How to prove a ring (with unity) with prime number of elements is field

[MathNugget]

Say, $(R, +, \cdot)$ is the ring (I'd like to prove this even when the operations aren't the usual sum and product, but I suppose the proofs are similar). I'll shorten the ring to R (and will mention when I talk about the set R without operations).

I searched it on google, I know (other) forums are full of proofs...unfortunately, I cannot follow up any of them.

$R$ is a ring with p elements where p is a prime and has no zero divisors. Prove that $R$ is a field and is isomorphic to $\mathbb{Z}_p$.

$R$ is a ring with p elements where p is a prime and has no zero divisors. Prove that $R$ is a field and is isomorphic to $\mathbb{Z}_p$. I can prove that $R$ is a division ring and I know that by

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Any ring of prime order commutative ?

Is any ring of prime order commutative ?

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Every finite ring of order $p$ , $p$ is prime, with identity is a field

I am trying to prove the below statement, Every finite ring of order $p$ , $p$ is prime, with identity is a field I made some attempts but did not succeed.

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The last one is the most elementary.
I know with Lagrange's theorem, since (R, +) is a group, and $0_R, 1_R \in (R,+)$, if I prove $1_R\neq 0_R$, then since the order of an element divides order of group, the elements of the set R would be $0_R, 1_R, 1_R+1_R, ..., 1_R+...+1_R$ p-1 times.

Say, $0_R=1_R$. Computing $a\cdot (0_R+0_R)=$ both ways (sum first, or distributivity), we get $a0_R=a0_R+a0_R \rightarrow 0_R=a0_R$.
Then I compute $1_R a=0_Ra \rightarrow a=0_R$. And $a \in R$ was an arbitrary element, so either the ring has just 2 elements.
If p>2, then $0_R \neq 1_R$.

With this, I think I roughly proved that $(R, +)\simeq (\mathbb{Z}_p, +)$.
Seems harder to prove that the elements are invertible under multiplication...

$a \cdot b=(a\cdot 1_R) \cdot (b\cdot 1_R)=(1_R+...+1_R)\cdot (1_R+...1_R)=1_R+....+1_R=r$, with $r \equiv ab (mod p)$. At this point, where would I go with the proof? It looks like the whole thing maps very clearly to $(\mathbb{Z}_p,+, \cdot)$, but with all the notations, I also feel like half of the proof is circular and I used somewhere in the proof that the rings are izomorphic...

 

[blamocur]

unfortunately, I cannot follow up any of them.

What do you mean? You can't read them? The very first link has good answers to you question.

Proving that $1\neq 0$ in $R$ is pretty straightforward by contradiction: if $1 = 0$ then for every $x\in R: x = 1\cdot x = 0$.

Personally, I think you proof is complete since you've proven ring isomorphism between $R$ and $\mathbb Z_p$.

 

[MathNugget]

What do you mean? You can't read them? The very first link has good answers to you question.

Even worse, I can read those posts, but they raise more questions .

The first comment says things like:

If $R$ is a ring with unit $1_R$, then there’s a map $\mathbb{Z} \rightarrow R$ given by $1 \rightarrow 1_R$ uniquely extendend to a ring homomorphism, whose kernel must be $n\mathbb{Z}$ for some $n\in \mathbb{N_0}$.

This is basically the conclusion (the question), reformulated a little and put as a premise. Or at least, I don't know how to explain why this is so (if it's supposed to be obvious)... That's why I was trying to find a proof that uses as little theoretical results as possible, I am also working on my final project, and it's supposed to be as close to 'self-sufficient' or 'standalone', so to say, as possible.

 

[blamocur]

This is basically the conclusion (the question), reformulated a little and put as a premise.

Why? What is missing? Can't you fill in all the details, like how to extend the homomorphism? You've pretty much done that in your post.
Personally, I like the proof which says that if a finite ring has no divisors of zero then for any non-zero $a\in R$ mapping $x\rightarrow ax$ is bijective.

