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The number of irreducible elements in $\Bbb Z_{36}$ ?

I was only able to eliminate 13 elements which were a zero and 12 units. Then I found the following facts :

$(a)$ An element $[r]$ in $\Bbb Z_n$ is irreducible iff $\gcd(r,n)=p$, where $p$ is a prime integer.

$(b)$ The number of irreducible elements in $\Bbb Z_n$ is $\psi(n)=\sum_{p|n}\phi \big(\frac np \big) $ where the sum runs over the prime divisors of n.

However I have not seen the proof anywhere. For $(a)$, I tried the following:

Since $\gcd(r,n)=p$ , then $rx+ny=p$ for some $x,y \in \mathbb{Z}$.

If possible, let $[r]=[u][v]$ for some non-units $[u],[v]$ in $\Bbb Z_n$, then how to show a contradiction ??

Any help or suggestions is appreciated. Thanks for your time.

Chris
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user-492177
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1 Answers1

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Since $\mathbb{Z}_n$ is not a domain there are irreducible elements which are not prime. The proper result is that an element

$[a]$ in $\mathbb{Z}_n$ is irreducible iff $(a,n)=p$ and $p^{2} \mid n$.

Let's prove it:

Let $a\in\mathbb{Z}_n$ such that $(a,n)\ne 1$ and $n\nmid a$ because it is both a nonzero and nonunit element being irreducible. We can have three cases:

1) There are two primes $p$ and $q$ (that can also be the same $p=q$) such that $pq \mid(a,n)$.
In this case we can let $a=bc$ where $p \mid b $ and $q \mid c$ and so we can see that $[a]$ is the product of nonzero nonunit elements ($[b],[c]$) so it is not irreducible.

2) There is a prime $p$ such that $(a,n)=p$ and also $p^{2} \nmid n$.
In this case we have $(n,p^{2})=p$ and $p \mid a$, so there are two integers $r,s$ (Bezout) such that $a=nr+p^{2}s$. So $[a]=[p][sp]$ which are nonzero nonunit elements so again $[a]$ is not irreducible.

3) There exists a prime $p$ such that $(a, n) = p$ and $p^{2} \mid n$.
In this case we show that $[a]$ is irreducible. Let $[b], [c] \in \mathbb{Z}_n$ and $[a] = [b][c]$. Then there is an integer $k$ such that $a = bc + kn$. By hypothesis (irreducibility of $[a]$), $p \mid bc$ and so $p$ must divide at least one of $b, c$. It cannot divide both, else, since $p^{2} \mid n$ we would have $p^{2} \mid a$ a contradiction. Let us say $p \mid b$ and $p \nmid c$ (the other way is symmetric). Now if there exists a prime $q \ne p$, then $q \nmid c$, for $(a, n) = p$. Thus $(c, n) = 1$ and so $[c]$ is a unit.

This ends the proof.

user26857
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AndreaBaleani
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  • Sorry for the late response .Apparently my question is wrong!! but I wonder whether there is some generalisation of irreducible elements beyond integral domain to which the above formula is relevant? – user-492177 May 01 '20 at 14:12
  • More generally it really depends on where you are working, The definition of irreducible is the same. A result is that in an integer domain every prime element is irreducible while for the other implication you need to be in a GCD which is a special type of integer domain in which every nonzero nonunit element can be written as a product of prime elements uniquely (up to order and units). For example the ring of integers is GCD (fundamental theorem of arithmetic). – AndreaBaleani May 01 '20 at 14:47
  • The above formula holds if you are in Zn for some n and doesn't hold in any integer domain – AndreaBaleani May 01 '20 at 14:47
  • You mean what you did is true in integral domains only, right? Also in the last case of your proof there seem to be a slight misunderstanding of the hypothesis since we are proving the converse, aren't we? – user-492177 May 01 '20 at 15:36
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    @Kishore & answerer: There is not standard definition of irreducible, associate, unique-factorization, etc for non-domains, so you need to define your terms, e.g. see here. – Bill Dubuque May 01 '20 at 15:49