# If $u\in F$ is Transcendental over $K$, $F$ an extension field of $K$, Show every element in $K(u)$ not in $K$ is transcendental over $K$.

Doing some problems out of Beachy’s Algebra text, I came across that problem, and I’m at a loss how to show it without a bit of hand waving. Do I make some statement about spaces, and prove by contradiction?

Ie. Suppose there exists an element $$v\in K(u): v\not \in K$$, such that $$\exists$$ minimal $$q(x)\in F[x], s.t. q(v)=0$$. Since $$v\notin K$$, $$v$$ may be expressed as the linear combination of $$au+b$$, where $$b$$ is some element in $$K$$, and $$a\ne 0$$. Then the minimal polynomial in $$K(u)$$ has form: $$q(t)=(t-v)=(t-(au+b)u),$$ and $$K(v)=(v-v)=0$$.

This feels like possibly too much work or too little rigor for a relatively simple question, am I missing something?

• I'm sorry, is $F = K(u)$? – Boots Mar 14 at 17:08
• $F$ doesn't need to be $K(u)$ for the statement to be true. – Robert Shore Mar 14 at 17:10
• If you're going through with a proof by contradiction you might want to show that if you have a polynomial $f \in K[x]$ with $f(au + b) = 0$ then you have a polynomial $g \in K[x]$ with g(u) = 0 – Boots Mar 14 at 17:13
• Why does $v = au + b$? Not seeing it. Why can't $v = p(u)$, $p(x) \in K[x]$, $\deg p(x) \ge 2$? – Robert Lewis Mar 14 at 17:15
• u and b form a basis for K(u) over K since it is a simple extension. – Boots Mar 14 at 17:35

If $$\frac {m(u)}{d(u)}$$ solves a polynomial of degree $$n, p(x) \in K[x]$$, then multiply through by $$d(u)^n$$ to find a polynomial in $$K[x]$$ with $$u$$ as a root, contradicting the assumption that $$u$$ is transcendental over $$K$$.
• $p(\frac{m(u)}{d(u)})=\sum_{k=0}^na_k(\frac{m(u)}{d(u)})^k=0$ for some $a_k \in K$. – Robert Shore Mar 14 at 18:24
• And how are we guaranteed that $\frac{m(u)}{d(u)}$ solves a polynomial in $K[x]$? – Boots Mar 15 at 6:25
• @Boots We're assuming toward a contradiction that $\frac{m(u)}{d(u)}$ is algebraic over $K$. – Robert Shore Mar 15 at 6:29