Splitting field of an irreducible polynomial over $\mathbb{Q}$

Question

Let $P(x)$ be an irreducible polynomial over $\mathbb{Q}$; I am interested its splitting field. I know that $\mathbb{Q}[x]/\langle P(x)\rangle $ have one of the roots of $P(x)$, and that regardless of what root $\xi$ we take the field $\mathbb{Q}(\xi)$ obtained by adjoining $\xi$ to $\mathbb{Q}$ is isomorphic to $\mathbb{Q}[x]/\langle P(x)\rangle$

I think think that if two field are isomorphic then a polynomial is irreducible over one of them implies it's irreducible over the second (is this true ? is it an iff claim ?)

From here I think that adding one root implies we added them all and so we know the splitting field of $P(x)$.

Are some of my claims wrong ? (why ?)

Answer 1

You have to be careful about what you mean when you say "if two fields are isomorphic then a polynomial is irreducible over one of them if and only if it is irreducible over the second", since that statement may not even make sense as written.

Rather: suppose that $F$ and $K$ are two fields, and $\sigma\colon F\to K$ is a field isomorphism between them. Then $\sigma$ extends to an isomorphism of rings, $\sigma\colon F[x]\to K[x]$, by acting on the coefficients. Let $g(x)\in F[x]$ be a polynomial, and let $\sigma g(x)$ be the corresponding polynomial in $K[x]$. Then $g(x)$ is irreducible in $F[x]$ if and only if $\sigma g(x)$ is irreducible in $K[x]$. This is true because $\sigma$ is an isomorphism between $F[x]$ and $K[x]$, so any factorization in one of the two is transfered, via $\sigma$ or $\sigma^{-1}$, into a factorization on the other.

No, it does not follow that adding one root adds all of them.

For example, take $x^3-2$. There are two complex roots and one real root, $\sqrt[3]{2}$, $\omega\sqrt[3]{2}$, and $\omega^2\sqrt[3]{2}$, where $\omega=\frac{-1+\sqrt{-3}}{2}$.

We know that $\mathbb{Q}(\sqrt[3]{2})$ is isomorphic to $\mathbb{Q}[x]/\langle x^3-2\rangle$, and likewise $$\mathbb{Q}(\omega\sqrt[3]{2})\cong\mathbb{Q}(\omega^2\sqrt[3]{2})\cong\mathbb{Q}(\sqrt[3]{2})\cong\frac{\mathbb{Q}[x]}{\langle x^3-2\rangle}.$$ But $\mathbb{Q}(\sqrt[3]{2})$ cannot contain all roots of $x^3-2$: $\mathbb{Q}(\sqrt[3]{2})$ is contained in $\mathbb{R}$, and the "other" two roots are not.

The result just tells you that there is an isomorphism over $\mathbb{Q}$ between $\mathbb{Q}(\sqrt[3]{2})$ and $\mathbb{Q}(\omega\sqrt[3]{2})$ that maps $\sqrt[3]{2}$ to $\omega\sqrt[3]{2}$; but once you do that, you "lose" $\sqrt[3]{2}$. That is a root in the domain which is not in the codomain. So you don't necessarily get all roots by adding a single one.

Answer 2

No it does not mean that adding one root gives you all the roots. For example let us calculate the splitting field of $x^4 - 6 \in \Bbb{Q}[x]$. By Eisenstein's Criterion with $p= 2$ this polynomial is irreducible over $\Bbb{Q}$. Now we find the roots of this polynomial in $\Bbb{C}$ to be

$$\sqrt[4]{6}e^{2 k \pi i/3} \hspace{2mm} \text{for} \hspace{2mm} k=0, \ldots 5.$$

Now you may guess that adjoining one root, say $\sqrt[4]{6}e^{2 \pi i/3}$ is enough to give us the splitting field. But then we have a problem because

$$\Bbb{Q}(\sqrt[4]{6}e^{2 \pi i/3}) \cong \Bbb{Q}[x]/(x^4 - 6) \cong \Bbb{Q}[\sqrt[4]{6}].$$

The field on the right hand side only contains two of the roots of the polynomial $x^4 -6$, so there must be something wrong. In fact the splitting field is actually

$$\Bbb{Q}(\sqrt[4]{6},e^{2\pi i/3}) \cong \Bbb{Q}(\sqrt[4]{6}, \sqrt{3}i)$$

that has degree 8 over $\Bbb{Q}$. It is a good exercise which you should do to show why this is the splitting field.

In some polynomial rings over certain fields, adjoining one root to the field does give you all the roots. Consider the polynomial

$$f(x) = x^p - x - a$$

in $\big(\Bbb{Z}/p\Bbb{Z}\big)[x]$ with $a \neq 0$. Then in $\Bbb{Z}/p\Bbb{Z}$, no matter what value we substitute in for $x$ by Fermat's Little Theorem $f(b) = a$ for all $b \in \big(\Bbb{Z}/p\Bbb{Z}\big)$. Now create a root for this polynomial in the field extension

$$\big(\Bbb{Z}/p\Bbb{Z}[x]\big)/(x^p - x - a).$$

Call that root $\gamma$. Now we know that this field must contain $\big(\Bbb{Z}/p\Bbb{Z}\big)$ because

$$\big(\Bbb{Z}/p\Bbb{Z}\big) \subset \big(\Bbb{Z}/p\Bbb{Z}[\gamma]\big) \cong \big(\Bbb{Z}/p\Bbb{Z}[x]\big)/(x^p - x - a).$$

Now we can view the polynomial $x^p - x - a$ as sitting inside $$\big(\Bbb{Z}/p\Bbb{Z}[\gamma]\big)[x].$$ Furthermore, this is a polynomial ring over a field because the element $\gamma$ is algebraic over $\Bbb{Z}/p\Bbb{Z}$. We know any polynomial in a polynomial ring over a field can have at most $\deg f$ roots. In your case if we find $\deg f$ distinct roots for the polynomial $f(x) = x^p - x - a$ lying in that field, we must have found all of them. The roots of $f(x)$ are distinct because the greatest common divisor of $f(x)$ and its formal derivative $f'(X) = px^{p-1} - 1 = -1$ is 1.

