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I suppose one can define a "prime" element $p$ in the ring $\mathbb{Z}/n\mathbb{Z}$ (not necessarily with $n$ prime) if $p$ cannot be written as a product of two elements of the ring different from $1$. For the first cases, I found that $2$ is prime in $\mathbb{Z}/3\mathbb{Z}$, $3$ is prime in $\mathbb{Z}/4\mathbb{Z}$ and $5$ is prime in $\mathbb{Z}/6\mathbb{Z}$.

Checking up to $\mathbb{Z}/11\mathbb{Z}$, I couldn't find any other primes.

My question is,what are the "prime" elements in the rings $\mathbb{Z}/n\mathbb{Z}$?

And, as $n$ approaches $+\infty$, will the primes of $\mathbb{Z}/n\mathbb{Z}$ be the primes of $\mathbb{Z}$?

Bill Dubuque
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marco trevi
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    Each element of a residue class ring is either a unit or a zero divisor (besides 0). – Wuestenfux May 07 '22 at 15:43
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    Your examples are units not prime element. Here a prime element cannot be written as $ab$ with $a,b$ non units (the general definition is $pc = ab$ implies one of $a,b$ is $=pd$). Do you see why $3$ is a prime element of $\Bbb{Z}/36\Bbb{Z}$ ? – reuns May 07 '22 at 15:51
  • In a general commutative ring, we distinguish two different types - irreducibles, and primes. But we also never call an element with a multiplicative inverse either prime or irreducible.(Primes are always irreducible, but irreducibles are not always prime.) – Thomas Andrews May 07 '22 at 16:07
  • @Thomas prime $\Rightarrow$ irreducible is always true in an integral domain, but fails more generally, e.g. note that $,x,$ is prime in $,R = \Bbb Z[x]/(x^2-x),$ by $R/(x) \cong \Bbb Z,,$ but $,x = x^2,$ is reducible in $R.\ \ $ – Bill Dubuque May 07 '22 at 16:18
  • I guess it depends on the definition. I’d have guessed the definition of irreducible was “If $x=ab,$ then one of $a,b$ is a unit. But my memory is rusty, @BillDubuque – Thomas Andrews May 07 '22 at 16:53
  • @Thomas That (plus $x$ is nonzero, nonunit) is indeed a common definition of an irreducible. In the example in my prior comment $x$ does not satisfy that, since in $R,$ we have $x = x^2,,$ and $x$ is a nonunit (by $R/(x) \cong \Bbb Z\not\cong 1)$, so $,x,$ is a prime that is not irreducible. $\ \ $ – Bill Dubuque May 07 '22 at 17:04

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