I want to show that $\mathbb{Q}(i+2^{1/4})$ and $\mathbb{Q}(i,2^{1/2})$ are equal. It is enough to show that :

  1. there are two polynomials $g(x),h(x) \in \mathbb{Q}[x]$ such that for $\delta=i+2^{1/4}$, $\sqrt{2}=g(\delta)$ and $i=h(\delta)$, and

  2. there is a polynomial $p(x,y) \in \mathbb{Q}[x,y]$ of two variables for $x=\sqrt{2}, y=i$ such that $\delta=p(\sqrt{2},i)$.

All manipulation with $\delta$ like some combination of inverse of it, squaring it, etc doesn't solve the problem. Also I calculated ${\delta}^8 + 4 {\delta}^6 +2{\delta}^4 +28{\delta}^2 +1 = 0$ which doesn't solve the problem neither.

So how $\mathbb{Q}(i+2^{1/4})=\mathbb{Q}(i,2^{1/2})$?

The trick in here works for that specific example. Is there a general algorithm?

Added : Based on the comments below, if $\mathbb{Q}(i+2^{1/4})=\mathbb{Q}(i,2^{1/2})$ is not true so what the primitive element $\delta$ such that $\mathbb{Q}(\delta)=\mathbb{Q}(i,2^{1/2})$ and in what way one calculates it ?

  • $\begingroup$ Why do you believe the two rings are equal? What do you think their degrees over $\mathbb{Q}$ are? $\endgroup$ Nov 10, 2021 at 3:04
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    $\begingroup$ It seems unlikely they are. If $i\in\mathbb Q(i +2^{1/4}),$ then you’d need $2^{1/4}\in \mathbb Q(i ,2^{1/2}),$ and that isn’t true. $\endgroup$ Nov 10, 2021 at 3:04
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    $\begingroup$ $2^{1/4} \not\in \Bbb Q(i, 2^{1/2})$ $\endgroup$
    – jjagmath
    Nov 10, 2021 at 3:05

2 Answers 2


You're wrong. $\mathbb Q(i,\sqrt2)\ne\mathbb Q(i+\sqrt[4]2)$.

Let $i+\sqrt2=\alpha$, then $\alpha^{-1}=\frac{\sqrt2-i}3$. So $\frac12\alpha+\frac32\alpha^{-1}=\sqrt2, \frac12\alpha-\frac32\alpha^{-1}=i$.

So, $\mathbb Q(i,\sqrt2)=\mathbb Q(i+\sqrt2)$, not $\mathbb Q(i+\sqrt[4]2)$.

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    $\begingroup$ So $\mathbb Q(i+\sqrt[4]2)$ was a typo of the book and took lots of my time! Thanks :) $\endgroup$
    – user200918
    Nov 10, 2021 at 3:17

For you second question the following theorem and proposition can help you for have a intuition about decide in which situation you can find a primitive element in finite extensions (I don´t add the proofs, but it are easy of find and understand in any book of Galois Theory ).


Let $F\subset K$ finite extension, then $K$ is a simple extension (it means that there are $\alpha\in K$ such that $K=F(\alpha)$ ) iff there are finitely many intermediate fields between $F$ and $K$.

And the primitive element theorem states that:

Theorem (primitive element)

Let $K$ a finite separable extension of $F$ then there are $\alpha\in K$ such that $K=F(\alpha)$.

In the proof of the theorem states that in the particular case of $K=F(\gamma, \beta)$ we can choose $\alpha=\gamma+ a \beta$ where $a\in F$ (you can find the $a\in F$ such that the equality follows).


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