Saturday, 3 May 2008

ac.commutative algebra - Examples of finite local rings of length 2 or 3

I'm just going to consider the local rings with residue field mathbfFp.



Suppose A' is an extension of mathbfZ/p with ideal mathbfZ/p. If we take the fiber product with mathbfZ, we get an extension of mathbfZ with ideal mathbfZ/p. Up to isomorphism, there is only one such extension: B=mathbfZ[t]/(t2,pt). The kernel of the map from B' to A' is isomorphic to mathbfZ and reduces modulo t to the ideal generated by p. Therefore, the square-zero extensions of mathbfZ/p are all isomorphic to mathbfZ[t]/(t2,pt,p+lambdat) for some lambdanot=0. If lambda is not a multiple of p, we get mathbfZ/p2; if lambda is a multiple of p, we get mathbfZ[t]/(p,tp,t2)=mathbfFp[t]/t2. So these are all the length 2 finite local rings with residue field mathbfFp.



For length 3, we'll look for extensions of mathbfZ/p2 by mathbfZ/p. The same analysis shows that these are all of the form mathbfZ[t]/(t2,pt,p2+lambdat). If lambda is not divisible by p, we get mathbfZ/p3 and if lambda is divisible by p we get mathbfZ[t]/(p2,pt,t2).



We also have to look for extensions of mathbfFp[t]/t2. By base change, any such extension A' gives an extension B' of mathbfZ[t] with ideal mathbfZ/p. Once again, there is only one of these up to isomorphism (since mathbfZ[t] is projective, for example) and it is given by mathbfZ[t,u]/(u2,pu,tu). The ideal of the map from B' to A' generates the ideal of the map from mathbfZ[t] to mathbfFp[t]/t2. Since this is generated by p and t^2 the ideal of A' in B' is generated by (p+lambdau,t2+muu) and A' is of the form



mathbfZ[t,u]/(u2,pu,tu,p+lambdau,t2+muu)



for some polynomials lambda,muinmathbfZ[t].




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



mathbfZ[t,u]/(p2,tp,t2+mulambda1p).



There are two possibilities up to isomorphism here, depending on whether mulambda1 is a quadratic residue modulo p.



If lambda is in (p,t) we get



mathbfZ[t,u]/(u2,pu,tu,p,t2+muu)=mathbfFp[t,u]/(u2,tu,t2+muu).



The mu is also not in (p,t) then we get mathbfF[t]/t3. If mu is in (p,t) we get mathbfFp[t,u]/(u2,tu,t2).



Modulo any mistake I made above, I think a complete list is of length 3 finite local rings with residue field mathbfFp is



  1. mathbfZ/p3,

  2. mathbfZ[t]/(p2,pt,t2),

  3. mathbfZ[t]/(t2p,t3),

  4. mathbfZ[t]/(t2alphap,t3) where alpha is a non-quadratic residue modulo p,

  5. mathbfFp[t]/t3, and

  6. mathbfFp[t,u]/(t,u)2.

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