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?

## Best Answer

Maybe this answer comes a bit late, hope it helps you or someone else.

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) $.