# about the local ring of $\mathbb{Z}_p[T]/(pT^2+T+1)$ at the prime p

is the localisation of the ring $$A:=\mathbb{Z}_p[T]/(pT^2+T+1)$$ at the prime ideal (p) isomorphic to $\mathbb{Z}_p$?

If not, how to understand this ring very explicitly?

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EGA II 6.2.5, 6.2.6 give a decomposition to quai-finite schemes, I am looking for a concrete example in order to understand such decomposition, the above is my example. According to EGA, $A_p$ is finite over $\mathbb{Z}_p$, and it seems very likely $A_p=\mathbb{Z}_p$ – Heer May 16 '12 at 20:54

## 1 Answer

That equation has a solution in $\mathbb Z_p$. Proof: $f(a)=-pa^2-1$ is a contraction mapping for the $p$-adic metric, and a fixed point must be a solution.

Let $\alpha$ be that solution, then $(T-\alpha)$ divides your equation. The other factor must be $pT-\alpha^{-1}$. So we have:

$A = \mathbb Z_p[T]/(T-\alpha) \times \mathbb Z_p [T]/(pT-\alpha^{-1})=\mathbb Z_p \times \mathbb Q_p$

The localization at $p$ is $\mathbb Z_p$.

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Another way to see immediately that the equation has a solution is via Newton's polygon: it is formed by two sides with slopes 0 and 1, and each corresponds to a non-trivial factor, having degree 1. – Maurizio Monge May 16 '12 at 21:45
Although I always prefer Newton poly., factorization follows also from Hensel’s Lemma, with the one polynomial in char. $p$ being $1+x$ and the other being $1$. HL guarantees a factor of degree $1$. – Lubin May 17 '12 at 2:27
Thanks a lot !!! – Heer May 17 '12 at 10:34