Why does $\lvert \Bbb{Q} [\sqrt[n]{2}] : \Bbb{Q} \rvert = n$ if $ \Bbb{Q} [\sqrt[n]{2}]$ is formed by adding only one element?

Question

I know that the degree of an extension the dimension of, in this case, $\Bbb{Q} [\sqrt[n]{2}]$ seen as a vector space over $\Bbb{Q} $. But I thought that the degree of an extension field formed by adding just one element to a given field was one. So, I find it confusing that

$\lvert \Bbb{Q} [\sqrt[n]{2}] : \Bbb{Q} \rvert = n$

Given that $ \Bbb{Q} [\sqrt[n]{2}]$ is formed by adding only one element, namely $\sqrt[n]{2}$, to the field of rational numbers. Any help would be very much appreciated. Thank you.

Answer 1

You are confusing the vector space obtained by adding one element with the field obtained by adding only one element.

Since $\mathbb Q[\sqrt[n]{2}]$ is a field, when we add $\sqrt[n]{2}$ we are also adding $\sqrt[n]{2}\cdot \sqrt[n]{2},...,\sqrt[n]{2}\cdot...\cdot\sqrt[n]{2}$ where in the last term there are $n-1$ products. It is enough to stop here since the product of n terms $$\sqrt[n]{2}\cdot...\cdot\sqrt[n]{2}=2 \in \mathbb Q$$ is already here.

Answer 2

$\mathbb{Q}[\sqrt[n]{2}]$ is a vector space over the field $\mathbb{Q}$. The equation $$[\mathbb{Q}[\sqrt[n]{2}]:\mathbb{Q}]=n$$ says that the dimesnion of this vector space over $\mathbb{Q}$ is $n$.

In general, if $K$ and $L$ are fields such that $K\subseteq L$, then $L$ can be considered as vector space over $K$. Then $[L:K]$ denotes the dimension of this vector space.