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Background: I'm looking at old exams in abstract algebra. The factor ring described was described in one question and I'd like to understand it better.

Question: Let $F = \mathbb{Z}_2[x]/(x^4+x+1)$. As the polynomial $x^4+x+1$ is irreducible over $\mathbb{Z}_2$, we know that $F$ is a field. But what does it look like? By that I am asking if there exists some isomorphism from $F$ into a well-known field (or where it is straightforward to represent the elements) and about the order of $F$.

In addition: is there something we can in general say about the order of fields of the type $\mathbb{Z}_2[x]/p(x)$ (with $p(x)$ being irreducible in $\mathbb{Z}_2[x]$)?

Sid
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  • All such fields are vector spaces over the field $\mathbb Z_2$. Can you prove they are finite-dimensional? – Thomas Andrews Aug 25 '14 at 23:15
  • @ThomasAndrews No, not without a hint at least, I'm not knowledgeable enough! But I have a suspicion that they are always of order $2^n$ (as the order of finite fields are always of prime power) with $n$ being somehow tied to the degree of $p(x)$. – Sid Aug 25 '14 at 23:21
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    Then where did you come across this question? Are you taking an algebra course? Do you know what $\mathbb Z_2[x]/p(x)$ means? Do you know the division algorithm works in $F[x]$ when $F$ is a field? – Thomas Andrews Aug 25 '14 at 23:24
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    @ThomasAndrews I have taken algebra courses so I hope I don't come across as an idiot. But I can be slow so if you want to lead me to the solution to the problem pedagogically in the comment section, you need to bear that with me. I am aware of what $\mathbb{Z}_2[x]/p(x)$ means, with $\mathbb{Z}_2[x]$ being the ring of all polynomials with coefficients $1$ or $0$ etc, and I considered (p(x)) to be the ideal generated by the polynomial p(x). Then we get the factor ring in question. Is this correct? Please tell me how we can use the division algorithm in $F[x]$. – Sid Aug 25 '14 at 23:36
  • Sid, here you see elements of this field (and the smaller fields of 4 and 8 elements) listed and some arithmetic of these fields worked on. – Jyrki Lahtonen Aug 26 '14 at 05:19

3 Answers3

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The elements of $F$ are $\{ f(x) + (x^4 + x + 1) \mid f(x) \in \mathbb{Z}_2[x], \deg f < 4 \}$. There are $2^4$ of them. Any field of order $2^4$ is isomorphic to $F$.

In general, if $p(x) \in \mathbb{Z}_2[x]$ is irreducible of degree $k$, then $\mathbb{Z}_2[x]/(p(x))$ is a field of order $2^k$.

There is a notation that makes this field more convenient to work with. Let $\alpha = x + (x^4 + x + 1) \in F$. Then for $f(x) \in \mathbb{Z}_2[x]$, $f(\alpha) = f(x) + (x^4 + x + 1)$. So, for example, we can write the element $x^2 + 1 + (x^4 + x + 1)$ as $\alpha^2 + 1$. In this notation,

$$F = \{ f(\alpha) \mid f(x) \in \mathbb{Z}_2[x], \deg f < 4 \}.$$

An isomorphic field is the nimber field of nimbers less than 16. The representation of the elements is simpler, but I'm finding nim-multiplication to be harder than polynomial multiplication (maybe there's a trick to it that I don't know).

Sid
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Snowball
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  • Thanks Snowball, this looks very clear to me now. So I suppose one could generalize further and say that $\mathbb{Z}_n[x]/(p(x))$ is a field of order $n^k$ if $p$ is an irreducible of degree $k$, by the same argument? – Sid Aug 25 '14 at 23:51
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    @Sid: That generalization holds only if $\mathbb{Z}_n$ is a field, i.e., if $n$ is a prime number. – Snowball Aug 26 '14 at 00:01
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    Just to add a bit, your field also contains a field $\mathbb F_4$, whose elements are the three cube roots of unity, together with zero. And that is the only subfield between your field and the prime field $\mathbb F_2$. – Lubin Aug 26 '14 at 00:59
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In this ring, $x^4+x+1 = 0$, or equivalently $x^4=x+1$. So $F=\{ax^3+bx^2+cx+d \mid a,b,c,d\in\mathbb{Z}_2\}$. Any higher powers can be reduced using $x^4=x+1$. Note that this means that this field is isomorphic to the finite field of 16 elements.

paw88789
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You were asking for other fields that are isomorphic to $F=\Bbb{Z}_2[x]/\langle x^4+x+1\rangle=\Bbb{Z}_2[\alpha]$ with $\alpha=x+\langle x^4+x+1\rangle$. I proffer one such. Unfortunately it doesn't easily generalize to all finite fields. Thus it is a bit ad hoc.

Consider the ring $R=\Bbb{Z}[\zeta]$, where $\zeta=e^{2\pi i/5}$ is a primitive fifth root of unity. In $\Bbb{Z}_2[x]$ we have $$ x^{12}+x^9+x^6+x^3+1=(x^4+x+1)(x^8+x^4+x^2+x+1). $$ Therefore the element $\beta=\alpha^3\in F$ satisfies the equation $\beta^4+\beta^3+\beta^2+\beta+1$. As this coincides with the minimal polynomial of $\zeta$ there exists a homomorphism of rings $f:R\to F$ determined by $f(\zeta)=\beta$.

This homomorphism is surjective. One way of seeing that follows from the calculation $$ \begin{aligned} \alpha&=\alpha\cdot(1+0)^2=\alpha\cdot(1+1+\alpha+\alpha^4)^2\\ &=\alpha\cdot(\alpha+\alpha^4)^2=\alpha\cdot(\alpha^2+\alpha^8)\\ &=\alpha^3+\alpha^9=\beta+\beta^3. \end{aligned} $$ What about $\ker f$? Clearly $2R\subseteq \ker f$. Because $R$ is a free abelian group of rank four, we know that $|R/2R|=2^4=16$. This forces $2R=\ker f$.

Conclusion: $$ F\cong R/2R=\Bbb{Z}[\zeta]/\langle 2\rangle. $$

Jyrki Lahtonen
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