How to determine a formal root of a polynomial in power series over finite field?

Question

I am looking for whether there is an easy way to define $y=\sqrt{-1}$ in $A_v$, where $A_v$ is the local ring with $A$ is the polynomial ring $\mathbb{F}_q[x]$ and $v$ is a prime ideal generated by a irreducible polynomial $x^2+1 = 0$ in $\mathbb{F}_q[x]$. That is $y^2 = q-1 + (q-1)(x^2+1) + (q-1)(x^2+1)^2 + (q-1)(x^2+1)^2 +\cdots$. Then I would like to expand other power series in $\mathbb{F}_q[[x]]$ or just polynomial in $\mathbb{F}_q[x]$ in terms of a power series in $x - y$, i.e. $f(x) = a_0 + a_1(x-y) +a_2(x-y)^2+\cdots$, where $a_i\in\mathbb{F}_q[x]$ with degree less than $2$.

Moreover, is there an direct way to express $y$ in terms of $x$?

Answer 1

It is not entirely clear to me what you want to do. This is too long to be a comment, so an answer it is. Please comment, and may be we can locate the question you really wanted to ask.

• If $q\equiv1\pmod4$, then the field $\Bbb{F}_q$ already contains a zero of the polynomial $x^2+1$, call it $i$. Even if $q\equiv3\pmod4$, then adjoining a zero of $x^2+1$ will give you the ring of formal power series over the extension field $\Bbb{F}_q[i]\simeq\Bbb{F}_{q^2}$.
• The polynomial $x^2+1$ is a unit of $\Bbb{F}_q[[x]]$: $$(1+x^2)^{-1}=1-x^2+x^4-x^6+\cdots.$$ Therefore the ideal it generates is all of $\Bbb{F}_q[[x]].$
• The series (I replaced $q-1$ with $-1$ for the sake of brevity) $$-1-(x^2+1)-(x^2+1)^2-(x^2+1)^3-\cdots$$ does not make sense in the ring $\Bbb{F}_q[[x]]$. This is because all those terms have a non-zero constant term. And we cannot define the sum of infinitely many non-zero constants in $\Bbb{F}_q$ in any meaningful way. The only Hausdorff topology there is the discrete one, and the sum infinitely many non-zero terms simply never converges.
• The same objection applies to an infinite sum of powers of $(x-y)$. They all have a non-zero constant term, so the infinite sum does not make sense.