# There are $10$ commutative rings of order $8$

I have read that there are $10$ commutative rings of order $8$ (in this question, rings are unital and associative by definition), but I haven't found a concrete list of them. But I think that they are the following:

• $\mathbb{F}_2 \times \mathbb{Z}/4$
• $\mathbb{F}_2 \times \mathbb{F}_2 \times \mathbb{F}_2$
• $\mathbb{F}_2 \times \mathbb{F}_4$
• $\mathbb{F}_2 \times \mathbb{F}_2[x]/(x^2)$
• $\mathbb{Z}/8$
• $\mathbb{F}_8$
• $\mathbb{F}_2[x]/(x^3)$
• $\mathbb{Z}/4[x]/(2x,x^2)$
• $\mathbb{Z}/4[x]/(2x,x^2-2)$
• $\mathbb{F}_2[x,y]/(x^2,xy,y^2)$

Can someone confirm this? Also, is there a direct proof that these are all?

Since rings of order $2$ and $4$ are easy to classify and every finite commutative ring is a direct product of local commutative rings, we may restrict our attention to local commutative rings of order $8$. These will be the last $6$ in the list.

The paper "Associative rings of order p^3" by Gilmer and Mott is not very specific about the proofs, and the paper "Finite associative rings" by Raghavendran is rather confusing for me. The papers also consider non-unital and non-commutative rings.

Edit. Here is a proof that these rings $R_1,\dotsc,R_{10}$ are pairwise not isomorphic. Probably it is not optimal, but for the moment it just works.

• $R_1,\dotsc,R_4$ are not local, but $R_5,\dotsc,R_{10}$ are.
• $R_2,R_3$ are reduced, but $R_1,R_4$ are not.
• $R_1$ has characteristic $4$, but $R_4$ has characteristic $2$.
• $R_2$ has a trivial group of units, but $R_4$ does not.
• $R_5$ is the only ring of characteristic $8$.
• $R_6$ is the only field.
• $R_7,R_{10}$ have characteristic $2$, but $R_8,R_9$ have characteristic $4$.
• $R_{10}$ has a unique maximal ideal $\mathfrak{m}$ satisfying $\mathfrak{m}^2=0$, but the one of $R_7$ only satisfies $\mathfrak{m}^3=0$.
• $R_9$ contains a root of $2$, but $R_8$ does not, because $R_8/(x) = \mathbb{Z}/4$ does not.
• Hint: Such a ring is in particular an Artin ring, hence a direct product of local Artin rings.. – Juan Fran Oct 26 '16 at 14:06
• @JuanFran: Of course, I know that. ;) – HeinrichD Oct 26 '16 at 14:08
• Apparently not, because once you know this you can easily figure this out... if $A$ is such a ring then $A=A_1\times A_2\times A_3$, $8=\# A_1 \# A_2 \# A_3$... – Juan Fran Oct 26 '16 at 14:13
• Have you seen the list? Not every ring is decomposable, because it can be already local. (By the way, it is not respectful and not clever either to tell me that I "apparently not" know this decomposition result.) – HeinrichD Oct 26 '16 at 14:15
• This clearly does not answer the question. And yes, my question is exactly about this "some work". – HeinrichD Oct 26 '16 at 14:19

The hard part of this is showing that the only commutative rings with additive group $(\mathbb{Z}/2\mathbb{Z})^3$ are the six you mentioned; the other additive groups have at most two generators and I imagine you've completed that part.

Here I'll give two proofs. The second proof is much shorter than the first.

## First Proof

To this end, I will call the generators of $(\mathbb{Z}/2\mathbb{Z})^3$ $1,e_2$ and $e_3$ (WLOG the multiplicative identity is a generator, since any nonzero element can be a generator). And before fixing our perspective on $e_2$ and $e_3,$ I will split it into two cases: rings with a subring with additive group $(\mathbb{Z}/2\mathbb{Z})^2,$ and rings with no such thing.

