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

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January 8th 2010, 05:23 AM
Mollier

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

Hi,

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

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

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

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

attempt:

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

(b) $remainder\;of\;\frac{1+(x)}{5}=0$
Answer: 4.

(c) $remainder\;of\;\frac{3\cdot(x)}{7}=1$
Answer:5.

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

Thanks.
January 8th 2010, 06:31 AM
Black

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

For the other direction, assume $m$ is not prime. Then we have $m=nr$ , where $1<n,r<m$ . But $nr=0$ in $\mathbb{Z}_m$ , which contradicts the fact that $\mathbb{Z}_m$ is an integral domain.
January 8th 2010, 07:11 AM
HallsofIvy

Quote:

Originally Posted by Mollier

Hi,

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

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

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

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

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 $Z_m$ 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.

Quote:

(b) $remainder\;of\;\frac{1+(x)}{5}=0$
Answer: 4.

(c) $remainder\;of\;\frac{3\cdot(x)}{7}=1$
Answer:5.

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

Thanks.
January 8th 2010, 11:25 PM
Mollier

Quote:

Originally Posted by Black

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

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

I understand your explanation now that HallsOfIvy dumbed it down for me.
Thanks Black!

Quote:

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 $Z_m$ 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!
January 9th 2010, 04:43 AM
HallsofIvy

Quote:

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"!(Giggle)
("dumb" just comes naturally to me.)

Quote:

Thanks Black!

Very nice, thank you very much!
January 10th 2010, 12:05 AM
Mollier

Quote:

Originally Posted by HallsofIvy

There is nothing I am better at that "dumbing down"!(Giggle)
("dumb" just comes naturally to me.)

Dumb seems to be a common talent. :)