Ring of order $p^2$ is commutative.

Question

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 $p$?

Answer 1

Let $R$ be a ring with $p^2$ elements. Let $x \in R$, we have to show that $Z(x) := \{r \in R : xr=rx\}$ coincides with $R$. It is an additive subgroup, even a subring, and therefore has order $p$ or $p^2$. If it has order $p^2$, 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 $Z(x)=R$, which has order $p^2$, a contradiction.

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

Answer 2

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.

Answer 3

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.

Answer 4

Let $R$ be a ring with unity and $|R|= p^2,$ where $p$ is prime. Let $I$ be a principal ideal of $R$ generated by some $x.$ If $|I|=p^2$ then $I=R$ so if $(R,+) \cong \mathbb{Z}/p \oplus \mathbb{Z}/p$ then we get a contradiction since $\mathbb{Z}/p \oplus \mathbb{Z}/p$ does not contain such an $I.$ So $R\cong \mathbb{Z}/p^2$ so there exists $x\in R$ such that every element of is of the form $k.x$ for some $1\leq k\leq p^2$ and hence the product of any $2$ elements will commute as a result of being finite sums of $x.$

Hence every principal ideal must have order $p.$ Let $x,y\in R$ be arbitrary, $x \neq 1, 0$ and let $I_1, I_2$ be the principal ideal generated by $x$ and $y$ respectively. Then both $(I_1, +)$ and $(I_2, +)$ are cyclic of order $p$ and hence can be written as $I_1 = \{0,a, \ldots (p-1).a\}$ and $I_2=\{0,b, \ldots (p-1).b\}.$ By definition of an ideal, $ab, ba \in I_1\cap I_2 = \{0\}$ it follows that $ab =ba=0.$

Since $I_1+I_2$ is an ideal of size $|I_1|. |I_2|/|I_1\cap I_2| = p^2$ we have $R= I_1+I_2.$ Then for any $c, d \in R$ we have $c = m_1.a +n_1.b$ and $d=m_2.a +n_2.b$ so that $cd = (m_1m_2).a + (n_1n_2).b =dc$ and hence $R$ is commutative.

Answer 5

I am just posting the existing proofs but in more details. $\DeclareMathOperator{\Z}{\mathbb Z}$ $\textbf{Proof by isomorphism}\tag*{}$

Let $p$ be a prime number and $(R,+,\cdot)$ be a ring of order $p^2$ with identity.

We note that $(R,+)$ is abelian. By fundamental theorem of abelian groups, there are only two ways to express an abelian group of order $p^2$ as product of cyclic groups. (Cases 1 and 2).

$\\$ $\mathbf{Case\ 1: } \ (R,+)\cong \Z_p\times \Z_p\tag*{}$

$\Z_p\times \Z_p$ is generated by two elements of order $p$.

$\begin{align}\Z_p\times \Z_p&=\langle (1,0), (0,1)\rangle\\ &=\{m(1,0)+n(0,1) : m,n\in \Z\}\end{align}$

By isomorphism, $(R,+)$ can also be generated by two elements of order $p$.

By Lagrange's theorem, $\text{ord}(1_R)=p$. (Since $1_R\neq 0_R$, it is clear that $\text{ord}(1_R)\neq 1$. The group is not cyclic so it does not have an element of order $p^2$. This leaves the factor $p$.)

If we choose one of our generators to be $1_R$, there exists another element $x\in R$ such that $(R,+)=\langle 1_R, x\rangle$. Thus, any element $r\in R$ can be written in the form $(m\cdot 1_R+n\cdot x)$ where $m,n\in\Z$.

Pick $r_1=m_1\cdot 1_R + n_1\cdot x$ and $r_2=m_2\cdot 1_R + n_2\cdot x$ from $R$. Show that $r_1r_2=r_2r_1$.

$\mathbf{Case\ 2: } \ (R,+)\cong \Z_{p^2}\tag*{}$ It is cyclic so it has an additive generator $x\in R$ so that $R=\{n\cdot x: n\in \Z\}$.

Every element is of the form $n\cdot x$. Pick $r_1=n_1\cdot x$ and $r_2=n_2\cdot x$. Show that $r_1r_2=r_2r_1$.

Lemma: If the additive group $(R,+)$ is cyclic, the ring R is commutative.

Conclusion: A ring with identity of order $p^2$ is commutative.

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A ring without identity of order $p^2$ may or may not be commutative. Example: This ring of matrices over the field $\Z_p$. $\left\{\begin{pmatrix}a&b\\ 0& 0\end{pmatrix}: a,b\in \Z_p\right\}\tag*{}$ is a non-commutative ring of order $p^2$. It does not have an identity.

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$\textbf{Proof by considering the centre}\tag*{}$

Let $p$ be a prime number and $(R,+,\cdot)$ be a ring of order $p^2$ with identity.

The center $Z(R)$ is subring of $R$ which is also an additive subgroup.

Clearly, $0_R,1_R\in Z(R)$ so by Lagrange’s theorem, $|Z(R)|=p$ or $p^2$.

If $|Z(R)|=p^2$, we are done.

If $|Z(R)|=p$, consider the quotient group $R/Z(R)$ under coset addition operation.

$\displaystyle |R/Z(R)|=\frac{|R|}{|Z(R)|}=\frac{p^2}{p}=p\tag*{}$

Thus, $R/Z(R)$ is cyclic group under coset addition. Let $r\in R$ such that $r+Z(R)$ is the generator of $R/Z(R)$.

Every coset in $R/Z(R)$ is of the form $n\left(r+Z(R)\right)=nr+Z(R)\tag*{}$ for some integer $n\in \Z$.

Pick any $x\in R$ then $x$ must be present in one of the cosets.

Let $x\in mr+Z(R)$ for some integer $m$. $\implies x=mr+c$ for some $c\in Z(R)$. Pick another element $y\in R$ then $y=nr+c'$ for some $c'\in Z(R)$ and $n\in\Z$.

Now show that $xy=yx$ using the fact that $c,c'\in Z(R)$.

Note that, in this proof, we used the fact that $1_R$ exists and it does not equal $0_R$ so that $|Z(R)|\geq 2$.