# Transcendence degree of $F(x)$ over a field $F$

I'm studying Algebraic geometry and have been stuck at processing the concept of Transcendental degree of a polynomial and came across the following argument online.

The transcendence degree of the field of rational functions $$F(x)$$ over a field $$F$$ is one. This means that the field $$F(x)$$ can be thought of as a one-dimensional extension of the field $$F$$, even though it contains an infinite number of elements. The transcendence degree measures the "size" of the transcendental part of an extension field. In the case of $$F(x)$$, the transcendence degree is one because the field is generated by a single transcendental element, namely $$x$$. To show that the transcendence degree is one, we need to show that $$x$$ is algebraically independent over $$F$$ and that any other transcendental element can be expressed as a rational function in $$x$$. The fact that $$x$$ is transcendental over $$F$$ is clear because $$x$$ is not a root of any non-zero polynomial with coefficients in $$F$$. Any other transcendental element $$y$$ can be written as $$y = \frac{f(x)}{g(x)}$$ for some polynomials $$f$$ and $$g$$ with coefficients in $$F$$, which shows that $$y$$ is algebraic over $$F(x)$$.

I don't understand why

$$x$$ is algebraically independent over $$F$$

as I thought that there could be $$0$$ in $$F(x)$$ when $$f(x) = 0$$ but $$g(x) \neq 0$$, meaning that $$x$$ is algebraic (i.e., $$x$$ satisfies a polynomial relation being equal to $$0$$).

Can someone elaborate on why "$$x$$ is algebraically independent over $$F$$"?

• You are not a new user on this forum. Please use MathJax.
– Mark
Commented Mar 15 at 14:23
• If $x$ is not algebraically independent over $F$, the extension $F(x)$ is not trascendental, but algebraic. Commented Mar 15 at 14:24
• @ajotatxe sorry I think I’m almost seeing what you meant but if $F(x)$ is algebraic what’s bad? Like $\mathbb{Q}(\sqrt{2})$ is an algebraic extension and I don’t see what’s bad about this in the rational function field… Commented Mar 15 at 14:50
• $\mathbb Q(\sqrt{2})$ is not the rational function field of $\mathbb Q$ ! The only possible isomorphism would be $\mathbb Q(x) \to \mathbb Q(\sqrt{2})$, $x \mapsto \sqrt{2}$ but now it would imply that $x^2 = 2$ in $\mathbb Q(x)$. Commented Mar 15 at 15:01
• Note about the tags: just because you encountered this concept while studying algebraic geometry doesn't mean its algebraic geometry. I've changed the tags accordingly Commented Mar 15 at 15:28

In the field $$F(x)$$, the symbol $$x$$ is a formal variable with no presupposed relations. I don't think it is completely obvious that $$x$$ is then transcendental over $$F$$, but as in @NaNoS' answer it suffices to check that the obvious map $$F[x] \to F(x)$$ taking a polynomial $$P(x)$$ to the rational function $$P(x)/1$$ is injective. Then any non-zero polynomial $$P$$ with $$F$$ coefficients evaluated at the element $$x = x/1$$ of $$F(x)$$ equals $$P(x)/1$$, and by the previous observation is non-zero.
In $$\mathbb{Q}(\sqrt{2})$$, $$\sqrt{2}$$ may or may not be a formal variable: you could think of this field as either (1) the subfield of say $$\mathbb{R}$$ or $$\mathbb{C}$$ generated by $$\mathbb{Q}$$ and an actual element $$\sqrt{2}$$ or (2) shorthand for the quotient ring $$\mathbb{Q}[x]/(x^2 - 2)$$, with the image of $$x$$ being named $$\sqrt{2}$$, which happens to be a field (since $$x^2 - 2$$ is irreducible over $$\mathbb{Q}$$). It is also useful to know that these two definitions produce isomorphic objects.
So we have $$F \hookrightarrow F[x] \hookrightarrow F(x)$$. Saying that $$x \in F(x)$$ is algebraically independant over $$F$$ is, by definition, saying that for every polynomial $$P \in F[x]$$, we have $$P(x) = 0$$ as an element of $$F(x)$$ if and only if $$P = 0$$ as an element of $$F[x]$$ (pay attention to the belonging of the elements). But now $$P(x)$$ is just the image of $$P$$ under the inclusion $$F[x] \hookrightarrow F(x)$$ so it is clear that $$x$$ is algebraically free.
• Thank you for your answer! Sorry what’s the difference between $P(x)$ and $P$? Is P a polynomial and P(x) an evaluation at x? Commented Mar 15 at 15:06