Primitive elements of the rationals function extension

Asked Jun 07 '24 at 02:05

Active Jun 07 '24 at 02:05

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I want to prove this:

Let $k$ a field, $E:=k(x)$ the quotient field of $k[x]$ and $u\in E$. Show tha $E=k(u)$ if, and only if, $u=\frac{ax+d}{cx+d}$ for some $a,b,c,d\in k$ such that $ad-bc\neq 0$.

I already proved that if $u=\frac{ax+d}{cx+d}$ with $ad-bc\neq 0$, then $E=k(u)$. But have not been able to prove that the other direction. Any suggestions? Thanks.

asked Jun 07 '24 at 02:05

Diego Ramírez

Can you do the special case that $u$ is a polynomial, first? – Qiaochu Yuan Jun 07 '24 at 02:21
If $u$ is a polynommial, then $\deg(u)\geq 2$, hence every polynomial in $k(u)$ have degree$\geq 2$. So $k(u)\neq E$. That's true? – Diego Ramírez Jun 07 '24 at 02:28
1

$k(u)=k(x)$ iff $x\mapsto u$ induces a $k-$automorphism of $k(x)$. Then you may use https://math.stackexchange.com/questions/13129. – Asigan Jun 07 '24 at 03:58
Yes, thank you. – Diego Ramírez Jun 07 '24 at 10:07

Primitive elements of the rationals function extension

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