# Basic transcendental field extension question

• PsychonautQQ

#### PsychonautQQ

My textbook says:
If u is transcnedetal over F, it is routine to verify that:
F(u) = {f(u)g(u)^(-1) | f,g in F[x]; g /= 0}

However, me being the scrubbiest of all scrubs does not understand what they did here.

First of all, I don't understand why they needed to invert the g(u) function.

It goes on to say that F(u) is isomorphic to F(x), the field of quotients of the integral domain F[x].

Can somebody shed some light on this stuff for me?

By definition $F(u)$ is the smallest field containing $F$ as well as $u$. Since $F(u)$ is a field it is closed under multiplication and addition. So, if it contains $u$ then it contains $u^n$ for any $n$, hence it contains $au^n$ for any $a\in F$ and so must also contain all sums of these terms ie. every polynomial of the form $f(u)$. Further, a field must contain inverses of all elements, so if it contains an expression $f(u)$ then it must contain the inverse $f(u)^{-1}$ as well. Finally, since we've shown it contains $f(u)$ and $g(u)^{-1}$, for any polynomials $f,g$ it must contain $f(u)g(u)^{-1}$ since it is closed under multiplication. This shows that $F(u)$ must contain the collection
$$K=\left\{\frac{f(u)}{g(u)}~\bigg|~ f,g\in F[x], g\neq 0 \right\}$$
The fact that the $u$ is transcendental is used here in that $g\neq 0$ implies that $g(u)\neq 0$ so that it makes sense to divide by $g(u)$ in the larger field containing $u$.

To finish it off, you just need to check that this collection of quotients of polynomials is actually a field, and then it will necessarily be the smallest field containing both $F$ and $u$.

Intuitively, the field of quotients of $F[x]$ is the smallest field containing $F[x]$ in which every nonzero element has an inverse. So it should be the set of all quotients of $f(x)/g(x)$ where $g\neq 0$. But since $u$ is transcendental, it satisfies no polynomial relations and so we can relabel $x$ by $u$ and we should still get the same thing.

To make this somewhat rigorous, to construct the field of quotients of $F[x]$, you take the collection of all pairs $(f(x),g(x) )$ where $g(x)\neq 0$ and quotient by the equivalence relation $(f(x),g(x))\sim (f'(x),g'(x))$ when $f(x)g'(x)-f'(x)g(x)=0$. Of course the pair $(f(x),g(x) )$ is intuitively supposed to represent the quotient f(x)/g(x) so there is an obvious map $\phi:\mathrm{Quot}(F[x])\to K$ defined by
$$\phi(f(x),g(x) )=\frac{f(u)}{g(u)}$$
To prove that these two things are isomorphic, check that this is well defined and an isomorphism. The fact that this is well-defined follows from the fact that the equivalence relation here is the same as cross multiplying fractions and the fact that $u$ is transcendental so that $g(u)\neq 0$ for $g\neq 0$. Similarly, the fact that it is a ring homomorphism is because the addition and subtraction in the quotient field are essentially defined to copy the addition/multiplication of fractions. Finally, surjectivity is obvious and injectivity follows using the fact that $u$ is transcendental.

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• PsychonautQQ