The order and the structure of $R = \mathbb F_2[x]/(x^4 + x^2 +1)$

Question

Here is the idea I was told for the solution:

we will consider the cosets of the form $a + bx + cx^2 + dx^3 + J$ where $J = x^4 + x^2 +1$ because by adding a multiple of $(x^2 + x + 1)^2$ we can get rid of all the forth power terms (I want an example to see this please). Then $a + bx + cx^2 + dx^3$ can be written as $(a + bx) + (e + fx)(x^2 + x + 1)$ (I also do not know why is this correct, so if someone explain it to me I would appreciate it). Finally when is this element invertible in $R$ (this is also something that I need help in it please).

Answer 1

Question: "Here is the idea I was told for the solution:

we will consider the cosets of the form $a + bx + cx^2 + dx^3 + J$ where $J = x^4 + x^2 +1$ because by adding a multiple of $(x^2 + x + 1)^2$ we can get rid of all the forth power terms (I want an example to see this please). Then $a + bx + cx^2 + dx^3$ can be written as $(a + bx) + (e + fx)(x^2 + x + 1)$ (I also do not know why is this correct, so if someone explain it to me I would appreciate it). Finally when is this element invertible in $R$ (this is also something that I need help in it please)."

Answer: If $k:=\mathbb{F}_2$ and $A:=k[x]$. Let $f:=x^2+x+1 \in A$. It follows $\mathfrak{m}:=(f)$ is a maximal ideal (the polynomial $f$ is irreducible). Your ring $R:=k[x]/\mathfrak{m}^2$ is a local ring with maximal ideal $\mathfrak{n}:=\mathfrak{m}/\mathfrak{m}^2$ and there is an exact sequence

$$0 \rightarrow \mathfrak{m}/\mathfrak{m}^2 \rightarrow R \rightarrow k[x]/\mathfrak{m}:=K \rightarrow 0.$$

where $K$ is a field. Hence a unit in $R$ is an element $z$ not in the maximal ideal $\mathfrak{n}$. The ring $R$ has a basis $\{1,z,z^2,z^3\}$ where $z:=\overline{x}$ is the equivalence class of $x$ in $R$. The maximal ideal $\mathfrak{n}$ is the principal ideal generated by the element $t:=z^2+z+1$. Hence any element $s$ in $R$ may be written as

$$(1)s:= (a_1+a_2z)+ (a_3+a_4z)(z^2+z+1)$$

with $a_i \in k$. Hence a unit in $R$ is an element $w:=a_1+a_2z$ with $a_i \in k$ and $a_1\neq 0$.

You may prove $(1)$ as follows: Let $s:= a+bz+cz^2+dz^3$ with $a,b,c,d\in k$. It follows

$$ s=a+bz + (c+dz)(z^2+z+1-(z+1)) =$$

$$a+bz + (c+dz)(z^2+z+1)-(c+dz)(z+1).$$

It follows

$$ -(c+dz)(z+1)=d-c-cz-d(z^2+z+1)$$

and hence

$$s=a-c+d+(b-c)z +(c-d+dz)(z^2+z+1).$$

Hence any element $s\in R$ may be written on this form. The element $s$ is not in the maximal ideal iff $s=a_1+a_2z$ with $a_i\in k$ and $(a_1,a_2) \neq (0,0)$.

Note: If $(R, \mathfrak{n})$ is a local ring and $z \in R-\mathfrak{n}$ is any element, it follows $z$ is a unit: Else $(z)+\mathfrak{n}$ is a strict ideal containing $\mathfrak{n}$.

Answer 2

First notice that in characteristic $2$, $$ (x^2 + x + 1)^2 = x^4 + x^2 + 1. $$ Thus the ideal generated by this quadratic polynomial, $I = \langle x^2 + x + 1 \rangle$, is related to the given ideal $J = \langle x^4 + x^2 + 1 \rangle$ by $$ I^2 = J. $$

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Why can we reduce any terms of degree $4$ or greater in $R$? Since $J = 0 + J$ is the additive identity, all calculations are modulo $J$: $$ x^4 + x^2 + 1 \equiv 0 \pmod J, $$ so $$ x^4 \equiv -x^2 - 1 \equiv x^2 + 1 \pmod J. $$

This reduction identity in $R$ allows you to work with only polynomials of degree less than $4$.

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In order to prove that every cubic polynomial can be expressed in the given form, multiply out the expression on the right and equate coefficients, starting from the top. (You will find that the coefficients of degrees $0$ and $1$ are not the same $a$ and $b$ on the right.)

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The invertibility question comes down finding zero-divisors in $R$ and avoiding them. The calculation at the start should give you a hint.