I've got a doubt about a ring of polynomial functions. The problem starting doing this exercise of Fraleigh (The 30).enter image description here

Here I had to show that $P_F$ isn't necessarily isomorphic to $F[x]$. It's easy, since there are infinitely many polynomials in a finite ring, but (If $n$ is the order of the ring) at most $n^n$ polynomial functions in $P_F$ (And seing 31, exactly $n^n$ if $F$ is a field).

It's also not possible to make a field of quotientts of $P_F$, since $P_F$ has zero divisors (I can take $f,g\in P_{\mathbb{Z}_3}$ such that $f(x)=(x-1)(x-2)$ and $g(x)=x$, and see that $fg(x)=0$ for all $x$ in $\mathbb{Z}_3$). Here the problem starts.

Even in an infinite Field like $\mathbb{R}$, there are problems making the field of quotients of $P_\mathbb{R}$, since being there some points where the functions vanish, the fractions of this field would be not defined in some points. But in any other way, I see $P_\mathbb{R}$ equal to $\mathbb{R}[x]$. What would be the difference (I think there must be a difference if I can make all fractions of one and not all of the other)?


1 Answer 1


I now think it'd be the same in $\mathbb{R}$. The field of quotients of $\mathbb{R}[x]$ and the field of quotients of the polynomial functions in $\mathbb{R}$. Just that they wouldn't be functions, at least with domain $\mathbb{R}$.


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