How to find the degree of an extension field ?
Let $f:=T^3-T^2+2T+8\in\mathbb Z[T]$ and $\alpha$ be the real root of $f$. Why is then $\mathbb Q(\alpha)$ is a number field of degree $3$ ?
I've seen somewhere that $[\mathbb Q(r):\mathbb Q]\le n$ if $r$ is a root of an irreducible polynomial with coefficients in $\mathbb Q$ of degree $n$. What does it change in my case, if the extension field would contain also the other roots, they're also roots of the polynomial $f$, how does the degree increase ?
Obviously, by finding an element in $\mathbb Q(\alpha)$, which is not in $\mathbb Q$, the degree cannot be $1$, so it remains to show that, it is also not $2$. Or is there a better way, can we find $3$ field embeddings ?
Best Answer
The polynomial is irreducible over the rationals, because its possible rational roots are to be found among $\pm1$, $\pm2$, $\pm4$ and $\pm8$. A direct check shows these numbers are not roots.
Since the polynomial has degree $3$, reducibility over $\mathbb{Q}$ coincides with having a rational root.
So, if $\alpha$ is a root of the polynomial, $f$ is its minimum polynomial and it's a standard result that the degree of $\mathbb{Q}(\alpha)$ over $\mathbb{Q}$ equals the degree of the minimum polynomial.