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$α ^2 + 1 ∈ E$ over $\Bbb F_2$, where E is the field $\Bbb F_2[x]/\langle x^3 + 3x + 1\rangle$ and $α ∈ E$ denotes the coset $\overline{x}$.

What I do know is that $[\Bbb F_2[x]/\langle x^3 + 3x + 1\rangle:\Bbb F_2]=3$ because the polynomial is irreducible over $\Bbb F_2$. But how do I go about to find the minimal polynomial of $\alpha^2+1?$ (I have not learnt Galois Theory yet.)

Misha Lavrov
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  • You must have a typo: $x^3 + 2x +1 = x^3 + 1$ over $\mathbb{F}_2$, which is not irreducible, since $1$ is a root. – Viktor Vaughn Mar 26 '17 at 04:44
  • Okay, but $3 = 1$ in $\mathbb{F}_2$, so we can write the above polynomial as $x^3 + x + 1$. – Viktor Vaughn Mar 26 '17 at 05:02
  • Obviously $\Bbb{F}_2[\alpha^2+1]=E$, so the minimal polynomial has degree three. There are only two irreducible polynomials of degree three over $\Bbb{F}_2$, namely $x^3+x+1$ and $x^3+x^2+1$. All you need to is to determined which of those has $\alpha^2+1$ as a zero. Mind you, I would still use the reciprocal polynomial (doesn't need Galois theory(. – Jyrki Lahtonen Mar 29 '17 at 18:46

1 Answers1

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All right, here's an easier way. Note that $$ 0 = \alpha^3 + \alpha + 1 = \alpha(\alpha^2 + 1) +1 \implies \alpha(\alpha^2 + 1) = 1 \implies \alpha^2 + 1 = 1/\alpha \, . $$ Dividing $\alpha^3 + \alpha + 1 = 0$ through by $\alpha^3$, we find that \begin{align*} 0 = \frac{\alpha^3 + \alpha + 1}{\alpha^3} = \left(\frac{1}{\alpha}\right)^3 + \left(\frac{1}{\alpha}\right)^2 + 1 \, . \end{align*}

Since the extension $\DeclareMathOperator{\Gal}{Gal} E/\mathbb{F}_2$ is Galois and $\mathbb{F}_2(\alpha^2+1) = E$ by degree considerations, then the minimal polynomial of $\alpha^2+1$ is $$ m(x) = \prod_{\tau \in \Gal(E/\mathbb{F}_2)} (x - \tau(\alpha^2 + 1)) \, . $$ That is, the other roots of the minimal polynomial are exactly the Galois conjugates of $\alpha^2 + 1$. Since $\Gal(E/\mathbb{F}_2)$ is cyclic and generated by the Frobenius automorphism \begin{align*} \sigma : E &\to E\\ \beta &\mapsto \beta^2 \end{align*} then $$ m(x) = (x - (\alpha^2 + 1))(x - \sigma(\alpha^2 + 1))(x - \sigma^2(\alpha^2 + 1)) \, . $$

Here's one last answer that's not as clever as the first, but not as inefficient as the second. Consider the $E$ as a vector space over $\mathbb{F}_2$ with basis $1, \alpha, \alpha^2$. Multiplication by $\alpha^2 + 1$ is an $\mathbb{F}_2$-linear map which has matrix $$ \begin{pmatrix} 1 & 1 & 0\\ 0 & 0 & 1\\ 1 & 0 & 0 \end{pmatrix} $$ with respect to the above basis. This matrix has characteristic (and in fact, minimal) polynomial $x^3 + x^2 + 1$. Since $\alpha^2 + 1$ is a root of this polynomial, and it is irreducible (it has no roots over $\mathbb{F}_2$ and has degree $3$), then it is the minimal polynomial.

Viktor Vaughn
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  • Thanks so much for your answer. But I haven't learnt Galois Theory yet. Is there a simpler explanation? –  Mar 26 '17 at 04:57