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

Question

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 $F$"?

Answer 1

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.

Answer 2

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.