# Thread: Prove that Z_m is a field iff m is a prime.

1. ## Prove that Z_m is a field iff m is a prime.

Hi,

problem:
Let $\displaystyle m$ be an integer, $\displaystyle m\geq2$ and let $\displaystyle \mathbb{Z}_m$ be the set of all positive integers less than $\displaystyle m$, $\displaystyle \mathbb{Z}_m=\{0,1,\cdots,m-1\}$.
If $\displaystyle \alpha$ and $\displaystyle \beta$ are in $\displaystyle \mathbb{Z}_m$, let $\displaystyle \alpha+\beta$ be the least positive remainder obtained by dividing the (ordinary) sum of $\displaystyle \alpha$ and $\displaystyle \beta$ by $\displaystyle m$,
and similarly, let $\displaystyle \alpha\beta$ be the least positive remainder obtained by dividing the (ordinary) product of $\displaystyle \alpha$ and $\displaystyle \beta$ by $\displaystyle m$.

(a) Prove that $\displaystyle \mathbb{Z}_m$ is a field if and only if $\displaystyle m$ is a prime.

(b) What is -1 in $\displaystyle \mathbb{Z}_5$?

(c) What is $\displaystyle \frac{1}{3}$ in $\displaystyle \mathbb{Z}_7$?

attempt:

(a) I could really use a hint or two on this one..

(b) $\displaystyle remainder\;of\;\frac{1+(x)}{5}=0$

(c) $\displaystyle remainder\;of\;\frac{3\cdot(x)}{7}=1$

Btw, is there a nice math symbol for "the remainder of"?

Thanks.

2. Finite integral domains are fields, so to show that $\displaystyle \mathbb{Z}_m$ is a field for $\displaystyle m$ prime, you just need to show that it is an integral domain.

For the other direction, assume $\displaystyle m$ is not prime. Then we have $\displaystyle m=nr$, where $\displaystyle 1<n,r<m$. But $\displaystyle nr=0$ in $\displaystyle \mathbb{Z}_m$, which contradicts the fact that $\displaystyle \mathbb{Z}_m$ is an integral domain.

3. Originally Posted by Mollier
Hi,

problem:
Let $\displaystyle m$ be an integer, $\displaystyle m\geq2$ and let $\displaystyle \mathbb{Z}_m$ be the set of all positive integers less than $\displaystyle m$, $\displaystyle \mathbb{Z}_m=\{0,1,\cdots,m-1\}$.
If $\displaystyle \alpha$ and $\displaystyle \beta$ are in $\displaystyle \mathbb{Z}_m$, let $\displaystyle \alpha+\beta$ be the least positive remainder obtained by dividing the (ordinary) sum of $\displaystyle \alpha$ and $\displaystyle \beta$ by $\displaystyle m$,
and similarly, let $\displaystyle \alpha\beta$ be the least positive remainder obtained by dividing the (ordinary) product of $\displaystyle \alpha$ and $\displaystyle \beta$ by $\displaystyle m$.

(a) Prove that $\displaystyle \mathbb{Z}_m$ is a field if and only if $\displaystyle m$ is a prime.

(b) What is -1 in $\displaystyle \mathbb{Z}_5$?

(c) What is $\displaystyle \frac{1}{3}$ in $\displaystyle \mathbb{Z}_7$?

attempt:

(a) I could really use a hint or two on this one..
Use the fact that a field has no 0 divisors. That is, there are no members, x and y such that xy= 0. Now prove the contrapositive of each direction.

To prove "If $\displaystyle Z_m$ is a field the m is prime" start "if m is NOT prime, there exist a, b< m such that ab= m" and show that there are 0 divisors.

To prove "if m is prime, then Z_m is a field" start "if Z_m is not a field, then there must exist zero divisors- a, b< m such that ab= 0 (mod m)" and show that m= ab.

(b) $\displaystyle remainder\;of\;\frac{1+(x)}{5}=0$

(c) $\displaystyle remainder\;of\;\frac{3\cdot(x)}{7}=1$

Btw, is there a nice math symbol for "the remainder of"?

Thanks.

4. Originally Posted by Black
Finite integral domains are fields, so to show that $\displaystyle \mathbb{Z}_m$ is a field for $\displaystyle m$ prime, you just need to show that it is an integral domain.

For the other direction, assume $\displaystyle m$ is not prime. Then we have $\displaystyle m=nr$, where $\displaystyle 1<n,r<m$. But $\displaystyle nr=0$ in $\displaystyle \mathbb{Z}_m$, which contradicts the fact that $\displaystyle \mathbb{Z}_m$ is an integral domain.
I understand your explanation now that HallsOfIvy dumbed it down for me.
Thanks Black!

Originally Posted by HallsofIvy
Use the fact that a field has no 0 divisors. That is, there are no members, x and y such that xy= 0. Now prove the contrapositive of each direction.

To prove "If $\displaystyle Z_m$ is a field the m is prime" start "if m is NOT prime, there exist a, b< m such that ab= m" and show that there are 0 divisors.

To prove "if m is prime, then Z_m is a field" start "if Z_m is not a field, then there must exist zero divisors- a, b< m such that ab= 0 (mod m)" and show that m= ab.
Very nice, thank you very much!

5. Originally Posted by Mollier
I understand your explanation now that HallsOfIvy dumbed it down for me.
There is nothing I am better at that "dumbing down"!
("dumb" just comes naturally to me.)

Thanks Black!

Very nice, thank you very much!

6. Originally Posted by HallsofIvy
There is nothing I am better at that "dumbing down"!
("dumb" just comes naturally to me.)
Dumb seems to be a common talent.