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# a finite extension of fields is an algebraic extension

###### Theorem 1.

Let $L/K$ be a finite field extension. Then $L/K$ is an algebraic extension.

###### Proof.

In order to prove that $L/K$ is an algebraic extension, we need to show that any element $\alpha\in L$ is algebraic, i.e., there exists a non-zero polynomial $p(x)\in K[x]$ such that $p(\alpha)=0$.

Recall that $L/K$ is a finite extension of fields, by definition, it means that $L$ is a finite dimensional vector space over $K$. Let the dimension be

$[L\colon K]=n$ |

for some $n\in\mathbb{N}$.

Consider the following set of “vectors” in $L$:

$\mathcal{S}=\{1,\alpha,\alpha^{2},\alpha^{3},\ldots,\alpha^{n}\}$ |

Note that the cardinality of $S$ is $n+1$, one more than the dimension of the vector space. Therefore, the elements of $S$ must be linearly dependent over $K$, otherwise the dimension of $S$ would be greater than $n$. Hence, there exist $k_{i}\in K,\ 0\leq i\leq n$, not all zero, such that

$k_{0}+k_{1}\alpha+k_{2}\alpha^{2}+k_{3}\alpha^{3}+\ldots+k_{n}\alpha^{n}=0$ |

Thus, if we define

$p(X)=k_{0}+k_{1}X+k_{2}X^{2}+k_{3}X^{3}+\ldots+k_{n}X^{n}$ |

then $p(X)\in K[X]$ and $p(\alpha)=0$, as desired.

∎

NOTE: The converse is not true. See the entry “algebraic extension” for details.

## Mathematics Subject Classification

12F05*no label found*

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