Let's do the second case first. Notice that there are only three multiplications to fix to determine the entire ring: $e_2^2, e_2e_3,$ and $e_3^2.$ Now since we said that there is no subring with additive group $(\mathbb{Z}/2\mathbb{Z})^2,$ it must be true that $1, e_2,$ and $e_2^2$ generate the whole group; thus WLOG $e_3=e_2^2.$

Since now $e_3^2=e_2(e_2e_3),$ there is in fact only one multiplication to set, $e_2e_3.$ Of the eight possibilities, four immediately give errors: $e_2e_3$ cannot be equal to $0, e_2, e_3,$ or $e_2+e_3+1,$ or we would obtain that $\{0,1,e_3,e_3+1\}$ is a subring. Thus we are left with:

• $e_2e_3=1$
• $e_2e_3=e_2+1$
• $e_2e_3=e_3+1$
• $e_2e_3=e_2+e_3.$

But in fact the first and fourth of these give errors as well, since $\{0,1,e_2+e_3,e_2+e_3+1\}$ is a subring in these cases. Another way to say the second and third of these is:

• $e_2^3-e_2-1=0$
• $e_2^3-e_2^2-1=0.$

Both of these clearly define $\mathbb{F}_8.$

Now we consider the case where there is a subring isomorphic to $\mathbb{F}_4, \mathbb{F}_2[x]/(x^2),$ or $\mathbb{F}_2^2.$ In each case let us say that that subring is $\{0,1,x,x+1\},$ and the other four elements are $\{y,y+1,x+y,x+y+1\}.$ Either $xy$ is to be found in the former subring or it isn't. In the first case by choosing $y$ appropriately we may have $xy=0$ in the case of $\mathbb{F}_4,$ and we may have either $xy=0$ or $xy=1$ in the cases of $\mathbb{F}_2[x]/(x^2)$ and $\mathbb{F}_2^2.$ Since the value of $y^2$ hasn't been set yet, this leaves us a priori with 40 possibilities. But in case $xy=0,$ we must have that $y^2$ is annihilated by $x,$ which is 2/8 for $\mathbb{F}_4$ and 4/8 for $\mathbb{F}_2[x]/(x^2), \mathbb{F}_2^2.$ And in case $xy=1,$ we must have $xy^2=y,$ which has been rendered impossible. So there are only 10 possibilities, in each of which $xy=0$:

• $\mathbb{F}_4,~y^2=0$
• $\mathbb{F}_4,~y^2=y$
• $\mathbb{F}_2[x]/(x^2),~y^2=0$
• $\mathbb{F}_2[x]/(x^2),~y^2=x$
• $\mathbb{F}_2[x]/(x^2),~y^2=y$
• $\mathbb{F}_2[x]/(x^2),~y^2=x+y$
• $\mathbb{F}_2^2,~y^2=0$
• $\mathbb{F}_2^2,~y^2=x+1$
• $\mathbb{F}_2^2,~y^2=y$
• $\mathbb{F}_2^2,~y^2=x+y+1.$

But the sixth case repeats the fifth under the replacement $z=x+y,$ and the eighth repeats the seventh under the replacement $z=x+y+1.$

The first ring is $\mathbb{F}_2[z]/(z^3),$ where the generator of the $\mathbb{F}_4$ subring is $z+1$ and $y=z^2.$ The second is clearly $\mathbb{F}_4\times\mathbb{F}_2.$ The third is $\mathbb{F}_2[x,y]/(x^2,xy,y^2)$ verbatim, the fifth is $\mathbb{F}_2[x]/(x^2)\times\mathbb{F}_2,$ and the fourth is $\mathbb{F}_2[y]/(y^3).$ The seventh repeats the fifth under $x\leftrightarrow y,$ the ninth is clearly $\mathbb{F}_2^3,$ and the last is $\mathbb{F}_4\times\mathbb{F}_2,$ where $(\omega,0)=y$ and $(0,1)=x.$

If $xy$ is one of the "upper four," then all possibilities for the value of $xy$ in the cases of the subrings $\mathbb{F}_4$ or $\mathbb{F}_2[1+i]$ lead to contradictions with the associativity of $x^2y$ (here $x=\omega,1+i$ resp.)

In the case of $\mathbb{F}_2^2,$ if $xy$ is one of the "upper four" and $y^2$ is one of the "lower four," then we can say WLOG either $y^2=0$ or $y^2=x.$ This makes a priori 8 possibilities. But only the following two do not lead to contradictions:

• $\mathbb{F}_2[e],~y^2=0,~xy=y$
• $\mathbb{F}_2[e],~y^2=x,~xy=y.$

(I have written $\mathbb{F}_2^2$ using an idempotent $e=x.$)

The former is $\mathbb{F}_2[i]\times\mathbb{F}_2,$ where $y=(1+i,0)$ and $x=(0,1),$ and the latter is the former under the transformation $z=x+y.$

If $xy$ and $y^2$ are in the "upper four," we still have a priori 16 cases to consider. But in fact we must have either $xy=y$ or $xy=x+y+1$ for the associativity of $x^2y,$ thus there are only eight.

