The Stacks project

Example 9.7.6 (Degree of a simple algebraic extension). Consider a monogenic field extension $E/k$ of the form discussed in Example 9.6.4. In other words, $E = k[t]/(P)$ for $P \in k[t]$ an irreducible polynomial. Then the degree $[E : k]$ is just the degree $d = \deg (P)$ of the polynomial $P$. Indeed, say
\begin{equation} \label{fields-equation-P} P = a_ d t^ d + a_{d - 1} t^{d - 1} + \ldots + a_0. \end{equation}

with $a_ d \not= 0$. Then the images of $1, t, \ldots , t^{d - 1}$ in $k[t]/(P)$ are linearly independent over $k$, because any relation involving them would have degree strictly smaller than that of $P$, and $P$ is the element of smallest degree in the ideal $(P)$.

Conversely, the set $S = \{ 1, t, \ldots , t^{d - 1}\} $ (or more properly their images) spans $k[t]/(P)$ as a vector space. Indeed, we have by ( that $a_ d t^ d$ lies in the span of $S$. Since $a_ d$ is invertible, we see that $t^ d$ is in the span of $S$. Similarly, the relation $t P(t) = 0$ shows that the image of $t^{d + 1}$ lies in the span of $\{ 1, t, \ldots , t^ d\} $ — by what was just shown, thus in the span of $S$. Working upward inductively, we find that the image of $t^ n$ for $n \geq d$ lies in the span of $S$.

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