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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 $F$"?

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    $\begingroup$ You are not a new user on this forum. Please use MathJax. $\endgroup$
    – Mark
    Commented Mar 15 at 14:23
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    $\begingroup$ If $x$ is not algebraically independent over $F$, the extension $F(x)$ is not trascendental, but algebraic. $\endgroup$
    – ajotatxe
    Commented Mar 15 at 14:24
  • $\begingroup$ @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… $\endgroup$
    – Rowing0914
    Commented Mar 15 at 14:50
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    $\begingroup$ $\mathbb Q(\sqrt{2})$ 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 $\mathbb Q(x)$. $\endgroup$
    – NaNoS
    Commented Mar 15 at 15:01
  • $\begingroup$ 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 $\endgroup$
    – Jakobian
    Commented Mar 15 at 15:28

2 Answers 2

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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.

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Rowing0914
    Commented Mar 15 at 15:06
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    $\begingroup$ @Rowing0914 I think that ronno's excellent answer is exactly what you ask. $\endgroup$
    – NaNoS
    Commented Mar 15 at 15:08

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