# Classifying Unital Commutative Rings of Order $p^2$

I'm trying to classify unital commutative rings of order $p^2$, where $p$ is a prime. At first, I happened to neglect the 'unital' and 'commutative' requirements, and after an arduous route I managed to show that there are $11$ such rings up to isomorphism. Some are non-commutative and non-unital, and the journey to that result is a bit ugly for a class on commutative algebra.

This should be easier if we only consider unital rings of order $p^2$. (It's easy to show that if it's unital, then it's commutative in this case). But so far, I haven't found a solution that doesn't use the fact that I found representations for the $11$ rings.

I am certain that a much less technical and messy method to find unital commutative rings of order $p^2$ is possible. For reference on the context, this question comes amidst a review of the Chinese Remainder Theorem, the Structure Theorem on Modules over a PID, tensor products, and algebras. I heavily suspect that if I were more fluent in applying the CRT, I would be able to get there.

Do you have any ideas?

By the way, I believe there are $4$. They should look like $\mathbb{F}_{p^2}, \mathbb{Z}_{p^2}, \mathbb{Z}_p [x]/(x^2),$ and (I don't know a convenient name for the fourth, but e.g., the Klein 4-group with standard ring structure on top, which I'm inclined to designate $\mathbb{Z}_{p \times p}$). In a more convenient designation, 1 is built on the additive group $C_{p^2}$ and 3 are built on $C_p \times C_p$. I would be content if I could enumerate how many rings are on each additive group, up to homomorphism, instead of actually classifying them.

• BTW, I'm not sure howyou got $K[x]/(x^2)$ with $K$ of characteristic zero. This is a $K$-algebra over a field of characteristic zero...hence infinite. However, if you take $K$ to be $\mathbb{Z}/p\mathbb{Z}$, your list becomes correct. Commented Feb 15, 2012 at 4:04
• @mixedmath: $K$ should not just be of characteristic $p$, but actually of order $p$. (So basically $\mathbb{F}_p$) Commented Feb 15, 2012 at 4:40
• @Cam: On thinking about it, you're right too. Thanks! Commented Feb 15, 2012 at 4:55
• The ring $R[\epsilon]/(\epsilon^2)$ with $R$ commutative is often called the "ring of dual numbers (over $R$)". So that would be a convenient name for the fourth example. Commented Feb 15, 2012 at 5:30
• Related: oeis.org/A127707/list Commented Sep 6, 2016 at 9:33

In a unital ring $R$ of order $p^2$ the additive order of $1$ can only be $p$ or $p^2$. In the second case, $1$ generates the additive group, so it is clear that the ring itself is isomorphic to $\mathbb Z/(p^2)$. So we concentrate in the first option.

In that case all non-zero elements have additive order $p$, so the ring is in fact a $2$-dimensional $\mathbb F_p$-algebra. Pick any element $x\in R$ which is linearly independent with $1$, so that $\{1,x\}$ is a basis. We must have $x^2=a+bx$ for some $a$, $b\in\mathbb F_p$, and then $R$ is isomorphic to $\mathbb F_p[X]/(X^2-aX-b)$.

If $X^2-aX-b$ is irreducible over $\mathbb F_p$, then $R$ is a field, and there is only one field of order $p^2$: $\mathbb F_{p^2}$. If not, then it factors as $(X-\alpha)(X-\beta)$ over $\mathbb F_p$. It is easy to see now that if $\alpha\neq\beta$ we have $R\cong\mathbb F_p\times\mathbb F_p$ and if $\alpha=\beta$ then $R\cong\mathbb F_p[X]/(X^2)$.

Interestingly, the fact that $R$ is commutative plays no role here, and there are no non-commutative rings of order $p^2$.

For fun, let us suppose now that $R$ does not have a unit, and let us see what happens.

First, suppose the addititive group is cyclic, so that there is an additive generator $x\in R$ of order $p^2$. Then there is an $n\in\{0,\dots,p^2-1\}$ such that $x^2=nx$. It is clear that the isoclass of $R$ is determined by $n$. But there is an ambiguity, as there are other generators: any other generator of the additive group is of the form $ax$ with $a$ a unit of $\mathbb Z/(p^2)$. If instead of $x$ we had started with $y=ax$, then as $y^2=a^2x^2=a^2nx=any$, instead of $n$ we would have found $an$. It follows that the isomorphism classes of rings of this form are parametrized by the quotient of $\mathbb Z/(p^2)$ under the action of its group of units given by left multiplication. If I did not get this too wrong, there are three orbits: the one for $0$, the one for $1$ and the one for $p$, so there are three rings of this type.

Second, let us suppose that the additive group is not cyclic, so that all non-zero elements have order $p$, and $R$ is in fact a $\mathbb F_p$-vector space.

