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Given $f(x)\in\Bbb F_5[x]$ with $\deg f(x)=11$ for example. Suppose that $f(x)$ is irreducible. We know that the splitting field of $f(x)$ over $\Bbb F_5$ must exist, and its degree is $\le11!$. Generally speaking, $\Bbb F_{5^3}$ may not be big enough to contain all the roots of $f(x)$, and so does, say $\Bbb F_{5^{11}}$. So my question is that, what is the "biggest" possible order of the finite field we need to contains all of its root? I can only bound it by $5^{11!}$, but I think it may be strictly smaller. Are there some theorems related to this?

Update: Let $K$ be the splitting field of $f(x)$ over $\Bbb F_5$. Then by a theorem of finite field, $K=\Bbb F_5(u)$. If we can choose such $u$ as a root of $f(x)$, then since the degree of the minimal polynomial of $u$ is less then $11$, $[K:\Bbb F_5]\leq 11$. Then it seems that $\Bbb F_{5^{11}}$ is quite enough?

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  • $\begingroup$ If $k=\Bbb F_q$ is a finite field, and $f$ an irreducible polynomial of degree $d$ over $k$, then its splitting field is $\Bbb F_{q^d}$. $\endgroup$ – Lord Shark the Unknown Apr 28 at 11:28
  • $\begingroup$ @LordSharktheUnknown So is the theoretic upper bound $5^{11!}$? (And there must exist such $f(x)$ for instance.) $\endgroup$ – Eric Apr 28 at 11:30
  • $\begingroup$ Yes, over every finite field there are irreducible polynomials of every degree. $\endgroup$ – Lord Shark the Unknown Apr 28 at 11:34
  • $\begingroup$ @LordSharktheUnknown Let $K$ be the splitting field of $f(x)$ over $\Bbb F_5$. Then $K=\Bbb F_5(u)$. If we can choose such $u$ as a root of $f(x)$, then since the degree of the minimal polynomial of $u$ is less then $11$, $[K:\Bbb F_5]\leq 11$. Then it seems that $\Bbb F_{5^{11}}$ is quite enough to contain all the roots of $f(x)$? $\endgroup$ – Eric Apr 28 at 12:26
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One knows that every extension of finite fields $\Bbb F_q\subset K$ is Galois with Galois group cyclic generated by the Frobenius morphism $x\mapsto x^q$.

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  • $\begingroup$ Make that $x^q$. $\endgroup$ – Lord Shark the Unknown Apr 28 at 11:28
  • $\begingroup$ @LordSharktheUnknown: Wow, that was quick! :) Corrected $\endgroup$ – Andrea Mori Apr 28 at 11:29

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