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Find with proof the number of units in the ring $$R=\mathbb{Z}_8 \times \mathbb{Z}_9 \times \mathbb{Z}_5$$

Since $\gcd(5,8,9)=1$, by Chinese Remainder Theorem, I get $R=\mathbb{Z}_{360}$

Since $\mathbb{Z}_{360}$ is not a field, not all elements has inverse. Then I get nowhere from here. Can anyone guide me ?

Idonknow
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    Hint: How many units in each of $\mathbb Z_8, \mathbb Z_9,,\mathbb Z_5$? – Thomas Andrews Apr 26 '13 at 13:22
  • @ThomasAndrews: Do we need ring tag here? :) – Mikasa Apr 26 '13 at 13:25
  • $R$ is already in the best possible form. It does not really help to write it $\mathbb{Z}_{360}$. It is easy to see that the unit group is are $\mathbb{Z}_8^\times\mathbb{Z}_9^\times \mathbb{Z}_5^*$. So you just have to compute $\phi(8)\phi(9)\phi(5)$. You might want to use or observe that $\phi(p^\alpha)=p^\alpha-p^{\alpha-1}$ for any prime $p$. – Julien Apr 26 '13 at 13:34

3 Answers3

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The number of units in $\Bbb Z_{360}$ is the number of positive integers less than $360$ that are coprime to it. That is, $\phi(360)$, where $\phi$ is Euler's totient function.

As an alternate approach, it is simpler to count the number of units in $\Bbb Z_8,\Bbb Z_9,\Bbb Z_5$. If $(x,y,z)$ is a unit of $R$, what can you say about $x,y,z$?

Cameron Buie
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Hint An element $a$ is invertible in $\mathbb Z_n$ if and only if gcd$(a,n)=1$.

It might be easier if you count the numbers of units in $\mathbb Z_8, \mathbb Z_9$ and $\mathbb Z_5$.

N. S.
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Hint: Show $(x,y,z)\in \mathbb{Z}_8 \times \mathbb{Z}_9 \times \mathbb{Z}_5$ is a unit if and only if $x$ is a unit in $\mathbb Z_8$, $y$ is a unit in $\mathbb Z_9$ and $z$ is a unit in $\mathbb Z_5$.

Thomas Andrews
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