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In an attempt to prove that polynomial ring of a regular ring is a regular ring, I encounter this beautiful result

"$R$ is a local Noetherian ring. $Q\in \operatorname{Spec}(R[X])$ and $P=Q\cap R$. Then $R_P\to R[X]_Q$ is flat local and $(R[X]/PR[X])_Q$ is a regular local ring"

I can only prove the above homomorphism is local, which is evident. Can you help me solve the other two properties? THank you

Bernard
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T C
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1 Answers1

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Hint 1:

$R[X]_\mathfrak{q}\simeq \bigl(R_\mathfrak{p}[X]\bigr)_\mathfrak{q}$, so the morphism $\;R_\mathfrak{p}\longrightarrow\bigl(R_\mathfrak{p}[X]\bigr)_\mathfrak{q} $ is the composition of $\;R_\mathfrak{p}\longrightarrow R_\mathfrak{p}[X] $, which is a free $R_\mathfrak{p}$-module, with the canonical morphism $\;R_\mathfrak{p}[X]\longrightarrow \bigl(R_\mathfrak{p}[X]\bigr)_\mathfrak{q} $, which is flat.

Hint 2:

$\bigl(R[X]/\mathfrak pR[X]\bigr)_{\mathfrak q}\simeq \bigl((R_\mathfrak{p}[X]/\mathfrak pR_\mathfrak{p})[X]\bigr)_{\mathfrak q}= \bigl(k(\mathfrak p)[X]\bigr)_\mathfrak{q} $ is a localisation of the polynomial ring over the residual field $k(\mathfrak p)$.

Bernard
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  • Clearly you used one property: $R[X]_q = (R_P[X])_q$. Can you prove this? – T C Mar 10 '18 at 00:10
  • The multiplicative set $R\setminus \mathfrak p$ is contained in the multiplicative set $R[X]\setminus\mathfrak q$, so making the elements in the first set invertible, then the elements in the second, is equivalent to making the elements of the second set invertible in a single stroke. – Bernard Mar 10 '18 at 00:22