# Thread: Prove or disprove the following set is a field.

1. ## Prove or disprove the following set is a field.

For a prime $p$, let $\mathbb{Z}_{p}$ be the set of all equivalence classes of the relation defined by:

$aRb \Leftrightarrow a \equiv b(mod p)$.

Prove or disprove: $\mathbb{Z}_{p}$ is a field.

So I already showed that the set has addition and multiplication well-defined. The only thing I'm having trouble with is showing that for any $a \in \mathbb{Z}_{p}$ there exists $a^{-1} \in \mathbb{Z}_{p}$ such that $aa^{-1} = 1$.

How would I go about showing that exactly?

2. Hint : show that the map $\pi : \mathbb{Z}_p \rightarrow \mathbb{Z}_p : x \mapsto ax$ is an injection, and hence a bijection. What is the inverse image of $1$?

3. For any $a\ne0$, of course...

This is because for any two positive natural numbers $m$ and $n$ there exist integers $a$ and $b$ such that $am+bn=\gcd(m,n)$, where $\gcd(m,n)$ is the greatest common divisor of $m$ and $n$. This fact is proved using Euclid's algorithm for finding the GCD.

Now for any $0, $\gcd(n,p)={}$..., so...

4. Hmm, thanks, but someone else told me about Fermat's little theorem, which I think I'll use. But again, thanks!

5. Sure, but what's the point of using another theorem if you can do without it? Can you prove Fermat's little theorem?

6. Originally Posted by Bruno J.
Sure, but what's the point of using another theorem if you can do without it? Can you prove Fermat's little theorem?
Not really, but it makes more sense to me than anything else; I am really struggling with this material. I do not understand how to show that the suggestion you provided is a bijection, let alone how it shows that there exists an inverse.

7. Originally Posted by Pinkk
Not really, but it makes more sense to me than anything else; I am really struggling with this material. I do not understand how to show that the suggestion you provided is a bijection, let alone how it shows that there exists an inverse.
Use what emarakov said! Asking whether $ax\equiv 1\text{ mod }n$ is solvable is exactly the same as asking whether the linear Diophantine equation $ax+ny=1$ is solvable, and as he stated this is precisely solvable when $(a,n)=1$. But! $(a,p)=1$ for any $0 if $p$ is prime

8. Originally Posted by Drexel28
Use what emarakov said! Asking whether $ax\equiv 1\text{ mod }n$ is solvable is exactly the same as asking whether the linear Diophantine equation $ax+ny=1$ is solvable, and as he stated this is precisely solvable when $(a,n)=1$. But! $(a,p)=1$ for any $0 if $p$ is prime
Okay, this makes sense now, but isn't that another statement I'd have to prove? Okay, so using Bruno's suggestion:

$f([x]) = f([y])$
$[ax] = [ay]$
$[x]=[y]$

I'm not sure how to show this function is surjective and how this leads to there being an inverse element.

9. Well you should recall that if $S$ is a finite set, and you have a function $f:S\rightarrow S$, then the following are equivalent :

1. $f$ is an injection;
2. $f$ is a surjection;
3. $f$ is a bijection.

So certainly if $\pi : \mathbb{Z}_p \rightarrow \mathbb{Z}_p$ is injective, then it is surjective and there exists $b \in \mathbb{Z}_p$ such that $\pi(b)=ab=1$.

So we show it's an injection.

Suppose $\pi(x_1)=\pi(x_2)$. Then $ax_1\equiv ax_2 \mod p$, so $a(x_1-x_2)\equiv 0 \mod p$. Now recall that this means $p | a(x_1-x_2)$, and that since $p$ is prime we must have either $p | a$ or $p|(x_1-x_2)$. By hypothesis, we have $p \nmid a$, therefore $p |(x_1-x_2)$ and hence $x_1 \equiv x_2 \mod p$, so $\pi$ is indeed an injection.

10. Bah, now I remember. Thanks again!

11. Originally Posted by Pinkk
Bah, now I remember. Thanks again!
You are welcome.

In my opinion it is much better to rely on as few theorems as you can, unless you have a very good understanding of them. A friend of mine who is a PhD student once gave me a good example, after I supplied him with a proof which he found too technical (and he was right). He said : "I'll prove to you that the cube root of 2 is irrational. By Fermat's Last Theorem, $a^3+a^3=b^3$ has no solution in integers."