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Which of the following rings are fields ?

$\mathbb{Z}_{2}$, $\mathbb{Z}_{3}$, $\mathbb{Z}_{4}$, $\mathbb{Z}_{5}$, $\mathbb{Z}_{6}$, $\mathbb{Z}_{7}$, $\mathbb{Z}_{8}$

And can you give a brief reason why?

Thanks

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  • $\begingroup$ Hint: We have $n = 0$ in each of the $\mathbb{Z}_n$. Consider when can two elements multiple to give $0$ in each of the $\mathbb{Z}_n$. $\endgroup$ – Alex Vong May 21 '17 at 7:43
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$\Bbb Z_n$ is a field iff $n$ is prime, here iff $n\in\{2,3,5,7\}$

The reson behind this is Bezout's identities :

Let $m\in\Bbb Z_n \setminus\{0\}$. If $n$ in prime we know there exists $u,v$ such that $mv + nu =1$ ie $mv = 1$ so $m$ is invertible in $\Bbb Z_n$.

On the other hand if $n$ is not prime, then there exists $k_1, k_2$ which multiply to $n$ and are both not equal to $n$. Then $k_1k_2 = 0$ in $\Bbb Z_n$ which means they are not invertible, and therefore $\Bbb Z_n$ is not a field.

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A ring ${\Bbb Z}_n$ is a field if $n$ is prime. An element $a\ne 0$ in ${\Bbb Z}_n$ is invertible if and only if the gcd of $a$ and $n$ is 1. This follows from the extended Euclidean algorithm which allows to write the gcd $d$ of two numbers $a$ and $n$ as a linear combination of these numbers. In a field all nonzero elements are invertible. So $n$ must be a prime.

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The ring you denote with $\mathbb{Z}_n$ is the quotient $\mathbb{Z}/n\mathbb{Z}$, with $n>0$.

In order the quotient ring is a field, $n\mathbb{Z}$ should be a maximal ideal in $\mathbb{Z}$. When can it be $n\mathbb{Z}\subsetneq k\mathbb{Z}\subsetneq \mathbb{Z}$?

Hint: look at divisibility.

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