well,if $[F_{q}(\alpha):F_{q}]=m$, can i prove that $F_{q}(\alpha)=F_{q^{m}}$ ? or i can only prove that $F_{q}(\alpha)$ is isomorphic to $F_{q^{m}}$ ?
thanks for advance.
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2$\begingroup$ Isomorphic.${}$ $\endgroup$– Gerry MyersonMar 22, 2016 at 6:25
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$\begingroup$ Not just isomorphic. If you fix an algebraic closure, then it is unique in the algebraic closure. $\endgroup$– ClaudiusMar 22, 2016 at 7:19
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$=$ can be sort of slippery. You really want $\cong$. For example, $\mathbb F_5[x] / (x^2+2)$, and $\mathbb F_5[x] / (x^2+3)$ are both $F_{25}$, but they're not "equal", they are different (but isomorphic) quotients of the same ring.
