Is $\sqrt{X-1}$ an element of the completed ring $\mathbb{C}[X]_\mathfrak{p}$ with $\mathfrak{p} = (X+1)$?

Question

Let's say we complete the ring $\mathbb{C}[X]$ respect to the ideal $\mathfrak{p} = (X+1)$, then we have the ring $\mathbb{C}[X]_{\mathfrak{p}}$. Can this ring have square roots of other polynomials? E.g. $$ \sqrt{X-a} \in \mathbb{C}[X]_{(X+1)} $$ for some $a \in \mathbb{C}$ ? This might be somewhat straightforward since we have the fundamental theorem of calculus available.

Answer 1

The $\mathbb{C}$-algebra morphism $T\mapsto X+1$ defines an isomorphism $\mathbb{C}[T]\cong \mathbb{C}[X]$ identifying the ideal $(T)$ with the ideal $(X+1)$. In particular, we obtain an induced isomorphism after completing, i.e.

$$\mathbb{C}[\![ T ]\!] \cong (X+1)-\text{adic completion of }\mathbb{C}[X].$$ (Here I am using that the $(T)$-adic completion of $\mathbb{C}[T]$ is the ring of formal power series). Now $X+a$ has a square root on the right-hand side if and only if $T+(a-1)$ has a square root on the left-hand side.

You can use the binomial formula to deduce that $T+b$ has a square root in $\mathbb{C}[\![ T ]\!]$ if and only if $b\neq 0$. Consequently, $X+a$ has a square root in your ring of interest if and only if $a\neq 1$.