Prove that $p$ is prime if and only if there are no zero divisors in the quotient ring.

$$R$$ is a factorial ring, and $$p\in R$$ is a non-zero non-invertible element. I need to prove that $$p$$ is prime if and only if there are no zero divisors in the quotient ring $$R/(p)$$.
Here is what I've done:
Since $$p\in R$$ where $$R$$ is a factorial ring, and $$p$$ is a non-zero non-invertible element, it is true that $$p=p_1\times\dots\times p_n$$ where $$p_i$$ is prime for $$1\leq i\leq n$$. Also, I know that $$a$$ is called a zero divisor of the ring $$R/(p)$$ if $$\exists b\ne0: ab=ba=0$$. The problem is I don't know what to do with the so-called quotient ring.

• $(p) \subset R$ is a prime ideal; what does that mean? When is an element equal to $0$ in the quotient? – ÍgjøgnumMeg May 7 at 21:02
• Just because no one else has actually explicitly stated this yet, but you don’t need that $R$ is a “factorial” ring, by which I assume you mean that every non-zero element of $R$ admits some decomposition into a unit and a product of prime elements. The result is true for all commutative rings $R$, and ideals $I$. I.e $I$ is prime if and only if $R/I$ is an integral domain (which is to say has no zero divisions). Note then that for $p\in R \backslash \left\{0\right\}$, $(p)$ is a prime ideal of $R$ if and only if $p$ is prime in $R$) – Adam Higgins May 7 at 23:53

Suppose $$p$$ is prime, and let's assume $$(a+(p))(b+(p))=(p)$$. By the definition of multiplication in the quotient ring it means that $$ab+(p)=(p)$$, and hence $$p|ab$$. Since $$p$$ is prime we conclude that $$p|a$$ or $$p|b$$ which implies $$a+(p)=(p)$$ or $$b+(p)=(p)$$. So $$R/(p)$$ has no zero divisors.

For the other direction suppose $$R/(p)$$ has no zero divisors and assume $$p|ab$$. Then $$(p)=ab+(p)=(a+(p))(b+(p))$$. Since there are no zero divisors we conclude that $$a+(p)=(p)$$ or $$b+(p)=(p)$$ which means that $$p|a$$ or $$p|b$$.

• Can you explain how does a+(p)=(p) or b+(p)=(p) imply that $R/(p)$ has no zero divisors? – Bonrey May 7 at 21:23
• A ring $R$ has no zero divisors if $ab=0$ implies $a=0$ or $b=0$. This is exactly what I used, just we have to remember that $(p)$ is the zero element in $R/(p)$. I assumed that $(a+(p))(b+(p))=(p)$ and proved that it implies $a+(p)=(p)$ or $b+(p)=(p)$. – Mark May 7 at 21:27
• Two questions: (1) Why $(p)$ is the zero element in the quotient ring if it generates the whole ring $R/(p)$? It would mean that $R=\{0\}$. Am I wrong? (2) Why should we first assume that $(a+(p))(b+(p))=(p)$ and not start directly with $ab+(p)=(p)$? – Bonrey May 7 at 21:37
• What is the definition of $R/(p)$? It is the set of cosets $\{a+(p):a\in R\}$ with operations $(a+(p))+(b+(p))=a+b+(p)$ and $(a+(p))(b+(p))=ab+(p)$. So what is the zero element? It is $0+(p)=(p)$, this follows from the definition of addition in $R/(p)$. Don't get confused: the element $p$ generates the ideal $(p)$, that doesn't mean $(p)$ generates the ring $R/(p)$. – Mark May 7 at 21:45
• As for the second question: if we want to show that a ring has no zero divisors then the way to do it is suppose a product of two elements is zero and show one of them must be zero. This is what we did. If we would just start directly from $ab+(p)=(p)$ it wouldn't be clear why we are assuming this. – Mark May 7 at 21:46

\begin{align} p\ \ \lnot \rm prime \iff&\ \ p\ \mid\ ab,\ \ p\ \nmid\ a,b,\ \ {\rm some}\ a,b\in R\\[.2em] \iff&\ \ 0 = \bar a\bar b,\ \ 0\neq \bar a,\bar b,\ \ {\rm some}\ \bar a,\bar b\in R/p\\[.2em] \iff&\ R/p\ \ \text{has a zero divisor}\end{align}\qquad\qquad

Suppose $$p$$ is not prime; what does this mean? It means there are elements $$a,b\in R$$ such that $$p$$ divides $$ab$$ but $$p$$ does not divide $$a$$ or $$b$$. Try to show that the condition that $$p$$ does not divide $$a$$ or $$b$$ means that the elements $$a+(p)$$ and $$b+(p)$$ are nonzero elements of $$R/(p)$$ [hint: $$p$$ divides $$a$$ iff $$a\in(p)$$], and then notice that $$(a+(p))(b+(p))=0$$ (why?), so $$R/(p)$$ has zero-divisors.

In the other direction, suppose $$R/(p)$$ has zero divisors; then you can find nonzero elements which multiply to zero, say $$a+(p)$$ and $$b+(p)$$, and try to reverse the logic from our first implication to show that $$p$$ divides $$ab$$ but $$p$$ does not divide $$a$$ or $$b$$.

Hint:

$$p\mid x\iff x\equiv 0\mod p\iff \bar x=\bar 0 \:\text{ in }\: R/(p)$$. You can use Euclid's lemma.

• Where $x$ is any element of the ring $R$? – Bonrey May 7 at 21:08
• Yes, $x$ is an element of $R$, and $\bar x$ is its class modulo $p$ (I've added a detail about this). – Bernard May 7 at 21:14