Lemma 106.4.4. 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 106.4.2 preserves multiplicities.

**Proof.**
By Lemma 106.2.8 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 106.4.2 (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

is a regular ring map of Artinian local rings. For such a ring map it is always the case that

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

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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