# Order of the splitting field of $f(x)$ with $\deg f(x)=11$ over $\Bbb F_p$?

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?

• If $k=\Bbb F_q$ is a finite field, and $f$ an irreducible polynomial of degree $d$ over $k$, then its splitting field is $\Bbb F_{q^d}$. – Lord Shark the Unknown Apr 28 at 11:28
• @LordSharktheUnknown So is the theoretic upper bound $5^{11!}$? (And there must exist such $f(x)$ for instance.) – Eric Apr 28 at 11:30
• Yes, over every finite field there are irreducible polynomials of every degree. – Lord Shark the Unknown Apr 28 at 11:34
• @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 $f(x)$? – Eric Apr 28 at 12:26

One knows that every extension of finite fields $$\Bbb F_q\subset K$$ is Galois with Galois group cyclic generated by the Frobenius morphism $$x\mapsto x^q$$.
• Make that $x^q$. – Lord Shark the Unknown Apr 28 at 11:28