne (1)$, then $(a,b)\subseteq M$, for some maximal ideal $M$ of $R$.

But the only maximal ideals of $R$ are $(p)$ and $(q)$.

If $M=(p)$, then $b\in M$ implies $p{\mid}b$, contradiction.

If $M=(q)$, then $a\in M$ implies $q{\mid}a$, contradiction.

It follows that $(a,b)=(1)$.

For the converse, suppose $a,b$ are zero-divisors such that $(a,b)=(1)$.

Can $a,b$ both be multiples of $p$? Can they both be multiples of $q$?

But since $a,b$ are zero-divisors, each is a multiple of at least one of $p,q$.

So what does that say about $ab$?

Answer 2

For $\implies$, consider that $p\mid ab$ and $q\mod ab$. So one of $a,b$ is a multiple of $p$, the other is a multiple of $q$ (since neither is $0$). Now use the Chinese remainder theorem to find $1\in (a)+(b)$.

For the other direction, think about what the different possible zero divisors in $\Bbb Z_{pq}$ are. And then realize that $(a)+(b)\not\subseteq (p)$ and $(a)+(b)\not\subseteq (q)$.

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"How can I derive that $a$ and $b$ are coprime?" First of all, "coprime" is an iffy notion in modular arithmetic. A number may be coprime to the base, but two numbers can't really be coprime to one another.

And if you're thinking in $\Bbb Z$ and reducing to $\Bbb Z_{pq}$ after the fact, then you still get trouble with the notion. Take, for instance, $a=3,b=10$ in $\Bbb Z_{15}$. They look coprime, but $a=18,b=10$ gives the exact same situation and now they're not coprime.

Of course, the definition that $a$ and $b$ are coprime iff $(a)+(b)=(1)$ makes sense. If this is the definition you're using, then you may disregard the above. But since this is also read by other people I felt that it was worth mentioning.