Suppose there is a non-zero idempotent $e\in R$. Let $\lambda:a\in R\mapsto ea\in R$ and $\rho:a\in R\mapsto ae\in R$. These are two idempotent $\mathbb F_p$-linear maps which commute. Linear algebra tells us then that there is a direct sum decomposition $$R=R_{00}\oplus R_{10}\oplus R_{01}\oplus R_{11}$$ with \begin{align} &R_{00}=\{x\in R:ex=xe=0\},\\ &R_{10}=\{x\in R:ex=x, xe=0\},\\ &R_{01}=\{x\in R:ex=0, xe=x\}, \\ &R_{11}=\{x\in R:ex=xe=x\}. \end{align} At most two of these subspaces can be non-zero, and we know that $R_{11}$ contains $e$. Also, $R_{11}$ cannot be all of $R$ because we are supposing there is no unit.

• If there is a non-zero element in $R_{10}$, call it $x$. Then $\{e,x\}$ is a basis, $ex=x$, $xe=0$ and $xx=x(ex)=(xe)x=0x=0$. The multiplication is therefore completely determined.

• Similarly, if there is a non-zero element $x\in R_{01}$, $\{e,x\}$ is again a basis, and we have $ex=0$, $xe=x$ and $xx=0$.

• Finally, if there is a non-zero element $x\in R_{00}$, we have $ex=xe=0$. Then $ex^2=(ex)x=0$ and similarly $x^2e=0$, so $x^2$ is too in $R_{00}$, and there is a scalar $a$ such that $x^2=ax$. If $a\neq0$, we let $y=a^{-1}x$, so that $y^2=a^{-2}x^2=a^{-1}x=y$, and since $\{e,y\}$ is a basis this determines the multiplication. If $a=0$, of course the multiplication is also fixed.

We are left with the case in which there are no non-zero idempotents in $R$. A classical theorem of Albert implies that all non-zero elements must be nilpotent. If there is an $x\in R$ which is non-zero and such that $x^2$ is linearly independent with $x$, then $\{x,x^2\}$ is a basis, and we must have $x^3=0$, for the map $a\in R\mapsto xa\in R$, being a nilpotent endomorphism of a vector space of dimension $2$, must have nilpotency index at most $2$. We see that the multiplication table is completely determined here too.

Finally, suppose all non-zero elements are nilpotent but for for each $x$ of them, we have that $x^2$ is a linear multiple of $x$. All elements of $R$ must therefore square to zero. Let $\{x,y\}$ be a basis, and suppose $xy=ax+by$ and $yx=cx+dy$. Then $0=(x+y)^2=(a+c)x+(b+d)y$, so $yx=-xy$. Now call $u=xy$. If $u$ is zero, then all products are zero. If $\{x,u\}$ is a basis, then we know its the multiplication, as $x^2=u^2=xu=ux=0$. If not, $\{y,u\}$ is a basis, and again all products are zero.

• Crap, I like your answer better than mine. (Mine has the stink of the Artinian principal rings I've been thinking about on it, rather unnecessarily as you show.) Commented Feb 15, 2012 at 4:30
• The usage of Albert's theorem that «a finite dimensional, power-associative algebra which is not a nilalgebra contains an idempotent» should be replaceable by a simpler argument... Commented Feb 15, 2012 at 5:49
• Also, the direct sum decomposition associated to the idempotent $e$ I used at the end is the so called Pierce decomposition, a simple idea which is behind tons of great classical algebra! Commented Feb 15, 2012 at 5:52
• @MarianoSuárez-Álvarez Sorry, but I'm new to abstract algebra. Could you explain this please: "In that case all non-zero elements have additive order p, so the ring is in fact a 2-dimensional Fp-algebra"? Commented Oct 19, 2023 at 22:29

What a strange coincidence: I've been wasting a bunch of time recently thinking about principal Artinian rings. Now every commutative, unital ring of order $p^2$ is principal: indeed, every nonzero proper ideal must in particular have order $p$ so is even generated as an additive subgroup by any one nonzero element.

Step 1: The only (commutative, unital: this will be omitted from now on) ring of order $p$ is $\mathbb{Z}/p\mathbb{Z}$. For instance, it must have a maximal ideal and a residue field, and the size constraints force the maximal ideal to be $(0)$ and the residue field to have order $p$.

Step 2: Every finite ring is an Artinian ring, hence isomorphic to a product of Artinian local rings. So for rings of order $p^2$, the only one which is a nontrivial product is $\mathbb{Z}/p\mathbb{Z} \times \mathbb{Z}/p \mathbb{Z}$. The others are local Artinian principal rings. Let $R$ be such a ring of order $p^2$.

Step 3: It is easy to see that the characteristic of $R$ is either $p^2$ or $p$ and in the former case we must have $R \cong \mathbb{Z}/p^2 \mathbb{Z}$.

Step 4: Suppose that $R$ has characteristic $p$. If the unique maximal ideal if $(0)$, then $R \cong \mathbb{F}_{p^2}$ is the finite field of order $p^2$. Otherwise the maximal ideal is generated by an element $t$ with $t^2 = 0$. From this it is easy to see that $R \cong (\mathbb{Z}/p \mathbb{Z})[t]/(t^2)$.

Final tally: $R$ is one of

$\mathbb{Z}/p \mathbb{Z} \times \mathbb{Z} / p \mathbb{Z}, \ \mathbb{Z}/p^2 \mathbb{Z}, \mathbb{F}_{p^2}, \ (\mathbb{Z}/p\mathbb{Z})[t]/(t^2)$.