On the prime spectrum of completion of local rings

Question

Let $(R, \mathfrak m)$ be the henselization of the local ring $\mathbb C[x,y]_{(x,y)}$ . Let $\hat R$ be the $\mathfrak m$-adic completion of $R$. Then there is a natural map $R \to \hat R$ which induces a natural map $\phi: \operatorname{Spec}(\hat R)\to \operatorname{Spec}(R)$ . Then is it true that $\phi$ is bijective ? If $\phi$ is not bijective, then what if we instead started with the strict henselization ?

Answer 1

Take $f(x)\in \Bbb{C}[[x]]$ non-algebraic over $\Bbb{C}(x)$,

$P=(y-xf(x))$ is a prime ideal of $\hat{R}=\Bbb{C}[[x,y]]$,

$R = \hat{R}\cap \overline{\Bbb{C}(x,y)}$,

Given $a\in R\cap P$ look at its field norm $$\frac{v}{w} = N_{\Bbb{C}(x,y)(a)/\Bbb{C}(x,y)}(a)\in Frac(\Bbb{C}[x,y]),\qquad v\in P\cap \Bbb{C}[x,y]$$

Assume $v\ne 0$. Since $P$ is a prime ideal, if $v=v_1v_2$ then one of the factor is in $P$ and we can repeat until $v\in \Bbb{C}[x,y]\cap P$ is irreducible.

If $v \in \Bbb{C}[x]$ or $\Bbb{C}[y]$ we get a contradiction since there are injections $\Bbb{C}[[x]]\to \hat{R}/P$, $\Bbb{C}[[y]]\to \hat{R}/P$. Otherwise $v\in \Bbb{C}[x,y]$ and $y$ is algebraic over $\Bbb{C}(x)$ in $\Bbb{C}[[x,y]]/(v)$ thus in $\hat{R}/P$, a contradiction since $y=xf(x)$ is not algebraic over $\Bbb{C}(x)$.

Whence $v=0$, $a=0$, $P\cap R = \{0\}$ and the map $Spec(\hat{R})\to Spec(R)$ is not injective.

Since $\hat{R}/\mathfrak{m} = \Bbb{C}$ is separably closed I don't see what you mean with the strict henselization.