Let $F$ be a commutative ring without zero divisors and $Q$ its field of fractions. How can I prove that field $Q(x)$ is a field of fractions of ring $F[x]$? And also why is it that field $Q((x))$ can't match with the field of fractions of ring $F[[x]]$? Thanks!
[Math] Prove that field $Q(x)$ is a field of fractions of ring $F[x]$
field-theorypolynomialsring-theory
Best Answer
I can show you an example of an element of $Q[[x]]$ that is not a fraction of two elements of $F[[x]]$, but I’m afraid it’s ridiculously advanced. I’m sure that others will give better examples.
I’m going to take $F=\Bbb Z_p$ and $Q=\Bbb Q_p$, the $p$-adic integers and numbers respectively. A basic tool for handling $\Bbb Z_p[[x]]$, or indeed power series over any complete local ring, is the Weierstrass Preparation Theorem. It’s incredibly powerful, but not at all deep, and my contention is that if $n$ mathematicians sit down to prove it, you’ll get $n$ different proofs, or more.
Weierstrass Preparation says this: Let $f(x)\in\Bbb Z_p[[x]]$, with first unit coefficient in degree $d$ (“unit” means that the coefficient has absolute value $1$). Then there is a monic polynomial $q(x)$ of degree $d$, with all coefficients in lower degrees divisible by $p$ (i.e. of absolute value less than $1$), and a unit power series $U(x)\in\Bbb Z_p[[x]]^*$ (so that $U(0)$ is a unit of $\Bbb Z_p)$, such that $f=qU$. Furthermore, $q$ and $U$ are unique with this property.
Notice that a power series in $\Bbb Z_p[[x]]$ may be evaluated at any element $\alpha$ in the (an) algebraic closure $\overline{\Bbb Q_p}$ of $\Bbb Q_p$ for which $|\alpha|<1$, because $\Bbb Q_p(\alpha)$ is a finite extension, and $|*|_p$ is uniquely extendible to such a field, which is even complete with respect to this absolute value. (So we don’t need to complete the algebraic closure for this purpose.) When we examine how W-Prep is involved here, we see that all the roots of $q$ have absolute value less than $1$, and that $U$ has no roots $\alpha$ at all with $|\alpha|<1$. Thus W-Prep says that a power series over $\Bbb Z_p$ has only finitely many zeros in the maximal ideal of the integers of the algebraic closure.
Once I exhibit a series $L(x)$ over $\Bbb Q_p$ that may be evaluated at any $\alpha\in\overline{\Bbb Q_p}$ for which $|\alpha|<1$, and which has infinitely many zeros of this type, it follows that $G$ is not the quotient of two $\Bbb Z_p$-power series $q$ and $r$. For, if $L=q/r$, we also have $rL=q$, infinitely many zeros on the left, only finitely many on the right.
My example of an $L$ as above is $\log(x)=x-x^2-2+x^3/3-x^4/4+\cdots$, just the ordinary logarithmic series. You easily check that it may be evaluated at any $\alpha$ with $|\alpha|<1$, and you can check with a little more difficulty that it vanishes at all the numbers $\zeta-1$, with $\zeta$ running through the $p$-power roots of unity in the algebraic closure, of which there are of course infinitely many.