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

Am I right to say that the field $\mathbb{Q}(\sqrt{3}, \sqrt{-1})$ is an algebraic extension of $\mathbb{Q}$?

Because $\mathbb{Q}\subset\mathbb{Q}(\sqrt{3})\subset\mathbb{Q}(\sqrt{3})( \sqrt{-1})=\mathbb{Q}(\sqrt{3}, \sqrt{-1})$.


share|cite|improve this question
All finite extensions are algebraic... – The Chaz 2.0 Sep 22 '11 at 19:38
"Algebraically" is an adverb. I think you mean "algebraic"! – The Chaz 2.0 Sep 22 '11 at 19:39
What do you mean by "algebraically"? Do you mean that $\mathbb{Q}(\sqrt{3},\sqrt{-1})$ is an algebraic extension of $\mathbb{Q}$? If so, then the answer is yes: A generic element in $\mathbb{Q}(\sqrt{3},\sqrt{-1})$ is of the form $a+b\sqrt{3}+c\sqrt{-1}$. Call this $\alpha$ and try to find a polynomial with $\alpha$ as zero. – Fredrik Meyer Sep 22 '11 at 19:40
@Fredrik, you need to add rational multiples of $\sqrt{-3}$, too. – Henning Makholm Sep 22 '11 at 19:42
Fred, how 'bout making that an answer? – The Chaz 2.0 Sep 22 '11 at 19:46
up vote 6 down vote accepted

HINT $\rm\ K =\: \mathbb Q(\sqrt 3, \sqrt{-1})\:$ is a $4$-dimensional vector space over $\rm\mathbb Q\:,\:$ viz. $\rm\: K =\: \mathbb Q\langle 1,\:\sqrt 3,\:\sqrt{-1},\:\sqrt{-3}\rangle\:.$

Hence $\rm\:\alpha\in K\ \Rightarrow\ 1,\ \alpha,\ \alpha^2,\ \alpha^3,\ \alpha^4\:$ are linearly dependent over $\rm\:\mathbb Q\:.\:$ This dependence relation yields a nonzero polynomial $\rm\:f(x)\in \mathbb Q[x]\:$ of degree $\le 4\:$ such that $\rm\:f(\alpha)=0\:.$

share|cite|improve this answer

Since a comment suggested that I add my comment as an answer:

To show that $K=\mathbb{Q}(\sqrt{3},\sqrt{-1})$ is an algebraic extension of $\mathbb{Q}$, you need to show that every element in K is the root of some polynomial $f(x)$ with coefficients in $\mathbb{Q}$.

A generic element in K has the form $a+b\sqrt{3}+c\sqrt{-1}+d\sqrt{-3}$ (thanks to Henning Makholm for pointing out the last term). Call this expression $\alpha$ and try to get rid of the square roots by repeatedly squaring. Use this to find a polynomial with $\alpha$ as root.

Since $\alpha$ was arbitrary, you would have shown that K is an algebraic extension of $\mathbb{Q}$. (alternatively, try to prove Chaz' assertion that every finite extension is algebraic)

share|cite|improve this answer
Depending on the context, I would have just said "It is finite, and hence algebraic"! – The Chaz 2.0 Sep 22 '11 at 22:34

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.