2

See this relevant post

Ring of order $p^2$ is commutative.

Either I don't understand the above or it isn't quite complete (presumably the former).

Let $R$ be a ring of order $p^2$. Then $char(R)=p$ and $Z(x)=p,p^2$ for any $x \in R$. In the latter case there is nothing to show. Hence, let $Z(x)=p$. Then $Z(x) \cong \mathbb{Z}/p\mathbb{Z}$ (as a group not necessarily a ring) and supposedly this is enough to say that $R$ is commutative but I do not readily see this. Especially since ring homomorphisms must send $1 \to 1$. A detailed explanation would be appreciated.

RhythmInk
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  • I'm also very interested to understand how we can express $R$ as the epimorphic image of $\mathbb{Z}[x]$ – RhythmInk Oct 31 '18 at 22:28
  • The isomorphism isn't a group isomorphism, it is a ring isomorphism. See theorem 1 of http://www.csc.villanova.edu/~pcesarz/PrimeOrderedRings.pdf – John Douma Oct 31 '18 at 22:38
  • Then under that isomorphism where does $x$ go? – RhythmInk Oct 31 '18 at 22:42
  • Did you mean where does $1$ go? If so, $\mathbb Z/p\mathbb Z={0, 1, ...,p-1}$ so it goes to $1$. – John Douma Oct 31 '18 at 22:45
  • No, $x \in Z(x)$ and so if $Z(x) \cong \mathbb{Z}/p\mathbb{Z}$ then where does $x$ map under this isomorphism – RhythmInk Oct 31 '18 at 22:47
  • Divide $x$ by $p$ and take the remainder. – John Douma Oct 31 '18 at 22:48
  • $x$ is an arbitrary ring element. That doesn't make any sense. – RhythmInk Oct 31 '18 at 22:50
  • I see. You should probably read the proof in the paper. It could be that the isomorphism isn't explicitly constructed since, as you say, we can invent a ring with arbitrary elements where division by $p$ makes no sense. – John Douma Oct 31 '18 at 22:54
  • Come to think of it, we can figure out where $x$ has to go. If your ring $R$ is of order $p$ then $x= 1_R+...+1_R$ for some number of $1_R$s. Therefore the isomorphism must take $x$ to $1_{\mathbb Z}+...+1_{\mathbb Z}$ mod $p$. – John Douma Nov 01 '18 at 01:09
  • And we don't need the mod $p$ because we are already mapping $R$ to $\mathbb Z_{p}$. Remove the mod $p$ and replace $1_{\mathbb Z}$ with $1_{\mathbb Z_{p}}$. – John Douma Nov 01 '18 at 01:17

1 Answers1

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We recall that $R$ has an additive subgroup of order $\text{char}(R)$; thus, since $\vert R \vert = p^2$, either $\text{char}(R) = p$ or $\text{char}(R) = p^2$.

If

$\text{char}(R) = \vert R \vert = p^2, \tag 1$

we are done, since every element of $r$ is a sum of $1$s; so suppose

$\text{char}(R) = p; \tag 2$

then

$\Bbb Z_p \simeq \Bbb F_p \subsetneq R, \tag 3$

that is, $\Bbb F_p$ is a sub-field of $R$, isomorphic to $\Bbb Z_p \simeq \Bbb Z/p\Bbb Z$; thus, $R$ is an $\Bbb F_p$-vector space; now let

$R \ni x \notin \Bbb F_p; \tag 4$

such an $x$ exists since $\vert R \vert = p^2 > p = \vert \Bbb F_p \vert$, so $R \setminus \Bbb F_p \ne \emptyset$. I claim $1_R$ and $x$ are linearly independent over $\Bbb F_p$; if not, then there are $a, b \in \Bbb F_p$, not both $0$, with

$a + bx = a1_R + bx = 0, \tag 5$

or

$x = -b^{-1}a \in \Bbb F_p, \tag 6$

contrary to hypothesis; we conclude that $1_R$ and $x$ are in fact linearly independent over $\Bbb F_p$; therefore $1_R$ and $x$ span a $2$-dimensional subspace of $R$; but a $2$-dimensional vector space over a $\Bbb F_p$ has precisely $p^2$ elements. Therefore $1_R$ and $x$ in fact span $R$; thus given $y, z \in R$ we may write

$y = a + bx, \; z = c + dx, \; a, b, c, d \in \Bbb F_p; \tag 7$

it then follows that

$yx = (a + bx)(c + dx) = ac + (ad + bc)x + bdx^2$ $= ca + (da + cb)x + dbx^2 = (c + dx)(a + bx) = zy, \tag 8$

and we see that $R$ is commutative.

Robert Lewis
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