Function Field over k --> k(x) - Rotman Definition and Proposition 3.29

[Math Amateur]

I am reading Joseph J. Rotman's book: A First Course in Abstract Algebra with Applications (Third Edition) ...

I am currently focused on Section 3.3 Polynomials ...

I need help with an aspect of the proof of Proposition 3.29 concerning elements of the function field \(\displaystyle k(x)\).

Rotman's definition of a function field over \(\displaystyle k\) and his Proposition 3.29 and its proof read as follows:View attachment 4695

In the proof of Proposition 3.29 displayed above, we read the following:

" ... ... elements in \(\displaystyle k(x)\) have the form \(\displaystyle f(x) {g(x)}^{-1}\) ... ... "I am perplexed by the above statement as it includes the polynomial \(\displaystyle {g(x)}^{-1}\) and I thought the only inverses in \(\displaystyle k[x]\) were the constant polynomials ... and that no other inverses existed in \(\displaystyle k[x]\) ... so how are we to make sense of the above statement ... unless we just regard \(\displaystyle f(x) {g(x)}^{-1}\) as the result of \(\displaystyle f(x)\) divided by \(\displaystyle g(x)\) ...

Can someone please clarify the above issue ...

Peter

 

[Euge]

Let's consider the rationals $\Bbb Q$, the fraction field of $\Bbb Z$. Its elements are represented in the form $m/n$, where $m\in \Bbb Z$ and $n\in \Bbb Z \setminus \{0\}$. Based on your reasoning, $1/n$ (or $n^{-1}$) is an integer. See where the problem is? Certainly, $1/2$ is not an integer. In fact, $1/n$ is only an integer for $n = \pm 1$.

 

[Deveno]

Recall that $k(x)$ isn't IN $k[x]$, it's "bigger".

Specifically, $k(x)$ is an equivalence over $k[x] \times (k[x] - \{0\})$, and $\dfrac{1}{g(x)}$ refers to the equivalence class of the pair:

$(1,g(x))$.

In $k(x)$ this is indeed a multiplicative inverse for $g(x)$ (which is identified with the equivalence class of $(g(x), 1)$ because the identification:

$f(x) \mapsto [(f(x),1)]$ is indeed an injective ring-homomorphism)

and the multiplication in $k(x)$ is given by:

$[(a(x),b(x))]\cdot[(c(x),d(x))] = [(a(x)c(x),b(x)d(x))]$, so:

$[(1,g(x))]\cdot[(g(x),1)] = [(g(x),g(x))] = [(1,1)]$.

(Recall that $[(a(x),b(x))] = [(c(x),d(x))]$ if and only if $a(x)d(x) = b(x)c(x)$, and surely $g(x)\cdot 1 = 1 \cdot g(x)$).

It is sometimes "easier" to visualize $\dfrac{1}{g(x)}$, as the function:

$f: K - S \to K$, where $S = \{a \in K: g(a) = 0\}$. given by:

$f(c) = \dfrac{1}{g(c)}$

Typically, when $K = \Bbb R$, these rational functions have vertical asymptotes at the zeroes of $g$.

For example, for $g(x) = x$, we get the hyperbola $y = \dfrac{1}{x}$, which has a vertical asymptote at $x = 0$ (Note this functions is NOT a polynomial function, for polynomial functions are defined at every point of $\Bbb R$).

 

[Math Amateur]

Recall that $k(x)$ isn't IN $k[x]$, it's "bigger".

Specifically, $k(x)$ is an equivalence over $k[x] \times (k[x] - \{0\})$, and $\dfrac{1}{g(x)}$ refers to the equivalence class of the pair:

$(1,g(x))$.

In $k(x)$ this is indeed a multiplicative inverse for $g(x)$ (which is identified with the equivalence class of $(g(x), 1)$ because the identification:

$f(x) \mapsto [(f(x),1)]$ is indeed an injective ring-homomorphism)

and the multiplication in $k(x)$ is given by:

$[(a(x),b(x))]\cdot[(c(x),d(x))] = [(a(x)c(x),b(x)d(x))]$, so:

$[(1,g(x))]\cdot[(g(x),1)] = [(g(x),g(x))] = [(1,1)]$.

(Recall that $[(a(x),b(x))] = [(c(x),d(x))]$ if and only if $a(x)d(x) = b(x)c(x)$, and surely $g(x)\cdot 1 = 1 \cdot g(x)$).

It is sometimes "easier" to visualize $\dfrac{1}{g(x)}$, as the function:

$f: K - S \to K$, where $S = \{a \in K: g(a) = 0\}$. given by:

$f(c) = \dfrac{1}{g(c)}$

Typically, when $K = \Bbb R$, these rational functions have vertical asymptotes at the zeroes of $g$.

For example, for $g(x) = x$, we get the hyperbola $y = \dfrac{1}{x}$, which has a vertical asymptote at $x = 0$ (Note this functions is NOT a polynomial function, for polynomial functions are defined at every point of $\Bbb R$).

Euge, Deveno

Thanks for the help ...

Still reflecting and to an extent puzzling over this ...

I think that you are both saying that a nonconstant polynomial g(x) does not have an inverse in k[x] but does have an inverse in k(x) ... namely the equivalence class of (1, g(x) ) ... is that correct?

Peter

 

[Deveno]

Yes, in exactly the same an integer $n$ does not have an inverse in the integers, but has one in the rationals.

Recall that in the rationals "fractions" are NOT unique:

$\dfrac{1}{2} = \dfrac{2}{4} = \dfrac{3}{6} =\cdots$ etc.,

but that all of these fractions have the same RATIO as each other, and it is this ratio that is captured by the equivalence class.

I urge you to read the "mathematics" section in this link, to better appreciate what this means:

https://en.wikipedia.org/wiki/Eudoxus_of_Cnidus