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

If $P(x)$ is an irreducible polynomial over $\mathbb{Q}$ and I am interested its splitting field : I know that $\mathbb{Q}/\langle P(x)\rangle$ have one of the roots of $P(x)$, and that regardless of what root we take the field obtained is isomorphic to $\mathbb{Q}/\langle P(x)\rangle$

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 ?)

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.

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

We know that $\mathbb{Q}(\sqrt{2})$ is isomorphic to $\mathbb{Q}[x]/\langle x^3-2\rangle$, and likewise $$\mathbb{Q}(\omega\sqrt{2})\cong\mathbb{Q}(\omega^2\sqrt{2})\cong\mathbb{Q}(\sqrt{2})\cong\frac{\mathbb{Q}[x]}{\langle x^3-2\rangle}.$$ But $\mathbb{Q}(\sqrt{2})$ cannot contain all roots of $x^3-2$: $\mathbb{Q}(\sqrt{2})$ 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 $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\omega\sqrt{2})$ that maps $\sqrt{2}$ to $\omega\sqrt{2}$; but once you do that, you "lose" $\sqrt{2}$. 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.

• Really clears things up, thanks! – Belgi Apr 20 '12 at 20:41

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

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

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

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

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

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

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!