Let $\Phi_5$ be the 5th cyclotomic polynomial and $\Phi_7$ the 7th. These polynomials are defined like this: $$ \Phi_n(X) = \prod_{\zeta\in\mathbb{C}^\ast:\ \text{order}(\zeta)=n} (X-\zeta)\qquad\in\mathbb{Z}[X] $$ I want to calculate the splitting field of $\Phi_5$ and the splitting field of $\Phi_7$ over $\mathbb{F}_2$. In $\mathbb{F}_2[X]$ we have $$ \Phi_5(X) = X^4 + X^3 + X^2+X+1 $$ and $$ \Phi_7(X) = (X^3+X+1)(X^3+X^2+1) $$ My question is: what are the splitting fields of the polynomials? I already know it should be of the form $\mathbb{F}_{2^k}$ for some $k\in\mathbb N$. Also the degree of every irreducible factor of a cyclotomic polynomial in $\mathbb{F}_q[X]$ is equal to the order of $q\in(\mathbb{Z}/n\ \mathbb{Z})^\ast$, assuming $(q,n)=1$.
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$k$ is eaxctly the order of $q$ in $\Bbb{Z}_n^*$. I'm sure this has been done many times on this site. – Jyrki Lahtonen May 22 '14 at 07:27
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@JyrkiLahtonen I'm sorry, I searched on the website but I couldn't find the answer. – Jolien May 22 '14 at 07:30
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2See here, here or here, – Jyrki Lahtonen May 22 '14 at 07:37
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@Praphulla: I'm afraid I don't understand at all where you got the first equality of polynomials (which is false). – Jyrki Lahtonen May 22 '14 at 08:41
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@JyrkiLahtonen : In $\mathbb{F}_4$ we have $x^4=x$ so i thought i would replace $x^4$ by $x$.... :O It does sound like non sense to me.... I should not have written $x^2=x$ as i do not know if that $x$ is in $\mathbb{F}_2$ or not... On the whole, I agree it is a non sense... – May 22 '14 at 08:46
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@Praphulla: We have, indeed, $x^4=x$ for all $x\in\Bbb{F}_4$. But when dealing with polynomials $x$ is an indeterminate - not a variable ranging over $\Bbb{F}_4$. There are explanations of the difference between (formal) polynomials and polynomial functions on this site. – Jyrki Lahtonen May 22 '14 at 09:17
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You're being too hard on yourself on occasion, @Praphulla. We all have our blind spots. Having said that many of us are also in the habit of covering ourselves with ashes for not realizing something right away :-/ – Jyrki Lahtonen May 22 '14 at 10:33
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@JyrkiLahtonen : Thank you for those words... :) Thanks for your support.... – May 22 '14 at 10:42
2 Answers
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Since we want the degree of an irreducible factor to be equal to one, we want $$ \text{order} (2^k) =1 $$ in $(\mathbb{Z} / 5\mathbb{Z})^\ast$. The only element with this order is 1. Therefore we search the smallest $k$ such that $2^k\equiv 1\mod 5$. A bit puzzzling gives us $$ 2^1=2\\ 2^2=4\\ 2^3=8=3\\ 2^4=16=1. $$ Therefore the splitting field of $\Phi_5$ should be $\mathbb{F}_{2^4}$.
Is this correct?
Jolien
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Well, we already have a method to get the degree of irreducible factors and we want to split the polynomial completely in lineair factors. These factor have degree one. – Jolien May 22 '14 at 07:40
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To be more precise: You get the correct answer to the question about the degree of the splitting field. I don't understand the part about linear factors. Aren't these obviously the relevant $(x-\zeta)$:s with $\zeta$ of order $n$? The linear factors of an irreducible factor of $\Phi_n(x)$ correspond to $(x-\zeta^i)$ with $i$ ranging over a cyclotomic coset modulo $n$. See the linked questions for more details. – Jyrki Lahtonen May 22 '14 at 08:38
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Hint :
- If $f(X)$ is irreducible in $F[X]$ then $F[X]/(f(X))$ is a field.
- any polynomial $f(X)$ has root in $F[X]/(f(X))$ (which need not be a field in general)
- What is the cardinality of $\mathbb{F}_2[X]/(X^4 + X^3 + X^2+X+1)$.
- How many finite fields of cardinality $n$ can you list out for a given $n$.
- Splitting field of $f(X)g(X)$ is contains splitting field of $f(X)$
- Splitting field of $(X^3+X+1)(X^3+X^2+1)$ is contains splitting field of $(X^3+X^2+1)$
- As $(X^3+X^2+1)$ is irreducible in $\mathbb{F}_2[X]$ its splitting field would be (???)
- What is splitting field of $(X^3+X+1)$.
- Do you see some relation between splitting field of $(X^3+X+1)$ and of $(X^3+X^2+1)$.
Can you now conclude??