I'm currently studying ring and field theory and I'm a little confused about whats going on behind the scenes of Field extensions,as I missed some course work last semester.
say we want to adjoin $\sqrt{2}$ to the field of rationals $ \Bbb Q $
1.we know that $ \Bbb Q(\sqrt{2}):=\{a+b\sqrt{2}| a,b \in \Bbb Q \} $
2.we also know that $ \Bbb Q[x]/<x^2-2> \cong \Bbb Q(\sqrt{2})$
my question is how do we get the form of $\Bbb Q(\sqrt{2})$ to be $a+b\sqrt{2}$ from the relation given in 2. looking around stack exchange i have seen some people say that we can obtain the representatives of the coset through long division ? I don't understand what is meant by this , but this more mechanical approach to understanding would be preferable to a theorem heavy explanation. If anyone can give any insights it would be much appreciated.
The point is that in $(2)$, you have set the polynomial $x^2-2=0$. Let $f(x)\in \mathbb{Q}[x]$ be any polynomial. It may be written (uniquely) as $$ f(x) = a_0+a_1x+a_2x^2+\cdots +a_nx^n.$$
Note now that since $x^2 = 2$ (so $x^{2n} = 2^n$) in $\mathbb{Q}[x]/\langle x^2-2\rangle$, the image of $f$ in this quotient may be written as $f(x) = b_0+b_1x$, where $$b_0 = a_0+2a_2+\cdots 2^ka_{2k},$$ and similarly for $b_1$. Thus, every element in $\mathbb{Q}[x]/\langle x^2-2\rangle$ may be written uniquely as $a+bx$ where $x^2=2$. This gives an isomorphism $$\mathbb{Q}[x]/\langle x^2-2\rangle\cong \mathbb{Q}(\sqrt{2}).$$