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If $K$ is a finite field, and $E/K$ is a finite degree extension, is $E$ finite? This seems very obvious, but I'm having trouble justifying it.

What I know: $E/K$ must be finitely generated, so $E = K(\alpha_1, \ldots, \alpha_n)$. Then every element of $E$ can be written as a rational function $$ \frac{x_1 \alpha_1^{m_1} + \ldots x_n \alpha_n^{m_n}}{y_1 \alpha_1^{m_1} + \ldots y_n \alpha_n^{m_n}} $$ where $y_i,x_i \in K$. Since there are only finitely many such $x_i,y_i$, I suppose there are only finitely many such rational functions? Is this correct?

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Yes. A finite extension of a field $K$ is isomorphic to $K^n$ as a $K$-vector space, where $n$ is the degree. If $K$ is finite, $|K^n|=|K|^n$ is finite.

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