Lemma 107.4.4. Let \mathcal{X} be an algebraic stack locally of finite type over a locally Noetherian scheme S. Let x_0 : \mathop{\mathrm{Spec}}(k) \to \mathcal{X} is a morphism where k is a field of finite type over S with image s \in S. If \mathcal{O}_{S, s} is a G-ring, then the map of Lemma 107.4.2 preserves multiplicities.
Proof. By Lemma 107.2.8 we may assume there is a smooth morphism U \to \mathcal{X} where U is a scheme and a k-valued point u_0 of U such that \mathcal{O}_{U, u_0}^\wedge is a versal ring to \mathcal{X} at x_0. By construction of our map in the proof of Lemma 107.4.2 (which simplifies greatly because A = \mathcal{O}_{U, u_0}^\wedge ) we find that it suffices to show: the multiplicity of an irreducible component of U passing through u_0 is the same as the multiplicity of any irreducible component of \mathop{\mathrm{Spec}}(\mathcal{O}_{U, u_0}^\wedge ) mapping into it.
Translated into commutative algebra we find the following: Let C = \mathcal{O}_{U, u_0}. This is essentially of finite type over \mathcal{O}_{S, s} and hence is a G-ring (More on Algebra, Proposition 15.50.10). Then A = C^\wedge . Therefore C \to A is a regular ring map. Let \mathfrak q \subset C be a minimal prime and let \mathfrak p \subset A be a minimal prime lying over \mathfrak q. Then
R = C_\mathfrak p \longrightarrow A_\mathfrak p = R'
is a regular ring map of Artinian local rings. For such a ring map it is always the case that
\text{length}_ R R = \text{length}_{R'} R'
This is what we have to show because the left hand side is the multiplicity of our component on U and the right hand side is the multiplicity of our component on \mathop{\mathrm{Spec}}(A). To see the equality, first we use that
\text{length}_ R(R) \text{length}_{R'}(R'/\mathfrak m_ R R') = \text{length}_{R'}(R')
by Algebra, Lemma 10.52.13. Thus it suffices to show \mathfrak m_ R R' = \mathfrak m_{R'}, which is a consequence of being a regular homomorphism of zero dimensional local rings. \square
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