proving that a group is a subfield of ℝ

Question

can someone help me with this question:

$\Bbb Q[\sqrt2]: = ⟨a + b\sqrt2 \mid a,b ∈ ℚ ⟩$

prove that $\Bbb Q[\sqrt2]$ with the addition and multiplication of ℝ, and that it is a field.

I don't know how to write the accurate symbols, so if you're not sure of something please let me know.

i know that there's a quick way to show that Q[√2] is a sub field but i don't understand how to do it. (English is not my first language so please, try to explain your logic a bit slower than usual.. much appreciated!)

thank you

Answer 1

A subfield of a field L is a subset K of L that is a field with respect to the field operations inherited from L. Equivalently, a subfield is a subset that contains 1, and is closed under the operations of addition, subtraction, multiplication, and taking the inverse of a nonzero element of L.

$\mathbb Q[\sqrt 2]$ contains $1$. We can trivially show that it is closed for addition, subtraction, and multiplication.

That leaves whether it contains the multiplicative inverses of nonzero elements. For that we have: $$\frac{1}{a+b\sqrt 2}=\frac{a-b\sqrt 2}{(a+b\sqrt 2)(a-b\sqrt2)}=\frac{a}{a^2-2b^2} + \frac{-b}{a^2-2b^2}\sqrt 2$$ which is an element of $\mathbb Q[\sqrt 2]$.

QED

Answer 2

Here is is a short conceptual proof, which is not in the link given above:

As $\sqrt 2$ is a root of a polynomial in $\mathbf Q[X]$, the ring $\mathbf Q\bigl[\sqrt 2\,\bigr]$ is a finitely generated $\mathbf Q$-vector space, with basis $\{1,\sqrt 2\}$.

Now, multiplication by a nonzero element $x$ in this ring is a linear map, and it is injective since it is a subring of $\mathbf R$, hence an integral domain. And for an endomorphism of a finite-dimensional vector space, injective implies surjective, hence $1$ is attained, i.e. there exists $y\in \mathbf Q\bigl[\sqrt 2\,\bigr]$ such that $xy=1$.