• $\mathbb{F}_2^2,~xy=y,~y^2=y$
• $\mathbb{F}_2^2,~xy=y,~y^2=y+1$
• $\mathbb{F}_2^2,~xy=y,~y^2=x+y$
• $\mathbb{F}_2^2,~xy=y,~y^2=x+y+1.$
• $\mathbb{F}_2^2,~xy=x+y+1,~y^2=y$
• $\mathbb{F}_2^2,~xy=x+y+1,~y^2=y+1$
• $\mathbb{F}_2^2,~xy=x+y+1,~y^2=x+y$
• $\mathbb{F}_2^2,~xy=x+y+1,~y^2=x+y+1.$

($x$ is an idempotent nonunit of $\mathbb{F}_2^2$ in each case.)

But the second and fourth violate the associativity of $xy^2.$ I'll leave it to you to determine which of the first, third, and fifth through eighth are valid rings, and which rings in your list they correspond to.

## Second Proof

This begins as a converse to what you noted. We have either a field, a ring with a unique nonzero maximal ideal, or a nonlocal ring. In the case of a field we know it's $\mathbb{F}_8.$ In the case of a nonlocal ring we have Theorem 3.1.4 on page 40 of Finite Commutative Rings and Their Applications by Bini and Flamini:

A finite, commutative ring with identity, R, can be expressed as a direct sum of local rings. This decomposition is unique up to permutation of direct summands.

(It's a good read!) and so this leaves only the local case.

The maximal ideal $\mathfrak{m}$ has either two or four elements, and we'll rule out $|\mathfrak{m}|=2$ at the end.

In the case of four elements, let's call those elements $0,x,y,x+y.$ If $\mathfrak{m}^2=0$, we have $x^2=xy=y^2=0,$ so we get $\mathbb{F}_2[x,y]/(x^2,xy,y^2).$

If $\mathfrak{m}^2\neq0,$ then we can't have both $x^2=0$ and $y^2=0,$ or else $(x+y)^2=0,$ and thus $(xy)^2=0$ leads to a contradiction in any of the three choices $xy=x,~xy=y,~xy=x+y$ by multiplying through by $x$ or $y.$ But $xy\in\mathfrak{m}$ then implies $xy=0,$ which means $\mathfrak{m}^2=0,$ contradiction. So WLOG $y=x^2,$ our eight elements are the at-most-quadratic polynomials in $x,$ and we need only determine which of the four elements of $\mathfrak{m}$ is $x^3.$ The choice $x^3=0$ leads to $\mathbb{F}_2[x]/(x^3).$ The choices $x^3=x$ and $x^3=x^2$ lead to $x+1$ being uninvertible, which is a contradiction since any element outside of $\mathfrak{m}$ must be invertible. And the choice $x^3=x^2+x$ leads to $x^2+x+1$ being uninvertible.

Finally, if $|\mathfrak{m}|=2,$ then $R/\mathfrak{m}\cong\mathbb{F}_4$ and $R$ has six units. Take a unit $x$ other than 1. Then $\bar x=\omega$ by an isomorphism, $\bar x^2=\omega+1,$ and $\bar x^3=1,$ and $x,x^2,x^3$ are all units in $R.$ Say $\mathfrak{m}=\{0,y\}.$ We can't have $y=x+1$ because they go to different things in the quotient, and so we have a unique way to write all eight elements of $R.$ Now $xy\in\mathfrak{m}$ implies $xy=0$ or $xy=y.$ But note too that $(x+1)y=0$ or $(x+1)y=y.$ Only two of the four possibilities do not lead directly to a contradiction and they are symmetric, so we may say $xy=0.$ But then if $x^2=x+1,$ multiplying through by $y$ gives us $y=0,$ and if $x^2=x+y+1$ then multiplying through by $y$ gives us $y^2=y.$ These are the only two possibilities since $\bar x^2=\overline{x+1},$ the former is a contradiction since $y\neq0,$ and the latter gives us $R\cong\mathbb{F}_4\times\mathbb{F}_2,$ which is a contradiction since we said $R$ is local.

• How do you conclude $(x+y)^2=0$ from $x^2=y^2=0$? It seems that you assume that the char. is $2$? – HeinrichD Aug 28 '17 at 14:02
• As I mentioned at the beginning of the answer, the hard part of this is showing that the only commutative rings with additive group $(ℤ/2ℤ)^3$ are the six you mentioned; the other additive groups have at most two generators and I imagine you've completed that part. – Daniel Briggs Aug 29 '17 at 1:25