Let $F$ be a field and $p(x)$ be a nonconstant polynomial in $F[x]$. We already know that we can find a field extension of $F$ that contains a root of $p(x)$. However, we would like to know whether an extension $E$ of $F$ containing all of the roots of $p(x)$ exists. In other words, can we find a field extension of $F$ such that $p(x)$ factors into a product of linear polynomials? What is the “smallest” extension containing all the roots of $p(x)$?

Let $F$ be a field and $p(x) = a_0 + a_1 x + \cdots + a_n x^n$ be a nonconstant polynomial in $F[x]$. An extension field $E$ of $F$ is a splitting field of $p(x)$ if there exist elements $\alpha_1, \ldots, \alpha_n$ in $E$ such that $E = F( \alpha_1, \ldots, \alpha_n )$ and \begin{equation*}p(x) = ( x - \alpha_1 )(x - \alpha_2) \cdots (x - \alpha_n).\end{equation*} A polynomial $p(x) \in F[x]$ splits in $E$ if it is the product of linear factors in $E[x]$.

##### Example21.29

Let $p(x) = x^4 + 2x^2 - 8$ be in ${\mathbb Q}[x]$. Then $p(x)$ has irreducible factors $x^2 -2$ and $x^2 + 4$. Therefore, the field ${\mathbb Q}( \sqrt{2}, i )$ is a splitting field for $p(x)$.

##### Example21.30

Let $p(x) = x^3 -3$ be in ${\mathbb Q}[x]$. Then $p(x)$ has a root in the field ${\mathbb Q}( \sqrt[3]{3}\, )$. However, this field is not a splitting field for $p(x)$ since the complex cube roots of 3, \begin{equation*}\frac{ -\sqrt[3]{3} \pm (\sqrt[6]{3}\, )^5 i }{2},\end{equation*} are not in ${\mathbb Q}( \sqrt[3]{3}\, )$.

##### Proof

The question of uniqueness now arises for splitting fields. This question is answered in the affirmative. Given two splitting fields $K$ and $L$ of a polynomial $p(x) \in F[x]$, there exists a field isomorphism $\phi : K \rightarrow L$ that preserves $F$. In order to prove this result, we must first prove a lemma.