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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.

FreeMind
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2 Answers2

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How much do you know about groups of order $p$? Do you know the following facts?

Fact: Let $(G,+,0)$ be an additive group of order $p$. Then $G$ is commutative; $G$ is cyclic; every nonidentity element of $G$ is a (group) generator of $G$.

Let $1 \in R$ be the multiplicative identity. Every element of $R$ is $n1 = 1+1+\dotsb+1$ ($n$ $1$s) for some integer $n$, $0 \leq n \leq p-1$. We have $(n1)(m1) = (nm)1 = r1$ where $r \equiv nm \pmod{p}$.

Let $n1 \in R$, $1 \leq n \leq p-1$. Write $an+bp=1$. Then an inverse for $n1$ is $a1$. Indeed $(n1)(a1) = 1 - (b1)(p1) = 1 - b0 = 1$, in $R$.

Zach Teitler
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  • That's what I was wondering about, we can put it directly in a $\mathbb{Z}$-module, tks . In general to see all the possible ring with unity structures on an abelian group $A$, you'd write $A = {\sum_i n_i g_i, n_i \in \mathbb{Z}}$ and then look at the possible values of $g_i g_j$ ? – reuns Sep 24 '17 at 15:32
  • @reuns It's far from my area but yes it seems to me a direct approach is as you say. Sufficient to consider values of products of generators of cyclic factors of additive group (that's what you meant, the $g_i$ are independent?). I'd guess it bogs down very quickly. No doubt some sophisticated idea out there somewhere. – Zach Teitler Sep 24 '17 at 16:00
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If $C$ is the center of $R$, then $|C|$ is a divisor of $p$. Argue that $R$ is commutative.

Let $I$ be an ideal of $R$. Then $|I|$ is a divisor of $p$. Hence…

egreg
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  • Otherwise : a finite integral domain is a field, and any non-trivial zero divisor would generate $(R,+)$ – reuns Sep 24 '17 at 13:25
  • @reuns I don't see how it's simpler. Not a big deal, though. – egreg Sep 24 '17 at 13:28
  • @reuns Do you think the OP knows words like "integral domain", "zero divisor", or "generate"? And I could add "center" and maybe even "ideal". The question being asked sounds like it would come up within 1-2 days of defining rings and fields. – Zach Teitler Sep 24 '17 at 13:50
  • @ZachTeitler The center is a basic example of subring; ideals are among the first things to present, together with their connection with fields. – egreg Sep 24 '17 at 13:53
  • True. It would be nice if people asking questions like these could say a little bit more about what they do or don't have available to them. – Zach Teitler Sep 24 '17 at 13:54
  • @ZachTeitler To me (one and two sided) ideal is quite an abstract concept, integral domain is more concrete. But I realize I assumed in some way $R$ is commutative, which follows from egreg's answer. If you have a more "elementary" argument.. – reuns Sep 24 '17 at 14:23
  • @reuns Goodness yes. I can answer this without using the word isomorphism. – Zach Teitler Sep 24 '17 at 15:03