Factoring By An Irreducible Gives A Simple Extension

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Problem 70 (Factoring By An Irreducible Gives A Simple Extension)

Let $F$ be a field and $p(x)\in F[x]$ be irreducible over $F$. If $a$ is a zero of $p(x)$ in some extension $E$ of $F$, then $F(a)$ is isomorphic to $F[x]/\left<p(x)\right>$. Furthermore, if $\deg p(x) = n$, then every member of $F(a)$ can be uniquely expressed in the form $c_{n-1}a^{n-1} + c_{n-2}a^{n-2} + \cdots + c_{1}a + c_{0}$ where $c_i\in F$. (In other words, the set $\left\{1, a, a^2, \ldots, a^{n-1}\right\}$ is a basis for $F(a)$ over $F$).

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