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What is the invertible elements of $R$ where $R= \mathbb Z[x] / (1+x+x^2)$?

The quotient basically means that we are taking all polynomials in $\mathbb Z[x]$ under the modulo $1+x+x^2$. That is, the terms that are multiple of $(1+x+x^2)$ will be zero i.e. each element that are in the same class with $(1+x+x^2)$ is going to be $0$.

In this example, I managed to write structure of the quotient ring $$\mathbb Z[x] / (4,2x,x^2)=\{ \overline{a_0+a_1x+a_2x^2}:a_0=0,1,2,3;a_1=0,1 \}$$

However, I couldn't even write the structure of $\mathbb Z[x] / (1+x+x^2)$ as above. And how can I find $(\mathbb Z[x] / (1+x+x^2))^*?$

Bill Dubuque
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Fuat Ray
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2 Answers2

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One can show that the this ring is isomorphic to $\mathbb{Z}[\omega]$, where $\omega=exp(2\pi i/3)$ is a primitive 3rd root of unity. The inverse in $\mathbb{Q}[\omega]$ is given by the Galois conjugate divided by the field norm $N(a+b\omega) = (a+b\omega)(a+b\omega^2) = a^2 +b^2 -ab$. This element is invertible in $\mathbb{Z}[\omega]$ if and only if $N(a+b\omega)=\pm{1}$. To summarize, the invertible elements of $\mathbb{Z}[x]/(1+x+x^2)$ are precisely $\{a+bx:a^2 +b^2 -ab =\pm{1}\}$.

Edit: Thanks Yathi for the correction in the dimension of the quotient.

Bennie
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The only invertible elements are:

$$\{\pm1, \, \pm x, \, \pm(1+x)\}.$$

$\textbf{Proof:}$

Let $I =\langle x^2+x+1 \rangle$. Any element of $R$ can be written as $ax+b+I$, where $a$ and $b$ are integers (because one can apply division algorithm to division by the monic polynomial $x^2+x+1$).

Now suppose $ax+b+I$ is a unit in $R$. Then for some $cx+d+I$ in $R$ we would have $$(ax+b)(cx+d)+I =1+I.$$ This implies that $$(ax+b)(cx+d)-1\in I.$$ That is $x^2+x+1$ divides $acx^2+(ad+bc)x+bd$. That is $$acx^2+(ad+bc)x+bd-1 = (x^2+x+1)f(x),$$ where $f(x)$ must be a non zero constant. Say $f(x)=k$.

Now, we have $$ ac=k; ad+bc=k; bd-1=k.$$

Suppose $a=0$. Then $k=0$ and $bd=1$. Thus both $b$ and $d$ are non zeroes. Also, $bc=0$ implying that $c=0$. Thus, we have $b=d=\pm1$. Clearly, in this case $ax+b+I=cx+d+I=\pm1+I$ and $\pm1+I$ are both units.

Suppose $b=0$. Then, $bd-1=k$ implies $k=-1$. Moreover $ad=-1$ and $ac=-1$. These two imply that, $ad-ac=0$. As $a$ cannot be zero, we must have that $c=d$. Also, as $ac=-1$, the only integers which satisfy these are $a=1, c=-1$ or $a=-1, c=1$. Thsu the units that we get in this situation are:

  1. Taking $a=1$ and $c=d=-1$ Units are: $x, -1-x$.
  2. Taking $a=-1$ and $c=d=1$ units are: $-x, 1+x$.

Now, suppose both $a$ and $b$ are non zero. Then from the relations we have listed down, we can arrive at

$$(k+1)\frac{a}{b}+k\frac{b}{a}=k.$$ Suppose we take $p=\frac{a}{b}$, then

$$(k+1)p+\frac{k}{p}=k.$$

This implies

$$(k+1)p^2-kp+k=0.$$

Solving for $p$, which is rational, we see that $$p=\frac{k\pm\sqrt{k^2-4(k+1)k}}{2(k+1)}$$

For $p$ to be rational, we must have that the term inside square a perfect square. That is $$-(3k^2+4k).$$ We need this to be positive. Hence $3k^2+4k$ must be negative or zero. Which happens only for $k=0$ or $k=-1$.

Now if $k=0$ or $k=-1$ we repeat the same argument done above to come to conclusion that the only invertible elements are $$\{\pm 1, \pm x, \pm(1+x)\}.$$

Yathi
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