0

I came up to the example above in the title in order to show an element of a ring can be prime but it may fail to be irreducible.

$\mathbb{Z}_{6} = \lbrace [0], [1], [2], [3], [4], [5] \rbrace$ and I would like to show $[2] \in \mathbb{Z}_{6}$ is a prime element but it is not an irreducible element. Recall an element $p$ of a ring is prime if $p | ab$ for every other elements $a$ and $b$ in the ring, then $p | a$ or $p |b$. Also, an element $a$ of a ring is irreducible if whenever $a=bc$ then either $b$ or $c$ is a unit. In order to prove the fact $[2]$ is prime in $\mathbb{Z}_{6}$ we should assume there are $[a], [b] \in \mathbb{Z}_{6}$ such that $[2] \: | \: [a][b]=[ab]$. And from here I should proof $[2] | [a]$ or $[2] | [b]$. No problem proving a particular case, for instance $[a] = [4]$ and $[b] = [3]$. So, $[4] [3] = [12]= [0]$ and since $[2] [3] = [6] = [0]$ it yields that $[2]$ divides $[4][3]$ and obviously $[2]$ divides $[4]$. My question is that in order to proof $[2]$ is indeed a prime element of $\mathbb{Z}_{6}$ it suffices to prove it like I did for all elements $[a], [b] \in \mathbb{Z}_{6}$ such that $[2] \: | \: [a][b] $ or there is a general way to prove $[2]$ is prime? For the last one I tried the following:

If $[2] \: | \: [a] [b]= [ab]$ for $[a], [b] \in \mathbb{Z}_{6}$. Then there is an element $[k] \in \mathbb{Z}_{6}$ such that $[2][k] = [a][b]$. The last means $6t= 2k-ab$ for some $t \in \mathbb{Z}$ but I cannot come from here the existance of an $[s] \in \mathbb{Z}_{6}$ such that $[s][2]= [a]$ or $[s][2] = [b]$ :/

Also, no doubt convincing about $[2]$ is irreducible, because $[2]= [2] [4]$ and neither $[2]$ nor $[4]$ are units in $\mathbb{Z}_{6}$.

Brian Moehring
  • 21,088
Sok
  • 501
  • 2
    Your last sentence shows [2] is reducible, rather than what you wrote"irreducible". – coffeemath Aug 26 '22 at 19:55
  • 1
    For prime it might be easier to prove the contrapositive: If $2\not|a$ and $2\not|b$ then $2\not| ab$. Assuming the first two, we can write $a=2a'+1$ and $b=2b'+1$. Then $ab=2(2a'b'+a'+b')+1$. For $2$ to divide this, it would need to divide $1$. – plop Aug 26 '22 at 20:01
  • The terms "prime", "reducible", and "irreducible" are ordinarily intended to be used in the setting of integral domains, and the integers modulo 6 do not form an integral domain. Focus on understanding how these concepts work and don't work in integral domains. – KCd Aug 26 '22 at 20:01
  • @user85667 \nmid looks better than \not| (and \mid looks better than |):$$a\mid b\quad a\nmid b$$versus$$a| b\quad a\not|b$$ – Arthur Aug 26 '22 at 20:02
  • Going from what you had written, you have $ab = 2(k-3t)$. This allows us to transfer the question to $\mathbb{Z}$, answer it there, and then apply the quotient $\mathbb{Z} \to \mathbb{Z}_6$ to find your $[s] \in \mathbb{Z}_6$. – Brian Moehring Aug 26 '22 at 20:08
  • Factorization theory is much more complicated in non-domains, i.e. when nontrivial zero-divisors exist, e.g. $,x=(3+2x)(2−3x),$ in $\Bbb Z_6[x].,$ Basic notions such as associate and irreducible elements in a domain bifurcate into a few inequivalent notions in non-domains, e.g. see the papers cited here. – Bill Dubuque Aug 26 '22 at 20:40
  • Structurally: by the third isomorphism theorem $,\Bbb Z/6/(\Bbb 2/6)\cong \Bbb Z/2,$ is a domain thus $,2/6,$ is prime. – Bill Dubuque Aug 26 '22 at 21:05

1 Answers1

2

Assume that $[2] \ |\ [a][b]$, then there's $[c]$ such that $2c=ab$ mod $6$, thus $2c+6k=ab$ for some integer $k$. Therefore either $a$ or $b$ is even. The images of even integers in $\mathbb{Z}_6$ are $[0], [2], [4]$. All of those are divisible by $[2]$, hence either $[a]$ or $[b]$ will be divisible by $[2]$. Thus $[2]$ is prime.

Alessandro
  • 1,334