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?

  • $\begingroup$ I'm sorry, is $F = K(u)$? $\endgroup$ – Boots Mar 14 at 17:08
  • $\begingroup$ $F$ doesn't need to be $K(u)$ for the statement to be true. $\endgroup$ – Robert Shore Mar 14 at 17:10
  • $\begingroup$ 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 $\endgroup$ – Boots Mar 14 at 17:13
  • $\begingroup$ Why does $v = au + b$? Not seeing it. Why can't $v = p(u)$, $p(x) \in K[x]$, $\deg p(x) \ge 2$? $\endgroup$ – Robert Lewis Mar 14 at 17:15
  • $\begingroup$ u and b form a basis for K(u) over K since it is a simple extension. $\endgroup$ – 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 $K$.

  • $\begingroup$ I don't understand but the underlying idea is looking correct. What do you mean that fraction solves a polynomial of degree n? $\endgroup$ – Boots Mar 14 at 17:51
  • $\begingroup$ $p(\frac{m(u)}{d(u)})=\sum_{k=0}^na_k(\frac{m(u)}{d(u)})^k=0$ for some $a_k \in K$. $\endgroup$ – Robert Shore Mar 14 at 18:24
  • $\begingroup$ And how are we guaranteed that $\frac{m(u)}{d(u)}$ solves a polynomial in $K[x]$? $\endgroup$ – Boots Mar 15 at 6:25
  • $\begingroup$ @Boots We're assuming toward a contradiction that $\frac{m(u)}{d(u)}$ is algebraic over $K$. $\endgroup$ – Robert Shore Mar 15 at 6:29
  • $\begingroup$ So everything in K(u) not in K can be written in this form? $\endgroup$ – Boots Mar 15 at 23:23

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