# Flat maps and Zariski tangent spaces

Let $$f:A \to B$$ be a finite flat local homomorphism of noetherian local rings.

Are there some nice conditions on $$A$$ and $$B$$ which guarantee that the dimension of the Zariski tangent space of $$A$$ (at its maximal ideal) is smaller or equal to the dimension of the Zariski tangent space of $$B$$ (at its maximal ideal)?

For example, if $$B$$ is regular then $$A$$ is regular, so the above would hold, but I don't want to assume something so strong.

I know no example where it fails, but I am particularly interested in the case that $$A$$ and $$B$$ are both lci and Artin.

## 1 Answer

Lemma. Let $$A \to B$$ be a flat local homomorphism of Noetherian local rings with the same residue field $$k$$ such that $$A$$ and $$B$$ are both (local) complete intersections. Then $$\dim(B) - \dim(A) + \dim_k \mathfrak m_A/\mathfrak m_A^2 \leq \dim_k \mathfrak m_B/\mathfrak m_B^2$$ provided $$B$$ is essentially of finite type over $$A$$.

Proof. By a theorem of Avramov, the ring map $$A \to B$$ is a local complete intersection homomorphism! Consider the distinguished triangle $$L_{B/A} \otimes_B^\mathbf{L} k \to L_{k/A} \to L_{k/B} \to (L_{B/A} \otimes_B^\mathbf{L} k)$$ of cotangent complexes associated to the ring maps $$A \to B \to k$$. Observe that $$H^0(L_{k/A}) = 0$$ and $$H^{-1}(L_{k/A}) = \mathfrak m_A/\mathfrak m_A^2$$ and similarly for $$B$$. Since $$A \to B$$ is a flat local complete intersection of relative dimension $$\dim(B) - \dim(A)$$, the complex $$L_{B/A}$$ is isomorphic in $$D(B)$$ to a complex of the form $$B^{\oplus r} \to B^{\oplus r + \dim(B) - \dim(A)}$$ for some integer $$r \geq 0$$ with the two terms sitting in degrees $$-1$$ and $$0$$ and the map given by a matrix whose coefficients are in $$\mathfrak m_B$$. To see this we use that $$B$$ is essentially of finite type over $$A$$ and hence a localization of an algebra of the form $$A[x_1, \ldots, x_{r + \dim(B) - \dim(A)}]/(f_1, \ldots, f_r)$$ which is a relative global complete intersection. Putting everything together we see that we get an exact sequence $$k^{\oplus r} \to \mathfrak m_A/\mathfrak m_A^2 \to \mathfrak m_B/\mathfrak m_B^2 \to k^{\oplus r + \dim(B) - \dim(A)} \to 0$$ of $$k$$-vector spaces. We conclude by dimension theory for finite dimensional vector spaces.

Remark. The assumption of $$B$$ being essentially of finite type over $$A$$ can be removed, but it becomes a bit technical if you want to do this. Conversely, you can spell out what Avramov's theorem means in nontechnical terms and prove the lemma without using the cotangent complex (but since Avramov's theorem is kind of hard you don't gain a lot in doing so I think).

Note. I haven't considered what happens if the residue fields aren't the same (for example if the residue field extension is inseparable). I haven't tried to make a counter example if the rings aren't lci.

• Thanks a lot! The assumption that the residue fields are the same is good enough for me, but knowing the limits of the statement would of course be nice. – ulrich Aug 4 at 2:45