# Completion of a Polynomial Ring

I want to show that the ring of formal power series $$R[[X]]$$ is the completion of $$R[X]$$ with respect to the metric $$d\colon R[[X]] \times R[[X]] \rightarrow \mathbb{R}, d(f,g) = 2^{\min \{i|f_i\ne g_i\}}$$. Since, I can not find a strict definition for what completion even means, I have a problem to start with. Is the definition the same as when working with vector spaces in functional analysis?

• See the reference here; Eisenbud. Many people have the same question, e.g., here. – Dietrich Burde Oct 11 '18 at 12:43
• I think you meant $$d(f,g)=2^{-\min\{i\mid f_i\neq g_i\}}.$$ – Jyrki Lahtonen Oct 11 '18 at 16:07

In Algebra the completion of a ring $$R$$ with respect to an ideal $$I$$ (a.k.a. the $$I$$-adic completion of $$R$$) is the limit $$\hat R$$, in the algebraic sense, of the quotient rings $$R/I^n$$ with respect to the canonical maps $$\pi_n:R/I^{n+1}\longrightarrow R/I^n,$$ i.e., an element $$r\in\hat R$$ is but a sequence $$r_n\in R/I^n$$ such that $$\pi_n(r_{n+1})=r_n$$. It is a straightforward exercise to show that $$\hat R$$ is a ring and that there's a canonical map of rings $$\Phi:R\rightarrow \hat R$$.
When $$R$$ is a topological ring the quotient topology in each quotient $$R/I_n$$ defines a topology in $$\hat R$$. If $$R$$ is Hausdorff and $$\bigcap_nI^n=(0)$$ then
• $$\hat R$$ is Hausdorff,
• $$\Phi$$ is an embedding.
Moreover, if $$R$$ is metric and $$I^n$$ can be realized as the ball centered in $$0$$ of radius $$1/n$$ then $$\hat R$$ is the completion of $$R$$ in the "usual" sense, namely the set of equivalence classes of Cauchy sequences in $$R$$.
The ring of formal power series $$K[[X]]$$ is the $$(X)$$-adic completion of the ring of polynomials $$K[X]$$.
Another very important example is the ring of $$p$$-adic integers $$\Bbb Z_p$$ which is the $$p$$-adic completion of the ring integers $$\Bbb Z$$ with respect to the maximal ideal generated by the prime $$p$$. The field $$\Bbb Q_p$$ of $$p$$-adic numbers is the field of fractions of $$\Bbb Z_p$$ and can be obtianed directly as the completion (in the metric sense) of $$\Bbb Q$$ under the $$p$$-adic metric induced by the $$p$$-adic absolute value.