Let $k \subset K=k(a_1,\ldots,a_n)$ be a finite separable field extension (assume that $k$ is an infinite field). We know that such an extension is simple, with infinitely many primitive elements, namely, $K = k(a_1+ \lambda_2 a_2 + \cdots +\lambda_n a_n)$ for all but finitely many $k \ni \lambda$'s. Call such $\lambda$'s "good choices".

My question: Is it possible to find a condition that will guarantee that there exist two different good choices $\lambda_i , \mu_i \in k$ such that $a_1+ \lambda_2 a_2 + \cdots +\lambda_n a_n$ and $a_1+ \mu_2 a_2 + \cdots +\mu_n a_n$ are conjugate (= have the same minimal polynomial over $k$).

Edit: The answer below only deals with a specific field $k$; I will be very grateful if one can explain what happens for other fields.


1 Answer 1


This answer is about the case $k=$ the field of rational numbers.

Perhaps you are looking for a property of a function that sends the co-ordinates of an element to conjugate's co-ordinates (in your words," good choices"). Such a mapping will co-ordinatise the field embeddings (because all field embedding have to send a primitive element to a conjugate). I want to draw your attention to the fact that except identity and complex conjugation no field homomorphisms from a number field to the complex numbers can be restriction of a continuous function with domain complex/real numbers.

As field homomorphism has to be identity on the set of rationals, and the latter being a dense set it is not possible.

  • $\begingroup$ Interesting! thanks. Do you think that in case that $k \neq \mathbb{Q}$ there is a chance to have a positive answer, namely, find a condition as I wish? (for example, $k \neq \{\mathbb{Q}, \mathbb{R}\}$ which has an algebraic closure of infinite degree). $\endgroup$
    – user237522
    Commented Oct 24, 2016 at 23:38
  • 1
    $\begingroup$ I do not know much about other fields. If you want one root to generate other roots, (by expecting a formula for the conjugate of an element) you are forcing the extension to be Galois. Otherwise the formula will involve parameters outside the base field $k$. $\endgroup$ Commented Oct 25, 2016 at 0:22

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .