# Examples of finite local rings of length 2 or 3

What is an example of a finite local rings, that has length 2 or 3? I want something different from $F_{q}[x] / x^{i}$ for $i=2, 3$; I'm looking for something more interesting. If you can give me examples of higher length, yet have "simple structure" (e.g. $F_{q}[x]/x^{i}$), that would be nice too.

I know this is related to Classification of finite commutative rings, but I didn't completely understand the answer there.

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The rings you are looking at are all artinian. So I suppose if you take any noetherian domain of dimension 1 $A$ (e.g. Dedekind domain), take one maximal ideal $m$, and consider $A/m^k$, then you would have a large class of examples. (of similar type to the one you've given) –  Ho Chung Siu Dec 5 '09 at 5:24

There are a bunch of different notions of length/depth in ring theory: Projective length, Artinian length, local depth, etc. If we take length to mean Artinian length, then Charles is right: The Artinian length of a finite-dimensional commutative algebra is just its dimension. Every such algebra is a direct sum of local ones, and you can chip away at each local summand of the ring from the bottom end, one dimension at a time.

The local algebras that have a description that looks as nice as $\mathbb{F}[x]/(x^n)$ are the toric ones. These local algebras are $\mathbb{F}[\vec{x}]$ divided by an ideal generated by monomials and a basis of monomials. You can make a diagram of the exponents of the monomials that aren't killed by the ideal. If the ring has $n$ generators, then the diagram is a stable stack of blocks in the $n$-dimensional orthant. For example, the algebra $\mathbb{F}[x,y]/(x^3,x^2y,y^3)$ has a basis of seven monomials: 1, $x$, $x^2$, $y$, $xy$, $y^2$, $xy^2$. The diagram of these monomials looks like this:

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I have put a 1 at the corner in the diagram corresponding to the monomial 1.

What is easy to forget is that all finite-dimensional local algebras with one generator in $m/m^2$ are of this form, but with more variables these are just special examples.

I was looking at the second part of the question first, interesting higher-dimensional examples. Here are some non-isomorphic local rings $R$ (not necessarily algebras) that have length 2 or 3 and such that $R/m = \mathbb{Z}/p$:

Length 2:

1. $(\mathbb{Z}/p)[x]/(x^2)$
2. $\mathbb{Z}/p^2$

Length 3:

1. $(\mathbb{Z}/p)[x]/(x^3)$
2. $(\mathbb{Z}/p)[x,y]/(x^2,xy,y^2)$
3. $(\mathbb{Z}/p^2)[x]/(px,x^2)$
4.  $(\mathbb{Z}/p^2)[\sqrt{p}]$  $\mathbb{Z}[\sqrt{p}]/p^{3/2} = \mathbb{Z}[x]/(x^2-\lambda p,x^3)$
5. $\mathbb{Z}[x]/(x^2-\lambda p,x^3)$ where $\lambda \in \mathbb{Z}/p$ is not a square. (Noted by Jonathan Wise.)
6. $\mathbb{Z}[x]/(x^2+2x,4)$ (Similar idea to previous, in characteristic 2.)
7. $\mathbb{Z}/p^3$

(Edit: My notation for #4 was not strictly correct.)

I think, although I can't really speak with authority, that these are all of them. I thought that I knew all of these rings, but that was naive. One point is that among algebras over $\mathbb{Z}/p$, the length is too small to see anything non-toric. But you can also have local rings that look like these toric local algebras (which I listed first), but have carries. The most creative one is the fourth one of length 3, namely $(\mathbb{Z}/p^2)[\sqrt{p}]$. You can express an element of this ring as three digits in base $p$, say $d_2d_1d_0$. Then addition carries from $d_0$ to $d_2$.

I would also guess that all of these generalize to $\mathbb{F}_q$, using the Witt vector construction in the cases with carries. And maybe it is again all of them.

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They are not all, for example $Z_p[[x]]/(x^™,xp,p^2)$. See below for a complete characterization. –  Hailong Dao Dec 5 '09 at 6:50
How is that example different from my #3? –  Greg Kuperberg Dec 5 '09 at 6:57
Oops, you are right, sorry! How about something like $Z_p[\sqrt{p}]/(p\sqrt{p})$? This is not 1) because we can't embed $Z/p$ in it. –  Hailong Dao Dec 5 '09 at 16:04
I threw confusion onto the question by using the wrong notation for #4. Your new example is the same as that one. –  Greg Kuperberg Dec 5 '09 at 18:48
This seems to miss at least one example: $\mathbf{Z}[t] / (t^2 - \lambda p, t^3)$ where $\lambda$ is a non-quadratic residue modulo $p$. –  Jonathan Wise Dec 5 '09 at 20:07
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Any local rings of length 2 are of the form $R/n^2$ with $(R,n)$ a regular local ring of dimension 1.

Any local rings of length 3 are of the form $R/n^2$ with $(R,n)$ a regular local ring of dimension 2 or $R/n^3$ with $(R,n)$ a regular local ring of dimension 1.

Proof for the length 3 case: Let $A,m,k$ be our ring. Then the exact sequence: $$0 \to m \to A \to k \to 0$$

shows that $length(m)=2$. So the number of generators of $m$ is at most 2. If it is 2, then $length(m/m^2)=2$, thus $m^2=0$. So $A=R/n^2$ with $(R,n)$ a regular local ring of dimension 2. If it is 1, then $A$ is a quotient of a regular local ring of dimension 1, and counting length shows that we need to kill the cube of the (principal) maximal ideal.

