1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $K$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.