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The question is:

Prove that if $p \in \mathbb{Z}$ is irreducible, then $p$ is also prime.

  • Irreducible is defined as follows: $n \in \mathbb{Z} $ is irreducible iff the only way to write $n=ab$ for some integers $a$ and $b$ is if $a = \pm1$ or $b = \pm1$.
  • Prime is defined as follows: an integer $n$ is prime iff whenever $a,b \in \mathbb{Z} $ and $p\mid ab$ then it must be true that $p\mid a$ or $p\mid b$ (or both).

Hint: Prove that if $n\mid ab$ but $n$ does not divide $a$, then $\gcd(a,n) = \pm1$. Then use Bézout's theorem and prove $n\mid b$.

user26857
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    That's a very good hint. If $p \nmid a$, are you able to prove that $\gcd(p, a) = 1$? – Dylan Moreland Sep 11 '12 at 17:51
  • @DylanMoreland I have no idea how to, I'm really bad at this stuff. –  Sep 11 '12 at 17:52
  • @JasonDeVito I keep trying this thing but I think it's completely wrong, I say let p=ab where a,b in Z. Then either p = +/- a or p= +/-b. Either way, p divides ab. I'm sure this is a terrible proof but why is it bad? –  Sep 11 '12 at 17:54
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    Minor point, definition of prime slightly off, need to exclude $\pm 1$, $0$. – André Nicolas Sep 11 '12 at 18:13

5 Answers5

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I hope this is a subtle enough hint.

Suppose that $p$ is irreducible and that $p\mid ab$, but $p\nmid a$. If we can show that $p\mid b$, then we have shown that $p$ is prime.

Let $g=(p,a)$. Note that $g\mid p$ and use the irreducibility of $p$ to show that $g=1$.

Therefore, by Bezout's Lemma, we can find $x$ and $y$ so that $$ ax+py=1\tag{1} $$ multiply both sides of $(1)$ by $b$. Can you now show that $p\mid b$ ?

robjohn
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  • How do you know that g=1? –  Sep 11 '12 at 18:19
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    @AllisonCameron: "use the irreducibility of $p$". Since $g,|,p$, what does the irreducibility of $p$ tell us? Since $g,|,a$ and $p!\not|;a$, what does that leave us with? – robjohn Sep 11 '12 at 18:23
  • I still don't get it. Sorry! Your explanations are great, I'm just really slow. –  Sep 11 '12 at 18:33
  • @AllisonCameron: assume we are working in positive integers. Look at the definition of irreducibility. If $g,|,p$ then either $g=1$ or $g=p$. – robjohn Sep 11 '12 at 18:36
  • I just played around with it and I think I get it now, thanks a ton! –  Sep 11 '12 at 18:49
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$\rm\begin{eqnarray} By\ \ GCDs:\ \, atom\,\ p\nmid a\:&\Rightarrow&\rm\ (p\ \,,\ a)=1,\ &\rm so&\rm\ p\:|\:pb,ab\:&\Rightarrow&\rm\:p\:|\:(pb\ ,\, \ ab) &=&\rm\ (p\, \ ,\ \ a)b &=&\rm b\\ \rm By\ Bezout:\ \, atom\,\ p\nmid a\:&\Rightarrow&\rm\:jp\!+\!ka\,=1,\ &\rm so&\rm\ p\:|\:pb,ab\:&\Rightarrow&\rm \:p\:|\ jpb\!+\!kab &=&\rm (jp\!+\!ka)b &=&\rm b\end{eqnarray}$

Remark $\ $ Note how the GCD proof eliminates the Bezout coefficients $\rm\:j,k\:$ (which only serve to obfuscate the proof) and, further, highlights the key role played by the distributive law for gcds.

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Bill Dubuque
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I will translate my previous answer to a more elementary proof so that the OP can understand.

Let $p \in \mathbb{Z}$ be irreducible. Let $a \in \mathbb{Z}$ such that $a$ is not divisible by $p$. Let $I = \{ax + py; x \in \mathbb{Z}, y \in \mathbb{Z}\}$. Let $c > 0$ be the least positive integer belonging to $I$. We claim that every element of $I$ is divisible by $c$. Let $d \in I$. $d$ can be written as $d = cq + r$, where $q, r \in \mathbb{Z}$, $0 \le r < c$. Since $cq \in I$, $r = d - cq \in I$. Hence $r = 0$ as claimed.

Hence, in particular, $a$ and $p$ are divisible by $c$. Since $p$ is irreducible, $c = 1$. Hence there exist integers $x, y$ such that $ax + py = 1$. Hence $ax \equiv 1$ (mod $p$).

Suppose $ab \equiv 0$ (mod $p$). Then $b \equiv xab \equiv 0$. Hence $p$ is prime.

Makoto Kato
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If $p$ is an integer that is irreducible then suppose has $p$ has factorization $p = ab$. As $p$ is irreducible then either $a=\pm1$ or $b=\pm1$. Suppose $a=\pm1$ then $p=\pm b$ $\implies$ $p|b$. Similarly, if $b=\pm1$ then $p |a$ thus $p$ is prime.

user41442
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Let $p \in \mathbb{Z}$ be irreducible. Since $\mathbb{Z}$ is a principal ideal domain, $p\mathbb{Z}$ is a maximal ideal. Hence $p\mathbb{Z}$ is a prime ideal. Hence $p$ is prime.

Makoto Kato
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    This might be a bit more advanced than the level of the question. – robjohn Sep 11 '12 at 18:25
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    @robjohn But the question is tagged as "abstract algebra". My proof is very basic in abstract algebra. – Makoto Kato Sep 11 '12 at 18:31
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    It's the first week of the course, I don't even know what ideals are yet. Thanks anyway for the help @makotokato :) –  Sep 11 '12 at 18:43