Example 15.11.14 (Moret-Bailly). Lemma 15.11.13 is wrong if the colimit isn't filtered. For example, if we take the coproduct of the henselian pairs $(\mathbf{Z}_ p, (p))$ and $(\mathbf{Z}_ p, (p))$, then we obtain $(A, pA)$ with $A = \mathbf{Z}_ p \otimes _\mathbf {Z} \mathbf{Z}_ p$. This isn't a henselian pair: $A/pA = \mathbf{F}_ p$ hence if $(A, pA)$ where henselian, then $A$ would have to be local. However, $\mathop{\mathrm{Spec}}(A)$ is disconnected; for example for odd primes $p$ we have the nontrivial idempotent

$(1/2 \otimes 1) \left(1 \otimes 1 - (1 + p)^{-1}u \otimes u\right)$

where $u \in \mathbf{Z}_ p$ is a square root of $1 + p$. Some details omitted.

## Comments (6)

Comment #4907 by Laurent Moret-Bailly on

This element has square 1; an idempotent derived from it is $\frac{1}{2}\left[(1+p)^{-1}u\otimes u-1\right]$.

Comment #5022 by on

I agree with these comments and I shall fix this the next time I go through all the comments.

Comment #5400 by Takagi Benseki（高城 辨積） on

a typo: "where henselian" should be "were Henselian" ?

Comment #5633 by on

OK, I am going to leave this as is for now.

There are also:

• 2 comment(s) on Section 15.11: Henselian pairs

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