Now notice for any $k \in \Bbb{Z}/p\Bbb{Z}$,

$$\begin{eqnarray*} f(\gamma + k) &=& (\gamma + k)^p - \gamma - k -a\\ &=& \gamma^k + k^p - \gamma - k - a \hspace{2mm} \text{(Using the fact that $p| \binom{p}{i}$ for $1 \leq i \leq p-1$})\\ &=& \gamma^k - \gamma - a \hspace{6mm} \text{(Fermat's Little Theorem shows that $k^p - k = 0$)}\\ &=& 0 \end{eqnarray*}$$

because $\gamma$ is a root of $f$. Letting $k$ vary from $0$ to $p-1$ shows that we have $p$ distinct roots for $f(x)$ lying in the field extension $\big(\Bbb{Z}/p\Bbb{Z}\big)[\gamma]$. So we have just seen an example where adjoining one root of a polynomial gives us all of the roots.

I hope this helps!

Answer 3

This is too long for a comment but is not a complete answer to the entire question.

Regarding the misconception that adjoining one root implies adjoining all of them, I initially guessed the same wrong conclusion by thinking about the standard examples $P(X) = X^2 + 1$ and $P(X) = X^2 - 2$.

In case it helps explain why degree 2 irreducible polynomial examples are misleading: notice that for any irreducible polynomial $P(x)$ of degree $d$, the field produced by adjoining any $(d-1)$ of the roots is always equal to the splitting field (i.e. produced by adjoining all $d$ roots).

This is because any root can always be written as a rational function of the remaining $(d-1)$ roots.

Take $(-1)^d P(0)$ (which is in $\mathbb{Q}$ because $P(x) \in \mathbb{Q}[x]$) and divide it by the product of the remaining $(d-1)$ roots and the result will always equal the given root. This is because of the factorization of the polynomial $P(X)$ when $X = 0$. (Because $P(X)$ is assumed to be irreducible, we know that zero is not a root, because otherwise we would be able to factor out a factor of $X$. Equivalently we would have that there is no constant term, noticing that the constant term of any polynomial equals $P(0)$.)

More concisely, letting $\theta_1, \theta_2, \dots, \theta_{(d-1)}$, $\theta_d$ denote an arbitrary ordering of the roots of an irreducible polynomial $P(x)$ (so again in particular we can assume that $\theta_i \not= 0$ for all $i=1,\dots,d$) we have $$P(0) = \left( \prod_{i=1}^{(d-1)} (0 - \theta_i) \right) (0 - \theta_d)$$ so $$\frac{P(0)}{(-1)^{(d-1)} \prod_{i=1}^{(d-1)} \theta_i} = -\theta_d$$ from which the claim follows using $((-1)^m)((-1)^m) = 1$ for any integer $m$.

As a result, we always have that $\mathbb{Q}(\theta_1, \dots, \theta_{(d-1)}) = \mathbb{Q}(\theta_1, \dots, \theta_{(d-1)}, \theta_d)$.

So in particular $\mathbb{Q}(i) = \mathbb{Q}(i, -i) = \mathbb{Q}(-i)$, and $\mathbb{Q}(\sqrt{2}) = \mathbb{Q}(\sqrt{2}, -\sqrt{2}) = \mathbb{Q}(-\sqrt{2})$. Likewise for the $X^3-2$ example, we have $\mathbb{Q}(\sqrt[^3]{2}, \omega\sqrt[^3]{2}) = \mathbb{Q}(\sqrt[^3]{2}, \omega^2\sqrt[^3]{2}) = \mathbb{Q}(\omega\sqrt[^3]{2}, \omega^2\sqrt[^3]{2}) = \mathbb{Q}(\sqrt[^3]{2}, \omega\sqrt[^3]{2}, \omega^2\sqrt[^3]{2})$, with $\omega$ a non-trivial third root of unity. (Test out the formula above to confirm it works.)

Notice also that this observation is definitely not novel, instead just being a special-case of the "well-known" Vieta's formulas. (Well-known in the sense of not usually being taught in school but often showing up in math competitions.)

Notice also that although we always have $\mathbb{Q}(\theta_1, \dots, \theta_{(d-1)}) = \mathbb{Q}(\theta_1, \dots, \theta_{(d-1)}, \theta_d)$, it is still possible for some examples for an even smaller subset of the roots (combined with the numbers in $\mathbb{Q}$) to generate all of the roots (and by extension the whole splitting field).

The reason why it's possible in some cases for an even smaller subset of the roots to generate the splitting field is related to why not all irreducible polynomials of degree $d$ have $S_d$ (symmetric group on $d$ characters) as their Galois group.

Compare the $X^4 - 6$ example given in another answer -- the degree of the field extension is $8 < 24 = 4!$, so the order of the Galois group is $8$, and thus the Galois group is not $S_4$. (In fact the Galois group has to be $D4$, the dihedral group of the square, because that is the only transitive subgroup of $S4$ with order $8$.)

Notice how this obersvation also points to the $d=2$ case being misleading / confusing due to the "Law of Small Numbers", given that the Galois group of any Quadratic polynomial is always $S_2 \cong \mathbb{Z}/2\mathbb{Z}$.