How to find the degree of an extension field?

Question

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 ?

Answer 1

Fact: Consider two polynomials $f$ and $p$ over $\Bbb{Q}$, with $p$ irreducible. It can be proved that if $f$ and $p$ share a root, then $p$ divides $f$.

How does this help? Suppose that $\alpha$ is a root of an irreducible polynomial $f \in \Bbb{Q}[X]$ of degree $n$ and that $[\Bbb{Q}(\alpha):\Bbb{Q}] = m < n$. Since $[\Bbb{Q}(\alpha):\Bbb{Q}]$ is the dimension of $\Bbb{Q}(\alpha)$ over $\Bbb{Q}$ as a vector space, this means that there are $q_0,\dotsc,q_{m-1} \in \Bbb{Q}$ such that $$ \alpha^m = q_0 + q_1 \alpha + \dotsb + q_{m-1} \alpha^{m-1} $$ hence $\alpha$ is a root of $g(X) = X^m - q_{m-1} X^{m-1} - \dotsb - q_0$. By the aforementioned fact it follows that $g$ divides $f$ in $\Bbb{Q}(X)$, but this is absurd because $f$ is irreducible.

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As for what happens when adjoining other roots, that depends on how much your field differs from being Galois.

For example, consider $f(X) = X^3 - d$, with $d$ a cube-free integer. Then its roots are $\sqrt[3]{d},\zeta_3\sqrt[3]{d},\zeta_3\sqrt[3]{d}$, where $\zeta_3$ is a primitive root of unity, i.e. a root of $X^2 + X + 1$. Then $\Bbb{Q}(\alpha_1, \alpha_2) = \Bbb{Q}(\sqrt[3]{d},\zeta_3)$ has degree $6$ for any two distinct roots $\alpha_1,\alpha_2$ of $f$.

On the other hand, say, adding any root of $X^4 + X^3 + X^2 + X + 1$ to $\Bbb{Q}(\zeta_5)$ won't change the degree of the extension (which is $5$).

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Regarding your specific problem: The splitting field of an irreducible cubic polynomial $f$ with rational coefficients can have either degree $3$ or degree $6$. In particular if, like in your case, $f$ has only one real root, then its splitting field has degree $6$. For example, you can find the full treatment of this at PlanetMath, where there is a proof that the splitting field of $f$ is $$ \Bbb{Q}(\alpha, \sqrt{D}) $$ where $\alpha$ and $D$ are, respectively, a root and the discriminant of $f$.

Note: $f$ has one real and two complex roots iff its discriminant is negative, thus in this case it is always true that $\sqrt{D} \notin \Bbb{Q}$.

Answer 2

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.