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Find the splitting field of polynomial $f(X)=X^{15}-2\in \mathbb{F}_ 7[X]$.

Splitting field of this polynomial should be of the form $\mathbb{F}_{7^n}$ for some suitable $n$.

For example, I know how to solve the much easier problem if we consider the polynomial $g(X)=X^{15}-1\in \mathbb{F}_7[X]$. In this case we have to take minimal $n$ such that $15 \mid7^n-1$.

Can anyone give the detailed answer how to do this in the case $X^{15}-2$?

I have seen the similar post in MSE but all of them is very bried and without details.

Would be very grateful if anyone can give a detailed answer!

Torsten Schoeneberg
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RFZ
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1 Answers1

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Note that $15=3\times 5$ and that $5$ does not divide $p-1=6$. Because $4^5\equiv2\pmod{7}$ we see that $$X^{15}-2=(X^3)^5-4^5=(X^3-4)(X^{12}+4X^9+2X^6+X^3+4).$$ Hence the splitting field is the compositum of the splitting field of these two factors.

The factor $X^3-4$ has no roots in $\Bbb{F}_7$, so it is irreducible and its splitting field is isomorphic to $\Bbb{F}_{7^3}$.

The splitting field of $X^{12}+4X^9+2X^6+X^3+4$ is obtained by adjoining cube roots of the roots of $$P:=X^4+4X^3+2X^2+X+4,$$ to $\Bbb{F}_7$. This polynomial is irreducible because $$2P(4X)=X^4+X^3+X^2+X+1=\Phi_5,$$ which is the fifth cyclotomic polynomial, and $\Bbb{F}_7$ contains no primitive fifth roots of unity. So the splitting field of $P$ is isomorphic to $\Bbb{F}_{7^4}$. Hence the splitting field of the second factor is either $\Bbb{F}_{7^4}$ or $\Bbb{F}_{7^{12}}$, either way the splitting field of the original polynomial is $\Bbb{F}_{7^{12}}$.

Servaes
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  • It's obvious that $X^3-4$ is irreducible over $\mathbb{F}7$. But I have some difficulties proving that it's isomorphic to $\mathbb{F}{7^3}$. I can show that the field $\mathbb{F}_7[X]/\langle X^4+X+1\rangle$ is the field with $7^3$ elements and contains all the roots of $X^3-4$. But why it is the smallest field? Could you explain this point please in details? – RFZ Mar 01 '19 at 18:03
  • Your quotient field $\Bbb{F}_7[X]/(X^4+X+1)$ has $7^4$ elements and contains no roots of $X^3-4$. In general, for a finite commutative ring $R$ and a monic polynomial $f\in R[x]$ of degree $n$, the ring $R[X]/(f)$ has precisely $|R|^n$ elements; loosely speaking there are $|R|$ choices for each of the $n$ coefficients of a representative. – Servaes Mar 01 '19 at 18:05
  • Furthermore, a finite field of order $p^n$ contains a subfield of order $p^m$ if and only if $m\mid n$. So in particular, the only proper subfield of $\Bbb{F}_{7^3}$ is $\Bbb{F}_7$. – Servaes Mar 01 '19 at 18:08
  • Sorry I meant $\mathbb{F}_7[X]/(X^3-4)$! – RFZ Mar 01 '19 at 18:09
  • But could you show this in general case: namely if we consider irreducible polynomial $f(X)\in \mathbb{F}_p[X]$ then $K:=\mathbb{F}_p[X]/(f(X))$ is the splitting field for $f(X)$? For example, it is not obvious to me that $K$ contains ALL the roots of $f(X)$ and is the smallest field with this property. Because I was trying to find in MSE but no results. – RFZ Mar 01 '19 at 18:12
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    This is a standard result; if $f\in\Bbb{F}_p[X]$ is irreducible of degree $n$ then each root of $f$ in an algebraic closure generates a subfield of order $p^n$. Any field contains at most one field of order $p^n$, so all roots are contained in this one field. – Servaes Mar 01 '19 at 18:16
  • +1 Another route to the same destination is to observe that $2$ is a primitive third root of unity. Hence the zeros of $x^{15}-2$ are roots of unity of orders either $9$ or $45$. The former are roots of the cubic factor. The latter belong to $\Bbb{F}_{7^{12}}$ because $7$ has order $12$ modulo $45$. But, I've posted the above drill enough many times on this site already. It is refreshing to see a different take! – Jyrki Lahtonen Mar 01 '19 at 22:37
  • Could you clarify this moment please? "The splitting field of $X^{12}+4X^9+2X^6+X^3+4$ is obtained by adjoining cube roots of the roots of $$P:=X^4+4X^3+2X^2+X+4,$$ to $\Bbb{F}_7$." – RFZ Mar 01 '19 at 23:13
  • could you correct the last comment, please? – RFZ Mar 02 '19 at 17:32
  • If $\alpha$ is a root of $X^{12}+4X^9+2X^6+X^3+4$ then $\alpha^3$ is a root of $X^4+4X^3+2X^2+X+4$, and conversely, if $\beta$ is a root of $X^4+4X^3+2X^2+X+4$ then any root of $X^3-\beta$ is a root of $X^{12}+4X^9+2X^6+X^3+4$. – Servaes Mar 02 '19 at 18:06