Field extension over the reals

Question

I am studying these days filed theory and i'm new to this subject,so my question may be to trivial. The field extension $\mathbb{R}$(${\sqrt{5}}$)/$\mathbb{R}$ has degree $1$ or $2$? We see that $\mathbb{R}$=$\mathbb{R}$(${\sqrt{5}}$) so a basis ot this field extension extention must be the set {1} or is the set {$1$,${\sqrt{5}}$} ?

Answer 1

As you have observed, $\mathbb{R}(\sqrt{5})$ is just $\mathbb{R}$, since $\sqrt{5}\in\mathbb{R}$. For any field $K$, considered as a vector space over itself, $\{1\}$ is a basis, since every element $a\in K$ is in its span (namely as the scalar product $a\cdot 1$). Note that $\{1,\sqrt{5}\}$ is not linearly independent over $\mathbb{R}$, since $\sqrt{5}\cdot 1+(-1)\cdot \sqrt{5}=0$ (here the first parts of each term are coefficients in $\mathbb{R}$).