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I am writing an article on transcendental numbers and I'm wondering if it is possible to construct a "simple" example of a transcendental element in a field extension.

When thinking about this, I intitially thought to use finite fields, specifically chains of subfields of prime powers. However, I don't believe this will work as all transcendental extensions are of infinite degree.

My next idea is to use fraction fields. As I understand it, given a field $F$ and a polynomial ring $F[x]$, then the fraction field $F(x)$ is a purely transcendental extension of $F$. That is, given a rational function such as $f(x)=x^{-1}\in F(x)$, there is no $p\in F[x]$ for which $p(x^{-1})=0$. Therefore, $x^{-1}$ would be transcendental over $F$. When you take $F=R$ this example seems to hold.

I would be grateful for any other ideas or corrections to my lines of thinking if appropriate. The fraction fields example is somewhat abstract, so a more concrete example would be helpful.

diracsum
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  • See Lüroth's theorem – lhf Apr 22 '24 at 00:09
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    You said $x^{-1}$ is transcendental over $F$; couldn’t you have simply said $x$? – J. W. Tanner Apr 22 '24 at 00:16
  • there is no nonzero $p$ for which $…=0$ – J. W. Tanner Apr 22 '24 at 00:36
  • How about $e$ or $\pi$ over $\Bbb Q?$ – suckling pig Apr 22 '24 at 00:36
  • "Given $f(x)=x^{-1}\in F(x)$, there is no $p\in F[x]$ with $p(x)=0$." This is incorrect: $p(x)=x-1$ will do it. But in any case, that is not the correct statement for $f(x)$ to be transcendental: $f(x)$ is algebraic over $F$ if there exists $p(y)\in F[y]$, where $y\neq x$, such that $p(x)=0$. – Arturo Magidin Apr 22 '24 at 00:47
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    $e^x$ is transcendental over $\Bbb C(x)$. As usual, this is much easier to prove for this function field example (vs. said number field examples) since derivatives are available yielding e.g. Wronskian based measures of algebraic independence (e.g. a high-school level proof of FLT for polynomials). Follow the link for further discussion. – Bill Dubuque Apr 22 '24 at 01:05
  • @BillDubuque That is quite helpful, thank you. I will look into this. JW Tanner, I realize yes I could've just chosen $x$. Also, Arturo, I think in your example $p(x^{-1})=x^{-1}-1 \neq 0$, right? Also I thought that, given a field extension and element $x\in K/F$, $x$ is transcendental w.r.t $F$ if for all $p\in F[x]$, $p(x)=0$ implies $p=0$. In my example, $K=F(x),;F=F$, therefore $x\in F(x)$ works since $c_n x^n + ... + c_0 = 0$ implies that $c_n = ... = c_0 = 0$. (yes, this is a bit of a cheat, but it seems to techincally work...) – diracsum Apr 22 '24 at 02:26
  • And @calc II the problem I have with $\pi$ and $e$ is that the proofs, even if only using elementary methods (such as Ivan Niven's proof for pi), are nonethless highly intricate and dense, and not to mention long, I was looking for an example which could easily and quickly be demonstrated, just to reveal the structure of transcendence and fields to the reader. – diracsum Apr 22 '24 at 02:30
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    The proof for "Liouville's constant" $\sum\limits_{k=0}^\infty 10^{-k!}$ is arguably easier. @diracsum – Amateur_Algebraist Apr 22 '24 at 07:57

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Of course, not every irrational will work, but it's a theorem, a proof being attributed to Gelfond and Schneider in Lang, that $\pi$ and $e$ are transcendental over $\Bbb Q.$

suckling pig
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    It was von Lindemann that published the first complete proof that $\pi$ is transcendental in 1882, and Hermite who proved that $e$ is transcendenta in 1873l. Gelfond and Schneider proved that if $a$ and $b$ are algebraic, $a\notin{0,1}$, and $b$ is not rational, then $a^b$ is transcendental, but that wasn't until 1934. I doubt that Lang would attribute the first proof of transcendence of $\pi$ or $e$ to Gelfond. – Arturo Magidin Apr 22 '24 at 00:52
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    He attributes a proof to Gelfond and Schneider in the appendix, $p. 867.$ I should have been more careful. @ArturoMagidin – suckling pig Apr 22 '24 at 01:24
  • @icantry Tell me please what is the name of this reference? – Math Admiral Jun 01 '24 at 12:07
  • Lang's Algebra. – suckling pig Jun 01 '24 at 13:41