# Irreducible polynomial

• Apr 6th 2010, 05:48 AM
bram kierkels
Irreducible polynomial
I want to show that $\displaystyle x^2+i$ is irreducible over $\displaystyle \mathbb{Q}(i)$

My try;
If there is an expansion it should be lineair. Thus there exist a $\displaystyle x \in \mathbb{Q}(i)$ so that $\displaystyle x^2+i=0$
Than $\displaystyle x= \sqrt{-i}$ or $\displaystyle x=- \sqrt{-i} \Rightarrow x \notin \mathbb{Q}(i)$?
• Apr 6th 2010, 06:40 AM
tonio
Quote:

Originally Posted by bram kierkels
I want to show that $\displaystyle x^2+i$ is irreducible over $\displaystyle \mathbb{Q}(i)$

My try;
If there is an expansion it should be lineair. Thus there exist a $\displaystyle x \in \mathbb{Q}(i)$ so that $\displaystyle x^2+i=0$
Than $\displaystyle x= \sqrt{-i}$ or $\displaystyle x=- \sqrt{-i} \Rightarrow x \notin \mathbb{Q}(i)$?

Suppose $\displaystyle \sqrt{i}\in\mathbb{Q}(i)=\{a+bi\;;\;a,b\in\mathbb{ Q}\}$ $\displaystyle \Longrightarrow \sqrt{i}=a+bi\Longrightarrow i=a^2+2abi-b^2$ .

Check now that from the last equation we get that $\displaystyle i$ is a rational, and thus a real, number...( hint: distinguish the cases $\displaystyle ab=\frac{1}{2}\,,\,\,ab\neq \frac{1}{2}$ )

Tonio