# Is there a field extension $E$ of $\mathbb R$ such that every formal power series with coefficients in $\mathbb R$ has a root in $E$?

Is there a field extension $$E$$ of $$\mathbb R$$ such that every formal power series with coefficients in $$\mathbb R$$ has a root in $$E$$? If so, can we describe its elements explicitly?

The set of complex numbers $$\mathbb C$$ can be thought of as a field extension $$F$$ of $$\mathbb R$$ such that every polynomial with coefficients in $$\mathbb R$$ has a root in $$F$$. A formal power series generalizes a polynomial in that the number of nonzero coefficients is allowed to be infinite–put another way, a polynomial is a special case of a formal power series in which only finitely many of the coefficients can be nonzero. Thus, it seems natural to ask if there are any field extensions of $$\mathbb R$$ such that all elements of $$\mathbb R[[x]]$$–the ring of formal power series with coefficients in $$\mathbb R$$–have roots in the extension.

We can see that if such a field $$E$$ exists, it will contain the complex numbers as a subfield for the same reasons as before. It is also evident that $$\exp(X):=\sum_{n=0}^\infty\frac{X^n}{n!}$$ has a root in this field, which I find somewhat interesting.

Important: in this context, we say that $$r$$ is a root of $$f(x)\in\mathbb R[[x]]$$ if and only if $$x-r$$ divides $$f(x)$$. This avoids the troubles of having to define statements like $$a_0+a_1r+a_2r^2+\cdots=0$$.

• Your first bullet point creates some troubles of its own: if $r$ is invertible, then $x - r = -r(1 - x/r)$, and this has the multiplicative inverse $-(1 + x/r + (x/r)^2 + ...)/r$. In particular, every element of $\mathbb R[[x]]$ is divisible by $x-r$ for all $r \neq 0$. Oct 9, 2022 at 5:16
• One alternative would be to restrict to converging power series. However, then your field extensions would also need to be subfields of the converging power series in order to make sense of roots. Oct 9, 2022 at 5:26

As Ravi points out in the comments, your definition of roots does not work. We can avoid this by working with a topology so that we can discuss convergence of convergent power series. A nice general setting is to work in a Banach algebra over $$\mathbb{R}$$: convergent power series behave very nicely in this setting (see the holomorphic functional calculus for details). Banach algebras are almost never fields but we can just drop the field requirement and consider general Banach algebras.
Unfortunately there are convergent power series which do not have roots in any Banach algebra, and $$\exp(x)$$ is itself an example: this is because it is always invertible, since its inverse is $$\exp(-x)$$.
Generally, using the holomorphic functional calculus (specifically, using the spectral mapping theorem) we can show that a convergent power series over $$\mathbb{R}$$ has a root in some Banach algebra over $$\mathbb{R}$$ if and only if it has a root in $$\mathbb{C}$$. So $$\mathbb{C}$$ is already as "algebraically closed" with respect to convergent power series as possible.
• Your $\exp(x)\exp(-x)=1$ point is unclear, how is it different to $(1-x)\sum_{k\ge 0} x^k = 1$ ? Oct 9, 2022 at 9:43
• No, I mean $1-x$ has a root whereas $\exp(x)$ doesn't. Somewhat we are using that $\exp(-x)$ converges everywhere, which is for example not true in $\Bbb{R}((t))$. Oct 9, 2022 at 18:49
• @reuns: ah, I mean it works as a counterexample in the sense that $\sum_{k \ge 0} x^k$ is invertible if it converges since it converges to the inverse of $1 - x$, so $\sum_{k \ge 0} x^k$ also can't have roots in a Banach algebra. And yes, we are using that $\exp(-x)$ converges everywhere in a Banach algebra. Oct 9, 2022 at 18:51