# Math Help - fields and subfields

1. ## fields and subfields

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))

2. 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?

3. 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, $x\in K$.

4. 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)
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 $a+b,ab\in F$.