The following example showed up in my textbook, and I am struggling to understand a certain part of it.

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Theorem 17.5 simply states that if $F$ is a field and $f(x)$ exists in $F[x]$, and is of degree 2 or 3, then $f(x)$ is reducible if and only if $f(x)$ has a zero in $F$. The corollary then says $F[x]/<f(x)>$ is a field.

The part I do not follow is:

$Z_2[x]/<x^3 + x + 1>$ = { $ax^2 + bx + c + <x^3 + x + 1> | a,b,c \epsilon Z_2$ }

I understand that $x^3 + x + 1$ is irreducible over $Z_2[x]$ and therefore it is a field(directly from the theorem and corollary), and I understand that given the equality, the field has 8 elements, but I do not understand the equailty itself.

I am still getting a grasp of quotient rings and polynomial rings so I apologize in advance if I don't understand something that is rather obvious.

Thanks for any help.

edit: Perhaps I do not fully understand the reasoning behind picking a cubic polynomial either. But I think it's because if we picked a quadratic polynomial, then our quotient ring wouldn't contain(all?) quadratic polynomials, and our field would be too small, is that correct?

  • 1
    $\begingroup$ The remainder of a ploynomial in $\Bbb Z_2[x]$ when divided by $x^3 + x + 1$ will necessarily be a quadratic polynomial, or of lesser degree, and since we have just two choices for coefficients, and just three possible coefficients, this enumerates all $8$ possible cosets. $\endgroup$ – David Wheeler Nov 9 '16 at 23:35

$\mathbb Z_2[x]/\langle x^3+x+1 \rangle$ has, as its underlying set of elements, the set of cosets of the ideal $\langle x^3+x+1\rangle$.

Such a coset can always be written in the form $$ a_0+a_1x+\cdots +a_nx^n + \langle x^3+x+1\rangle $$ (where the bit on the left is an arbitrary element of $\mathbb Z_2[x]$). By polynomial long division, we may write $$ a_0+a_1x+\cdots+a_nx^n = q(x)(x^3+x+1)+r(x) $$ where $q(x)$ is some polynomial and $r(x)=ax^2+bx+c$ is a polynomial of degree less than $3$. But now we have $$ q(x)(x^3+x+1)\in \langle x^3+x+1\rangle $$ and so \begin{align} a_0+a_1x+\cdots +a_nx^n + \langle x^3+x+1\rangle&=q(x)(x^3+x+1)+r(x)+\langle x^3+x+1\rangle\\&=r(x)+\langle x^3+x+1\rangle\\&=ax^2+bx+c+\langle x^3+x+1\rangle \end{align} which is of the desired form.

  • $\begingroup$ Ok, so if I were to find an irreducible polynomial of degree 4, I would find myself a field of degree 16? And so on into higher degrees? $\endgroup$ – SenoritaChad Nov 9 '16 at 23:43
  • $\begingroup$ @SenoritaChad That is exactly right - though I would normally use the word order rather than degree to denote the number of elements of a field. $\endgroup$ – John Gowers Nov 9 '16 at 23:49
  • $\begingroup$ Woops, got my words mixed up, I usually do I promise! Another similar question to make sure I understand, if I took $Z_3[x]$ as my field and found an irreducible polynomial of degree 2, then the field I constructed would have 9 elements? $\endgroup$ – SenoritaChad Nov 9 '16 at 23:54
  • $\begingroup$ @SenoritaChad It would indeed have $9$ elements. $\endgroup$ – John Gowers Nov 10 '16 at 6:25

$\mathbb{F}:=\mathbb{Z}_2[X]/\langle X^3+X+1\rangle$ is a vector-space over $\mathbb{Z}_2$. Using euclidean divisions by $X^3+X+1$, it has dimension $3$, a basis being $\{1+\langle X^3+X+1\rangle,X+\langle X^3+X+1\rangle,X^2+\langle X^3+X+1\rangle\}$. Regarding its cardinality, one has $\mathbb{F}\cong{\mathbb{Z}_2}^3$ as vector-spaces, which implies $\#\mathbb{F}=2^3=8$.

  • $\begingroup$ I apologize for having a lack of understanding of the content in this answer. But you've yielded the basis the same way the other answer by Donkey has shown? Which is taking remainders when divided by $x^3 + x +1$ $\endgroup$ – SenoritaChad Nov 9 '16 at 23:41
  • $\begingroup$ No apologies needed! Regarding your question : yes, Donkey_2009 and I are exactly saying the same things. $\endgroup$ – C. Falcon Nov 9 '16 at 23:44

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