2

Let $p$ be a prime. Show that there exists only two non isomorphic rings with $p$ elements.

My definition of rings does not require that the ring must have an identity $1\neq 0.$

I feel that there are two rings of order $p.$ They are $(\Bbb Z_p,+,.)$ And $(\Bbb Z_p,+,×),$ where $.$ is multiplication modulo $p$ and $×$ is such an operation so that $a.b=0,\forall a,b\in \Bbb Z_p.$

I took an arbitrary commutative ring $R$ order $p.$ If $R$ has a unit element, then I tried to show $R\cong (\Bbb Z_p,+,.).$

I know that every cyclic group of prime order is isomorphic $\Bbb Z_p$ under $+.$ I tried defining a map $\phi:R\to \Bbb Z_p$ such that, $\phi(1)=1_p.$

By knowing where $1$ goes, I feel I can fully know what $\phi$ really is?

The choice that $1$ goes to $1_p$ under $\phi$ was just a heuristic intuition. Since, $R$ is a cyclic group of prime order under $+$ operation, so every non zero element of $R$ is a generator of $(R,+).$ This means, if $a\in R$ then, $\exists n\in\Bbb Z\setminus\Bbb Z^-$ such that $a=n(1).$ So, $\phi(a)=\phi(n(1))=n\phi(1)=n(1_p)=n_p.$ We note that $a$ is arbitrary. So, $\phi$ is completely determined.

We have, $\phi(n(1))=n_p,\forall n\in\Bbb Z.$

Now we must verify that $\phi$ is indeed a ring homomorphism.

Let $a,b\in R$ then there exists, $n_1,n_2\in \Bbb Z$ such that $a=n_1(1)$ and $b=n_2(1).$ Thus $$a+b=(n_1+n_2)(1)\implies \phi(a+b)=(n_1+n_2)_p=(n_1)_p+(n_2)_p=\phi(a)+\phi(b).$$ Again, $$ab=(n_1n_2)_p(1)\implies \phi(ab)=(n_1n_2)_p=(n_1)_p(n_2)_p\phi(a)\phi(b).$$

So, $\phi$ is a ring homomorphism.

Let $n_p\in \Bbb Z$ then, $\exists n(1)\in R$ and, $\phi(n(1))=n_p.$ This proves, that $\phi$ is onto.

If $$\phi(m(1))=\phi(n(1))\implies m_p=n_p\implies (m_p)(1)=m(1)=(n_p)(1)=n(1).$$ Thus, $\phi$ is one-to-one.

Finally, we can say that $R\cong (\Bbb Z_p,+,.).$

Is my work till now valid?

However, I don't know how to show that $R\cong (\Bbb Z_p,+,×)$ if $R$ does not have a unit element, i.e a $1\neq 0.$

Thomas Finley
  • 2,734
  • This is one of those "proofs by contradiction" where you can just strike the contradiction declaration. – ziggurism Dec 19 '23 at 12:25
  • 2
    No, I don't think so. In a ring with unit, $1\cdot a=a$ for all $a$. So your second ring is not actually a ring. Therefore something is wrong with your proof. Maybe you want to do rings without unit? – ziggurism Dec 19 '23 at 12:30
  • Are you sure this claim is true? There certainly is only one unital ring with $p = 2$ elements, and I believe there is only one when $p = 3$ as well. I feel there should be only one for any prime in general. – X-Rui Dec 19 '23 at 12:34
  • Ok I see so you do allow non-unital rings. I think your partial proof is correct! And I think the other half is provided in the link you shared. – X-Rui Dec 19 '23 at 12:46
  • @X-Rui That's the problem. I don't really get how the ither half is answered there. – Thomas Finley Dec 19 '23 at 12:49
  • An improvement might be to show $\phi$ is well-defined, since $a = n 1_R = (n + p) 1_R$, and $\phi(a)$ is kind of dependent on the representative you choose. Well, not actually, but you should prove it. – X-Rui Dec 19 '23 at 12:54
  • @X-Rui Ok, but how to prove the other half? Actually that's the part where I am stuck. – Thomas Finley Dec 19 '23 at 13:06
  • Why the downvote? I don't get the reason. – Thomas Finley Dec 19 '23 at 13:12
  • @ziggurism Yes, you are correct. So, till now, I was able to show only that if $R$ has a unit element then $R\cong (\Bbb Z_p,+,.)$. But I don't know how to show that $R\cong (\Bbb Z_p,+,×)$ if $R$ does not have a unit element, i.e $1\neq 0.$ Can you please help me with this ? – Thomas Finley Dec 19 '23 at 13:15
  • You wrote $0\neq 1$ which is incompatible with a ring where $0=1$ – ziggurism Dec 19 '23 at 14:36

1 Answers1

1

This answer is almost entirely based on the answer to a duplicate question https://math.stackexchange.com/a/1090955/1264774 by Andreas Caranti. I just try to provide a clearer explanation in better details.

If $R$ is not isomorphic to $(\mathbb{Z}_p, +, \times)$ where all products are zero, then there are some $a, b \in R$ such that $ab \neq 0$. We prove this $R$ is unital, so with your correct partial proof, we can have a complete proof.

The map $\psi: R \to R; x \mapsto ax$ is a group homomorphism. The image $\psi (R)$ is a subgroup of $(R, +)$. Since $|R| = p$, we have $|\psi (R)| = 1 \text{ or } p$. By assumption, $ab \neq 0$, so $|\psi (R)| \geq 2$. Hence, $|\psi (R)| = p$, and $\psi$ is bijective. In particular, there exists an $e \in R$ such that $a e = \psi(e) = a$. For all $x \in R$, we have $\psi(e x) = a (e x) = (a e) x = a x = \psi(x)$, so $e x = x$ by the bijectivity of $\psi$. By a similar argument with $R \to R; x \mapsto xb$, we will find an $e' \in R$ such that $\forall x \in R, x e' = x$. Now, we have found a left identity $e$ and a right identity $e'$ of $R$, so $e = e e' = e'$ is the two-sided multiplicative identity $R$.

Thus, $R$ is unital. By the proof in the description, this $R$ is isomorphic to $(\mathbb{Z}_p, +, .)$, i.e. the usual ring of integers modulo $p$.

X-Rui
  • 1,299