fields and subfields

**JCIR** · April 25th 2008, 11:38 AM

Assume Q[sqrt(2)] is a subfield of the Real field R, and prove that it is the smallest subfield of R that contains sqrt(2)

Q[sqrt(2)]= a + b(sqrt(2))

**Jhevon** · April 25th 2008, 11:41 AM

Originally Posted by JCIR

Assume Q[sqrt(2)] is a subfield of the Real field R, and prove that it is the smallest subfield of R that contains sqrt(2)

how is $Q[\sqrt{2}]$ defined?

**ThePerfectHacker** · April 25th 2008, 12:32 PM

Originally Posted by JCIR

Assume Q[sqrt(2)] is a subfield of the Real field R, and prove that it is the smallest subfield of R that contains sqrt(2)

Q[sqrt(2)]= a + b(sqrt(2))

The way $\mathbb{Q}(\sqrt{2})$ is usually defined is to be the smallest subfield containing $\mathbb{Q}$ and $\sqrt{2}$ .

Look Here.

So I will assume you are defining $\mathbb{Q}(\sqrt{2}) = \{ a+b\sqrt{2}|a,b\in \mathbb{Q}\}$ and you want to show it is the smallest such field. Let $\mathbb{Q}\subseteq K\subseteq \mathbb{R}$ be a field containing $\mathbb{Q}$ and $\sqrt{2}$ . We need to show $\mathbb{Q}(\sqrt{2})\subseteq K$ . Let $x\in \mathbb{Q}(\sqrt{2})$ then $x=a+b\sqrt{2}$ but then $a+b\sqrt{2}\in K$ because it is closed under sums and products, thus, $x\in K$ .

**ThePerfectHacker** · April 25th 2008, 01:02 PM

This is JCIR you answered my question on fields and subfields, can you tell my why is it closed under sums and products ( i need to prove that too)
Thanks alot for your help.

Because a field has to be closed under sums and products (that is basically the definition of "field"). Meaning if $a,b\in F$ then $a+b,ab\in F$ .

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