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$R$ 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.

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  • $\begingroup$ $(p) \subset R$ is a prime ideal; what does that mean? When is an element equal to $0$ in the quotient? $\endgroup$ – ÍgjøgnumMeg May 7 at 21:02
  • $\begingroup$ 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 $R$) $\endgroup$ – Adam Higgins May 7 at 23:53
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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 $p|b$.

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  • $\begingroup$ Can you explain how does a+(p)=(p) or b+(p)=(p) imply that $R/(p)$ has no zero divisors? $\endgroup$ – Bonrey May 7 at 21:23
  • $\begingroup$ 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 $b+(p)=(p)$. $\endgroup$ – Mark May 7 at 21:27
  • $\begingroup$ 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 $ab+(p)=(p)$? $\endgroup$ – Bonrey May 7 at 21:37
  • $\begingroup$ 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 $R/(p)$. $\endgroup$ – Mark May 7 at 21:45
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    $\begingroup$ 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. $\endgroup$ – Mark May 7 at 21:46
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$$\begin{align} p\ \ \lnot \rm prime \iff&\ \ p\ \mid\ ab,\ \ p\ \nmid\ a,b,\ \ {\rm some}\ a,b\in R\\[.2em] \iff&\ \ 0 = \bar a\bar b,\ \ 0\neq \bar a,\bar b,\ \ {\rm some}\ \bar a,\bar b\in R/p\\[.2em] \iff&\ R/p\ \ \text{has a zero divisor}\end{align}\qquad\qquad$$

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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 $b$.

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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.

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

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