$x\in K(t)$ is algebraic over $K$ if and only if $x\in K$ (with $t$ transcendental over $K$)

Question

I already proved the following:

Lemma A: Let $K$ be a field. Let $t$ be transcendental over $K$. Let $f \in K[X] \backslash K$. Then $f(t)$ is transcendental over $K$.

Proof: Because $t$ is transcendental over $K$, we have by definition: $\forall p\in K[X] \backslash \{ 0\} : p(t) \neq 0$. Suppose $f(t)$ is algebraic over $K$. Then there is a polynomial $q\in K[X] \backslash \{0 \}$ with $q(f(t))=0$. Define $g:=q(f(X)) \in K[X] \backslash \{ 0 \}$. Note: $g(t)=0$. Contradiction. Thus, $f(t)$ is transcendental over $K$. QED

Now I try to prove this.

Lemma B: Let $K$ be a field. Let $t$ be transcendental over $K$. Let $x\in K(t)$. Then $x$ is algebraic over $K$ if and only if $x\in K$.

My attempts: I write $x$ as $x=\frac{f(t)}{g(t)}$ with $f$ and $g$ being polynomials over $K$. If $x$ is algebraic over $K$, then there is a polynomial $p\in K[X]\backslash \{ 0 \} $ with $p(x)=0$. This is the same as $p\left (\frac{f(t)}{g(t)} \right ) = 0$. Now, I want to prove that if $x\not \in K$, this leads to a contradiction, but I don't know how to tackle this. Can anyone give me a hint?

Answer 1

Note that $g(t)^N p \left( \frac{f(t)}{g(t)} \right) = 0 $ is a monic polynomial in $t$ for a suitable large enough $N$

Answer 2

Lemma B Let $K$ be a field. Let $t$ be transcendental over $K$. Let $x\in K(t)$. Then: $x$ is algebraic over $K$ $\Leftrightarrow$ $x\in K$.

Proof "$\Leftarrow$" This is trivial.

"$\Rightarrow$" Suppose $x\not \in K$. Write $x=\frac{f(t)}{g(t)}$ with $f,g \in K[X]$ and $g\neq 0$. Because $x$ is algebraic over $K$, there is a polynomial $p=\sum _{i=0} ^N c_i X^i \in K[X] \backslash \{0\}$ with $p(x)=0$. We have: $$0=p(x)=p\left (\frac{f(t)}{g(t)} \right )= \sum _{i=0} ^N c_i \left (\frac{f(t)}{g(t)} \right )^i = \sum _{i=0} ^N c_i \frac{(f(t))^i}{(g(t))^i}$$ Now it follows that $$0= (g(t))^N \cdot \sum _{i=0} ^N c_i \frac{(f(t))^i}{(g(t))^i} = \sum _{i=0} ^N c_i (f(t))^i (g(t))^{N-i} \in K[t]$$ But then $t$ is algebraic over $K$. Contradiction. Therefore, we must have $x\in K$. QED