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I'm trying to find the splitting field of $t^6+t^5+t^4+t^3+1$ over $\mathbb{F}_2[t]$.

My attempt: This polynomial factors into the irreducibles $(t^2+t+1)(t^4+t+1)$. The first one has roots which we may (suggestively) call $\omega, \omega^2$.

What I'm stuck at: Is $t^4+t+1$ irreducible over $\mathbb{F}_2[\omega]$? If yes, the degree of the splitting field is $8$; else, it's $4$.

Observations:

$1)$ It's easy enough to check that neither $\omega, \omega^2$ are roots of this polynomial.

$2)$ If $a$ is a root of $t^4+t+1$, then the other roots are $a^2, a+1, a^2+1$.

tychonovs-scholar
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A.D.
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  • It's not too much to multiply a generic monic quadratic with another one, where each has coefficients from $0,1,\omega,\omega+1$ and see if it's possible to get $t^4+t+1$. – 2'5 9'2 Dec 08 '24 at 23:12
  • Nitpick: you are looking for a splitting field over $\Bbb{F}_2$ :-) – Jyrki Lahtonen Dec 09 '24 at 05:44

1 Answers1

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The easy answer relies on the facts that there exists exactly one finite extension of $\mathbb F_2$ of each degree $d\in\mathbb N_+$, denoted $\mathbb F_{2^d}$, and that $\mathbb F_{2^n}\subseteq\mathbb F_{2^m}$ iff $n\mid m$. For proofs of these facts see wikipedia or any book on field theory. Now since $t^4+t+1$ is irreducible over $\mathbb F_2$, your splitting field contains an extension of degree $4$, and by the above criterion $\mathbb F_{16}$ contains $\mathbb F_4$, which is the splitting field of $t^2+t+1$. So the entire polynomial already splits over $\mathbb F_{16}$.

More generally, we see that if $f=f_1\cdots f_r$ is a factorisation of $f\in\mathbb F_p[t]$ in irreducibles, then the splitting field of $f$ has degree $\operatorname{lcm}(\deg f_1 ,\ldots,\deg f_r)$.

If you don't want to rely on these facts, you can explicitly factor $t^4+t+1$ over $\mathbb F_4$. You already noted that it has no roots, so we must have $$t^4+t+1=(t^2+at+b)(t^2+ct+d)$$ for some $a,b,c,d\in\mathbb F_4=\{0,1,\omega,\omega+1\}$. Multiplying out the right hand side and solving the system of equations yields $$t^4+t+1=(t^2+t+\omega)(t^2+t+\omega+1).$$

anankElpis
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  • Where can I find a proof of the fact that the splitting field has degree lcm(...)? – A.D. Dec 09 '24 at 01:47
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    That follows directly from the things I said above. Say $\deg f_i=n_i$, then the splitting field of $f_i$ is $\mathbb F_{p^{n_i}}$, The splitting field of $f$ is the smallest field containing all of them, so it is $\mathbb F_{p^n}$ for the smallest $n$ such that $n_i\mid n$ for all $n$. That's the definition of the $\operatorname{lcm}$. – anankElpis Dec 09 '24 at 03:16
  • Very well done (+1). The factorization in the end is not difficult to do this way. I happened to use it as an example for the use of Galois theory in this old answer. Only bring it up because that way generalizes to some other cases (after you have the discrete log table at hand). – Jyrki Lahtonen Dec 09 '24 at 05:50