Of course, $R$ may or may not contains a field, so one may have things like $k[[x,y]]/(x^2,xy,y^2)$ or $Z_p[[x]]/(p^2,xp,x^2)$.

MORE: We can assume our regular $R$ is complete, since $A$, being Artinian, is complete. Then the Cohen Structure Theorem completely described $R$. If $R$ contains a field, then $R \cong k[[x,y]]$. If $R$ is mixed charateristic and unramified, then $R\cong V[[x]]$, with $(V,pV)$ a discrete valuation ring. If $R$ is ramified, then $R=V[[x,y]]/(f)$, with $f=p+g$ such that $g \in (x,y)^2$. From those you can get $A$ accordingly by killing the square of the maximal ideal.

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I'm just going to consider the local rings with residue field $\mathbf{F}_p$.

Suppose A' is an extension of $\mathbf{Z}/p$ with ideal $\mathbf{Z}/p$. If we take the fiber product with $\mathbf{Z}$, we get an extension of $\mathbf{Z}$ with ideal $\mathbf{Z}/p$. Up to isomorphism, there is only one such extension: $B' = \mathbf{Z}[t] / (t^2, pt)$. The kernel of the map from B' to A' is isomorphic to $\mathbf{Z}$ and reduces modulo t to the ideal generated by p. Therefore, the square-zero extensions of $\mathbf{Z}/p$ are all isomorphic to $\mathbf{Z}[t] / (t^2, pt, p + \lambda t)$ for some $\lambda \not= 0$. If $\lambda$ is not a multiple of p, we get $\mathbf{Z} / p^2$; if $\lambda$ is a multiple of $p$, we get $\mathbf{Z}[t] / (p, tp, t^2) = \mathbf{F}_p[t] / t^2$. So these are all the length 2 finite local rings with residue field $\mathbf{F}_p$.

For length 3, we'll look for extensions of $\mathbf{Z} / p^2$ by $\mathbf{Z} / p$. The same analysis shows that these are all of the form $\mathbf{Z}[t] / (t^2, pt, p^2 + \lambda t)$. If $\lambda$ is not divisible by $p$, we get $\mathbf{Z} / p^3$ and if $\lambda$ is divisible by $p$ we get $\mathbf{Z}[t] / (p^2, pt, t^2)$.

We also have to look for extensions of $\mathbf{F}_p[t] / t^2$. By base change, any such extension A' gives an extension B' of $\mathbf{Z}[t]$ with ideal $\mathbf{Z} / p$. Once again, there is only one of these up to isomorphism (since $\mathbf{Z}[t]$ is projective, for example) and it is given by $\mathbf{Z}[t,u] / (u^2, pu, tu)$. The ideal of the map from B' to A' generates the ideal of the map from $\mathbf{Z}[t]$ to $\mathbf{F}_p[t] / t^2$. Since this is generated by p and t^2 the ideal of A' in B' is generated by $(p + \lambda u, t^2 + \mu u)$ and A' is of the form

$\mathbf{Z}[t,u] / (u^2, pu, tu, p + \lambda u, t^2 + \mu u)$

for some polynomials $\lambda, \mu \in \mathbf{Z}[t]$.

Edit: If $\lambda$ is not in $(p, t)$ then it is invertible in the quotient, so we get

$\mathbf{Z}[t,u] / (p^2, tp, t^2 + \mu \lambda^{-1} p)$.

There are two possibilities up to isomorphism here, depending on whether $- \mu \lambda^{-1}$ is a quadratic residue modulo p.

If $\lambda$ is in $(p,t)$ we get

$\mathbf{Z}[t,u] / (u^2, pu, tu, p, t^2 + \mu u) = \mathbf{F}_p[t,u] / (u^2, tu, t^2 + \mu u)$.

The $\mu$ is also not in $(p, t)$ then we get $\mathbf{F}[t] / t^3$. If $\mu$ is in $(p,t)$ we get $\mathbf{F}_p[t,u] / (u^2, tu, t^2)$.

Modulo any mistake I made above, I think a complete list is of length 3 finite local rings with residue field $\mathbf{F}_p$ is

1. $\mathbf{Z} / p^3$,
2. $\mathbf{Z}[t] / (p^2, pt, t^2)$,
3. $\mathbf{Z}[t] / (t^2 - p, t^3)$,
4. $\mathbf{Z}[t] / (t^2 - \alpha p, t^3)$ where $\alpha$ is a non-quadratic residue modulo $p$,
5. $\mathbf{F}_p[t] / t^3$, and
6. $\mathbf{F}_p[t,u] / (t,u)^2$.
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Did I miss any? –  Greg Kuperberg Dec 5 '09 at 7:28
Isn't that the only one of length two? And for length three, shouldn't $\mathbb{F}_q[x,y]/(x^2,xy,y^2)$ work? I believe (possibly incorrectly) that here length and dimension over the base field agree.
Ahh, yeah, as we've got a finite ground field, we'll also get things of the for $\mathbb{F}_q[x]/f$ where $f$ is an irreducible quadric or cubic, thus giving us $\mathbb{F}_{q^2}$ and $\mathbb{F}_{q^3}$, thanks for pointing that out. Is this list complete for length 2 and 3 things? I'm not immediately seeing how to construct more, but I'm mostly a complex numbers kind of guy. –  Charles Siegel Dec 5 '09 at 5:31