Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If I have a subfield $F$ of a field $E$, and an algebraic (over $F$) $\alpha\in E$, I can form $F(\alpha)$ which is isomorphic to $F[x]/\langle f(x)\rangle$ for $f(x) = irr(\alpha, F)$. That is, $f(x)$ is the minimal degree and monic element of $F[x]$ such that $f(\alpha) = 0$.

The book I'm using defines $F(\alpha)$ officially as the image of $F[x]$ under $\phi_{\alpha}$, the map $f(x)\mapsto f(\alpha)$, and the isomorphism mentioned above comes from the fact that its kernel is $\langle f(x)\rangle$.

My question is: if there are other roots of the polynomial $f(x)$ in $E$, then are they necessarily contained in $F(\alpha)$? My intuitive guess is yes, since the field $F[x]/\langle f(x)\rangle$ doesn't know the difference between distinct roots of $f(x)$. But if this is correct then why?

share|cite|improve this question
up vote 4 down vote accepted

No. Take $F = \mathbb{Q}, E = \mathbb{C}$ and $\alpha = \sqrt[3]{2}$. Then $F(\alpha)$ lies in $\mathbb{R}$ but $f(x) = x^3 - 2$ has two complex roots which do not lie in $\mathbb{R}$, hence cannot lie in $F(\alpha)$.

The fact that this can be false motivates the definition of a normal extension.

share|cite|improve this answer

The other roots are not necessarily contained in $F(\alpha)$. For example let $\alpha=2^{\frac{1}{3}}$. $F=Q$, the rationals, then $F(\alpha)$ is contained in the real numbers. However the other roots of the minimal polynomial are not real.

share|cite|improve this answer

Since others have provided nice examples, let me address your last point about $F[x]/ \langle f(x)\rangle$ not knowing the difference between the roots of $f$.

Suppose $a_1,a_2, \cdots ,a_n$ are roots of $f$, an irreducible polynomial in $F$. Since $F[x]/ \langle f(x)\rangle$ doesnt know the difference between the roots of $f$ (i.e. the $a_i$s) this will mean that all the fields $F(a_1),F(a_2), \cdots F(a_n)$s are isomorphic. This is as good as it gets though.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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