# Degree of an extension inside a function field of one variable

abstract-algebrafield-theory

In the book Field and Galois theory by Morandi, author mentions following example of a finite extension. Let $$t$$ be an indeterminate and $$k(t)$$ be the field of rational functions in $$t$$ over a field $$k$$.

Let $$u=\frac{f(t)}{g(t)}\in k(t)\setminus k$$ (where $$f(t),g(t)\in k[t]$$, $$g(t)\neq 0$$).

Then $$k(t)$$ is an extension of $$k(u)$$ of finite degree, which is equal to maximum of degrees of $$f(x)$$ and $$g(x)$$.

I did not understand his arguments, so tried to look at a concrete example.

As per his assertion, $$\mathbb{Q}(\pi)$$ is finite degree extension of $$\mathbb{Q}(\frac{1+\pi}{2+\pi^3})$$, and degree should be $$3$$.

When I tried to prove that degree of extension is $$3$$, I intentionally thought that $$1,\pi,\pi^2$$ would be linearly independent over $$\mathbb{Q}(\frac{1+\pi}{2+\pi^3})$$, but I could not arrive at contradiction by assuming dependence. Can one help to sort this problem to get clear picture of result in Morandi's book?

Suppose $$u=\frac{1+\pi}{2+\pi^3}$$, and $$a_2(u), a_1(u), a_0(u) \in \mathbb{Q}(u)$$ are all from the field of rational functions $$\mathbb{Q}(u)$$. The question is equivalent to the following one. (This is because if $$\{1,\pi,\pi^2\}$$ are linearly dependent, $$p(x)$$ must be reducible in $$\mathbb{Q}(u)[x]$$)
Suppose $$ux^3-x-2+u = p(x)$$, show that no polynomial $$a_2(u)x^2+a_1(u)x+a_0(u) = q(x)$$ of degree 2 exists such that $$q(x)|p(x)$$.
The trick here is to notice that $$p(x)\in \mathbb{Q}[u][x]$$, and if a $$p(x) = q(x)r(x)$$ for some $$q(x), r(x) \in \mathbb{Q}(u)[x]$$, then there must be some $$r_0(x),q_0(x)\in \mathbb{Q}[u][x]$$ such that $$p(x) = q_0(x)r_0(x)$$.