 

[MathNugget]

Why? What is missing? Can't you fill in all the details, like how to extend the homomorphism? You've pretty much done that in your post.

I guess I was hoping there's plenty more needed to make the proof complete...

Personally, I like the proof which says that if a finite ring has no divisors of zero then for any non-zero $a\in R$ mapping $x\rightarrow ax$ is bijective.

What does this prove? that the group with multiplication is cyclical too?



oh, nevermind... if $x\rightarrow ax$, and $a \neq 1$, then a different element is sent to 1. and since it happens for all nonzero a, it's done...

 

[MathNugget]

I actually like your proof, and I think I managed to prove that is a bijection.
$x \rightarrow ax$
the 2 sets, R and f(R) have same (finite) order. so injectivity=surjectivity=bijection (easy to prove because injection and surjection create an inequality between orders of groups, and here they're finite)

$ax=ay \Leftrightarrow ax-ay=0 \Leftrightarrow a(x-y)=0\Leftrightarrow x-y=0\Leftrightarrow x=y$ because there's no 0 divisors...

so then it's obvious.

 

[MathNugget]

Why? What is missing? Can't you fill in all the details, like how to extend the homomorphism? You've pretty much done that in your post.
Personally, I like the proof which says that if a finite ring has no divisors of zero then for any non-zero $a\in R$ mapping $x\rightarrow ax$ is bijective.

new problem...How would I prove the ring has no divisors of zero from what we have so far?

 

[Steven G]

What is the definition of a ring with/without no zero divisors? Do you understand that definition?
Also, it is the forum policy to start a new thread for a new question.
The reason is that someone has to read post 7 to know that there is a new question.

 

[blamocur]

Also, it is the forum policy to start a new thread for a new question.

To be fair, post #7 is part of the thread IMHO. When @MathNugget said "new problem" it meant new question about the thread, not a "real new problem".

 

[blamocur]

How would I prove the ring has no divisors of zero from what we have so far

Good point! I somehow assumed that rings with units have no zero divisors, but $\mathbb Z_4$ provides a good counterexample Time to use prime order of the ring. Here is a hint: if $x$ is a divisor of 0 then all its multiples are divisors too.

 

[MathNugget]

Good point! I somehow assumed that rings with units have no zero divisors, but $\mathbb Z_4$ provides a good counterexample Time to use prime order of the ring. Here is a hint: if $x$ is a divisor of 0 then all its multiples are divisors too.

True. But as far as I know, we cannot Lagrange's theorem on a monoid (like the set of the ring, with the 2nd operation). So, we'll have to prove it is cyclical (which I assume it's what you want to do) with a different technique than for the set with the first operation (which is a group).


I'll stick to the 1+ 1+....+1 solution, even though it looks weird.

But I learned to write this: $\underbrace{1+...+1}_{\text{p times}}$ in Latex, so the day was fairly productive. Thanks again for help; at this pace, your name should be on my final paper too


Funny enough though, all rings $(\mathbb{Z}_n, +, \cdot)$ are either fields, or have 0 divisors. There's no nuance for them...

 

[blamocur]

For any element $x$, zero divisor or not, the set of its multiples is a subgroup of $\mathbb Z_p^+$. How many elements can such a subgroup have?

 

[MathNugget]

I see, you're saying, based on what we proved so far, that ab+ac=a(b+c) , so it's a subgroup with addition... (which can only be proven by the same 1+ 1 +...+1 approach).

Then it either has order 1 (aa=a, all the time, so a = 0 or 1) or p. take an element a which isn't 0 or 1, and then I guess it's accessible to prove the group is cyclical...

 

[blamocur]

I see, you're saying, based on what we proved so far, that ab+ac=a(b+c) , so it's a subgroup with addition... (which can only be proven by the same 1+ 1 +...+1 approach).

Then it either has order 1 (aa=a, all the time, so a = 0 or 1) or p. take an element a which isn't 0 or 1, and then I guess it's accessible to prove the group is cyclical...

True. And 1 cannot be a zero divisor in a non-trivial ring.