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

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 ?

• Why do you believe the two rings are equal? What do you think their degrees over $\mathbb{Q}$ are? Nov 10, 2021 at 3:04
• 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. Nov 10, 2021 at 3:04
• $2^{1/4} \not\in \Bbb Q(i, 2^{1/2})$ Nov 10, 2021 at 3:05

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 $$\mathbb Q(i+\sqrt[4]2)$$.

• So $\mathbb Q(i+\sqrt[4]2)$ was a typo of the book and took lots of my time! Thanks :)
– 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 ).

Proposition

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).