Exercise 4.3.15 (Simple extension)

Question

Prove that $\mathbb{Q}(\sqrt 2, \sqrt 3) = \mathbb{Q}(\sqrt 2 + \sqrt 3)$.

Hassaan Naeem · Accepted Answer

extbf{Solution:}
We know that $\sqrt2 + \sqrt3 \in \mathbb{Q}(\sqrt2, \sqrt3)$ and hence $\mathbb{Q}(\sqrt2 + \sqrt3) \subseteq \mathbb{Q}(\sqrt2 , \sqrt3)$. Now we show the inclusion the other way.
We use the hint and get that $(\sqrt2 + \sqrt3)^3 = 11\sqrt2 + 9\sqrt3 \in \mathbb{Q}(\sqrt2+ \sqrt3)$. Then we have that:\ $11\sqrt2 + 9\sqrt3 - 9(\sqrt2 + \sqrt3 ) = 2\sqrt2 \in \mathbb{Q}(\sqrt2 + \sqrt3)$, 
hence $\sqrt2 \in \mathbb{Q}(\sqrt2 + \sqrt3)$. Similarly, we get that $\sqrt3 \in \mathbb{Q}(\sqrt2 + \sqrt3)$. Therefore, $\mathbb{Q}(\sqrt2, \sqrt3) \subseteq \mathbb{Q}(\sqrt2 + \sqrt3) \quad \square$

Exercise 4.3.15 (Simple extension)

Answers

Comments

Exercise 4.3.15 (Simple extension)

Answers

Comments

Add answer