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I would like to show that ring of order $p^2$ is commutative.

Taking $G=(R, +)$ as group, we have two possible isomorphism classes $\mathbb Z /p^2\mathbb Z$ and $\mathbb Z/ p\mathbb Z \times \mathbb Z /p\mathbb Z$.

Since characterstic must divide the size of the group then we have two possibilities $p$ and $p^2$.

Now IU don't understand how can I reason to say that the multiplication is commutative and how can I conclude for the case when characterstic is $p$?

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3 Answers 3

up vote 10 down vote accepted

Let $R$ be a ring with $p^2$ elements, let $0 \neq x \in R$, we have to show that $Z(x) = \{r \in R : xr=rx\}$ coincides with $R$. It is an additive subgroup, even a subring, contains $x$, and therefore has order $p$ or $p^2$. In the latter case we are done. Assume that it has order $p$. Every ring of order $p$ is canonically isomorphic to $\mathbb{Z}/p$. It follows that $x=z \cdot 1$ for some $z \in \mathbb{Z}$. But then obviously $Z(x)=R$.


For rings without unit, also called rngs, this fails: There are $11$ rngs with $p^2$ elements, see here. Two of these are non-commutative, namely $E=\langle a,b : pa=pb=0, a^2=a, b^2=b, ab=a, ba=b \rangle$ and $F = \langle a,b : pa=pb=0, a^2=a, b^2=b, ab=b, ba=a\rangle.$

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I hadn't read this while composing my answer (I never think about non unital rings !): +1 . –  Georges Elencwajg Feb 16 '13 at 15:20

Warning: I assume here that "ring" means "unital ring", not "rng" without unity.

There is a canonical ring morphism $f:\mathbb Z\to R$ (this is true for all rings).
Its image $f(\mathbb Z)\subset R$ has cardinality either $p^2$ or $p$.
$\bullet $ In the first case $f(\mathbb Z)=R$ and since $f(\mathbb Z)= \mathbb Z/p^2\mathbb Z$ (the only quotient of $\mathbb Z$ of cardinality $p^2$) we are done: $R= \mathbb Z/p^2\mathbb Z$, a commutative ring.

$\bullet \bullet$ In the second case $f(\mathbb Z)= \mathbb Z/p\mathbb Z$ (the only quotient of $\mathbb Z$ of cardinality $p$) and $R$ is a $\mathbb Z/p\mathbb Z$-algebra.
That algebra is then generated by any element $r\in R\setminus (\mathbb Z/p\mathbb Z)$, i.e. $R=\mathbb Z/p\mathbb Z[r]$, which immediately implies that $R$ is commutative, since $f(\mathbb Z)=\mathbb Z/p\mathbb Z$ is in the center of $R$ and since powers of $r$ commute with each other.

Complement
Actually, we can classify all the rings in $\bullet \bullet$.
If $m(x)=x^2+ax+b\in \mathbb Z/p\mathbb Z[x]$ is the minimal polynomial of $r$ over $\mathbb Z/p\mathbb Z$ we then have $R=\frac{ \mathbb Z/p\mathbb Z[x]}{\langle m(x)\rangle}$ and it follows that $$R=\mathbb F_{p^2} \;\text {(the field with} p^2 \text {elements)},\;\mathbb Z/p\mathbb Z\times \mathbb Z/p\mathbb Z \;\text{or} \;(\mathbb Z/p\mathbb Z)[x]/(x^2)$$ according as $m(x)$ is irreducible, reducible with distinct roots or reducible with a double root.

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Now $\mathbb{Z}/p^2$ is missing. Sorry ^^ –  Martin Brandenburg Feb 16 '13 at 15:25
    
@Martin, yes you are quite right: due to some stupid confusion of mine I had replaced the correct nilpotent algebra $(\mathbb Z/p\mathbb Z)[x]/(x^2)$ by the ring $\mathbb Z/p^2\mathbb Z$ already mentioned in $\bullet$, which isn't even a $\mathbb Z/p\mathbb Z$-algebra! Corrected now (your last comment has been teken care of by explicitly stating that the classification only concerns $\bullet \bullet)$. Thanks a lot for your vigilance. –  Georges Elencwajg Feb 16 '13 at 15:44
    
You're welcome. So there are $11$ rngs of order $p^2$, $4$ rings of order $p^2$, and they are all commutative. $p^3$ is far more complicated: Associative rings of order $P^3$ by Robert Gilmer and Joe Mott. –  Martin Brandenburg Feb 16 '13 at 16:04
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"Associative rings"? Good grief, can't we agree on any of the properties of a ring?? –  Pete L. Clark Feb 16 '13 at 19:25
    
I completely agree with you, @Pete. And "good grief" reminds me of Charlie Brown, who so amused and moved me in the golden days when I started reading in English, a long time ago. Sigh... (as he also often said) –  Georges Elencwajg Feb 16 '13 at 21:22

Recall that a ring which is generates by one element as a ring is commutative. Indeed, it is an epimorphic image of $\mathbb Z[X]$.

Let now $R$ be of order $p^2$. Then $R$ is generated as a ring by one element:

  • If the additive group is cyclic, then any additive generator will generate $R$ as a ring.

  • If the additive group is not cyclic, it is generated by any two $\mathbb F_p$-linearly independent elemements. Since $1\in R$ is not zero, we can pick a $x\in R$ such that $\{1,x\}$ generates the additive group. In particular, $x$ generates $R$ as a ring.

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My rings have a unit, of course. –  Mariano Suárez-Alvarez Feb 16 '13 at 19:36
    
i came to a situation something like saying that $x^2=kx$ , but i couldn't really justify if it was true . –  Theorem Feb 16 '13 at 19:38
    
I do not understand your comment, @Theorem. –  Mariano Suárez-Alvarez Feb 16 '13 at 19:44
    
Can you elaborate on your second point . when you say that if the additive group is not cyclic then it is generated by two $\mathbb F_p$- linearly independent elements ? –  Theorem Feb 16 '13 at 19:48
    
If the additive group is not cyclic, we know it is isomorphic to a vector space of dimension $2$ over the field $\mathbb F_p$. –  Mariano Suárez-Alvarez Feb 16 '13 at 19:52

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