My answer was True and this is my argument:

Since $\mathbb{Z}_{n}$ has got $2$ operations plus the other properties of a ring, I figured that it is indeed a ring. On the other hand, since $\mathbb{Z}_{n}$ is abelian (commutative) and has the multiplicative identity $1$, It can be concluded that it is a $Commutative$ Ring with an $identity$, hence implies that it is a Field.

I've also read a theorem which states that: "Every finite integral domain is a field."

Is my argument acceptable and can I also use the above theorem as an argument?

Your help would be really appreciated.

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    $\begingroup$ Well, $\mathbb{Z}_{28}$ is also a commutative ring with identity. To be a field, every nonzero element must have a multiplicative inverse. $\endgroup$ – Daniel Fischer May 20 '14 at 21:50
  • $\begingroup$ To show that no element is a zero divisor, the fact that the order of $k$ is $n/\gcd(k,n)$ may be helpful (In any finite cyclic group of order $n$ the order of $g^k$ is $o(g)/\gcd(k,n) $ ). $\endgroup$ – Stefan Hamcke May 20 '14 at 21:57
  • $\begingroup$ @Stefan No need for that when the simpler Lagrange Theorem suffices . $\endgroup$ – Bill Dubuque May 20 '14 at 22:05
  • $\begingroup$ @BillDubuque: Oops, forgot about that, thanks. Well at least in $\Bbb Z_n$ for arbitrary $n$ it implies that all coprime $k$ are invertible, as they have order $n$ and are thus non zero divisors. $\endgroup$ – Stefan Hamcke May 21 '14 at 0:42

Yes it is a field; more generally $\mathbf{Z}/n\mathbf{Z}$ is a field iff $n$ is a prime number. But no your argument is not sufficient, but it is if you argue that it is an integral domain (to see this notice that $(n)$ is a prime ideal iff $n$ is 0 or a prime number; in case $n=0$ this is $\mathbf{Z}$ which is not a field). It is certainly not the case that every commutative ring is a field, but a field is by definition a commutative ring.

To make the argument explicit for your convenience: First show that every finite integral domain $A$ is a field (hint: for $a\neq 0$ in $A$ consider the mapping $A\rightarrow A$ defined by $x\mapsto ax$; show that it is injective and hence bijective since $A$ is finite). Then show that $\mathbf{Z}/n\mathbf{Z}$ (a finite ring for $n$ at least $1$) is an integral domain iff $n=0$ or $n$ a prime (use the classification of prime ideals of $\mathbf{Z}$).

  • $\begingroup$ Ok, so perhaps not all $\mathbb{z}_{n}$ is a field? only $\mathbb{z}_{p}$, given that $p$ is a prime? $\endgroup$ – Ozahm May 20 '14 at 21:55
  • $\begingroup$ yes indeed, this is what I am saying $\endgroup$ – Alexander Grothendieck May 20 '14 at 21:56
  • $\begingroup$ @Ozahm P must be a prime power. $\endgroup$ – DanielV May 20 '14 at 22:21
  • $\begingroup$ @DanielV Could you give a simple example please. $\endgroup$ – Ozahm May 20 '14 at 22:30
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    $\begingroup$ @DanielV No, $\mathbb{Z}/n\mathbb{Z}$ is a field iff $n$ is prime. There are fields of prime power order $p^k$, but they are not isomorphic to $\mathbb{Z}/p^k \mathbb{Z}$. $\endgroup$ – André 3000 May 20 '14 at 22:31

A commutative ring with identity is not necessarily a field. You should check your definitions. For instance, $\Bbb Z_6$ has $2\cdot 3=0$, hence isn't even an integral domain.

You absolutely can use your theorem. Note that $\Bbb Z_{29}$ has no zero divisors (why?), and hence it is an integral domain, and hence by your theorem is a field.

However, one need not appeal to such advanced machinery. Bezout's theorem says that if $gcd(a,n)=1$ then there's a pair of integers $(p,q)$ such that $pa+qn=1$. Picking $n=29$ this shows that every nonzero element of $\Bbb Z_{29}$ is invertible (why?). Since $\Bbb Z_{29}$ is a nonzero commutative ring, it is a field.

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    $\begingroup$ Indeed I should take another look at my definitions. I have just realized that $29$ is $prime$ which is why $\mathbb{Z}_{29}$ it has no zero divisors. Thank You. $\endgroup$ – Ozahm May 20 '14 at 22:04

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