Field adjoined transcendental
Theorem
Let $E / F$ be a field extension.
Let $\alpha \in E$.
If $\alpha$ is transcendental over $F$, then:
- $F \left[{\alpha}\right] \ \cong_F \ F \left[{x}\right]$
and:
- $F \left({\alpha}\right) \ \cong_F \ F \left({x}\right)$
where $\cong_F$ denotes $F$-homomorphism.
Also:
- $F \left[{\alpha}\right] \ne F \left({\alpha}\right)$
The difference between $F \left[{\alpha}\right]$, $F \left[{x}\right]$, $F \left({\alpha}\right)$ and $F \left({x}\right)$, and indeed, what they actually mean in the first place.
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Proof
Let $\psi: F \left({x}\right) \to F \left({\alpha}\right)$ be the unique $F$-homomorphism which sends $x \mapsto \alpha$.
Thus, for $f \in F \left[{x}\right] \subseteq F \left({x}\right)$, say of degree $n$ where $\displaystyle f \left({x}\right) = \sum_{i=0}^n a_i x^i$.
Then :
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \psi \left({f \left({x}\right)}\right)\) | \(=\) | \(\displaystyle \psi \left({\sum_{i=0}^n a_i x^i}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i=0}^n \psi \left({a_i}\right) \psi \left({x}\right)^i\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | as $\psi$ is a ring homomorphism | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{i=0}^n a_i \alpha^i\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | as $\psi \restriction_F = 1_F$ | ||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle f \left({\alpha}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
Hence, for $f \left({x}\right), g \left({x}\right) \in F \left[{x}\right]$ such that $g \left({x}\right) \ne 0$, we have:
- $\psi \left({\dfrac{f \left({x}\right)}{g \left({x}\right)}}\right) = \dfrac{f \left({\alpha}\right)}{g \left({\alpha}\right)}$
Now, as the $\operatorname{Dom} \left({\psi}\right) = F \left({x}\right)$, a field, $\psi$ is injective by
to be continued(again) - this proof is incomplete
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