Consider the field $F_2[α] = F_2[x]/(x^3 + x + 1)$, where $α$ is the image of $x$ in $F_2[x]/(x^3 + x + 1)$. Write down the $8$ elements of this field in terms of $α$.
I have no clue how to start this question. Thanks!
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1Well, you have $c_0+c_1\alpha+c_2\alpha^2$ where $c_0,c_1,c_2\in F_2$ – J. W. Tanner Apr 08 '19 at 21:18
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@JeanMarie: did you mean $x^3+\pmb x=x+1+\pmb x=1$? – J. W. Tanner Apr 08 '19 at 21:28
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Thanks! @J.W.Tanner – RandomThinker Apr 08 '19 at 21:30
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J. W. Tanner Yes, thank you. I fix it in a new comment – Jean Marie Apr 08 '19 at 21:30
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Thanks! @JeanMarie I get it now! – RandomThinker Apr 08 '19 at 21:30
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1Your elements are the $ax^2+bx+c$ (for $a,b,c∈{0,1}$ : this gives indeed $8$ elements), and the main rule is $x^3+x+1=0⟺x^3=x+1$ . Thus, for example $x∗(x^2+1)=x^3+x=x+1+x=1$. Is this hint sufficient ? – Jean Marie Apr 08 '19 at 21:32
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Hello @JeanMarie I have one more question. Why ^3++1=0? – RandomThinker Apr 08 '19 at 21:36
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1This is the philosophy of quotienting... Take the comparison with $\mathbb{Z}/(3\mathbb{Z})$ which is the field with $3$ elements, still called $0+3\mathbb{Z}, 1+3\mathbb{Z}, 2+3\mathbb{Z}$, what happens when you multiply $(2+3\mathbb{Z})\times(2+3\mathbb{Z})=4+3\mathbb{Z}=1+3\mathbb{Z}$ because you have broken $4=1+3$ and have incorporated the $3$ term into $3\mathbb{Z}$. But in fact, one does not use these heavy notations, one computes directly 4=1 "because $3=0$" (with quotes) : ... (ctd) – Jean Marie Apr 08 '19 at 21:53
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1... (ctd) you see : what you place in the denominator of the quotient structure is considered as zero (more exactly the "zero class"). – Jean Marie Apr 08 '19 at 21:53
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If you want to do a lot of arithmetic with this field, you may want to have a discrete logarithm table at hand. – Jyrki Lahtonen Apr 10 '19 at 07:41
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The field $F_2$ has two elements, $0$ and $1$.
Elements of $F_2[\alpha]$ can be expressed as $c_0 + c_1\alpha + c_2\alpha^2,$ with $c_0, c_1,$ and $ c_2\in F_2$.
Higher powers of $\alpha$ are not needed because $\alpha^3= \alpha+1$
(by construction, $\alpha^3+\alpha+1= 0$ so $\alpha^3= -\alpha-1$, and $-1=1$ in $F_2,$ so $\alpha^3= \alpha+1$).
There are $2$ possibilities for each of $c_0, c_1,$ and $c_2$, so $8=2^3$ possibilities altogether:
$0, 1, \alpha, 1+\alpha, \alpha^2, 1+\alpha^2, \alpha+\alpha^2,$ and $1+\alpha+\alpha^2$.
J. W. Tanner
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This is a form of graphical representation complementing the solution by @J. W. Tanner : here are (Fig. 1) the addition table and (Fig. 2) the multiplication table of this field.
Fig. 1 : Addition table. Please note the diagonal of zeros (deep blue) due to the fact that $w+w=0$, whatever $w$... remark as well the partition into blocks $4 \times 4$ : why that ?
Fig. 2 : Multiplication table. Please note the natural bordering by zeros......
How did I obtain them ? Without entering into the details, I will just say that instead of working with polynomials having their coefficients in $\{0,1\}$, I have been using a $3 \times 3$ matrix $M$ such that $M^3+M+1=0 \ \color{'red'}{modulo 2}$. It is easy to find such a matrix (the so-called companion matrix of polynomial $x^3+x+1$). Here it is :
$$M=\begin{pmatrix}0&1&1\\1&0&0\\0&1&0\end{pmatrix}$$
The elements of the field are thus all the $aI_3+bM+cM^2$, $a,b,c \in \{0,1\}$.
Jean Marie
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