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Let's say we adjoin a second degree algebraic number to $F_3=\mathbb Z/3\mathbb Z$, it doesn't matter which. Then we get a field of $9$ elements, $F_{3^2}$. On the other hand, if we adjoin a fourth degree algebraic number, we get $F_{3^4}$.

But, $F_{3^2}$ contains a primitive element, say $z$, a primitive $8$-th root of unity. But then $z$ is a root of the irreducible (over $F_3$) polynomial $x^4+1$. But then isn't $F_3(z)$ isomorphic to $F_{3^4}$?

Jack M
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    The polynomial $x^4+1$ is reducible over $\mathbf F_3$, and in fact over $\mathbf F_p$ for every prime $p$. – KCd Mar 03 '15 at 13:33
  • See this question for a discussion of the fact brought up by KCd. An ad hoc calculation specific to $p=3$ is $$x^4+1=x^4-3x^2+1=(x^4-2x^2+1)-x^2=(x^2-1)^2-x^2$$ that you can factor like any difference of two squares. – Jyrki Lahtonen Mar 04 '15 at 05:13

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