Ideals of $\Bbb{Z}_4[X]/(X^2+X+1)$ I want to find (maximal) ideals of $\Bbb{Q}[X]/(X^2)$ and $\Bbb{Z}_4[X]/(X^2+X+1)$. 
for the 1st one I can do this, because $(x)$ is a maximal and $(x)^2=0$.

Now, I want to know how ideals (specially maximal ideals) of $\Bbb{Z}_4[X]/(X^2+X+1)$ looks like? there are many difficulties: $\Bbb{Z}$ is not field, I cant decompose $X^2+X+1$ ...
thnx for help
 A: For the first one, you are right that $(x)$ is a maximal ideal of $R=\mathbb{Q}[X]/(X^2)$, where $x$ is the image of $X$ under the canonical projection $\mathbb{Q}[X]\to R$.  But it is also the only ideal of $R$ that is proper and nonzero.  In other words, all ideals of $R$ are $\{0\}$, $(x)$, and $R$.
To show this, suppose $I$ is a ideal of $R$ such that $I\neq \{0\}$ and $I\neq R$.  Recall that every element of $R$ can be written as $ax+b$ for some $a,b\in\mathbb{Q}$.  If $I$ contains $ax+b$ for some $a,b\in\mathbb{Q}$ with $(a,b)\neq (0,0)$, then $a\neq 0$ (otherwise, $I$ contains $1$ and $I=R$).  That is, we can assume without loss of generality that $a=1$.  
Now, if $I$ contains $x+b$, then it contains $$(x-b)(x+b)=x^2-b^2=-b^2.$$  If $b\neq 0$, then $I$ contains $1$, and $I=R$ is a contradiction.  Hence, $b=0$, and so every nonzero element of $I$ is of the form $kx$ where $k\in\mathbb{Q}$ is such that $k\neq 0$.  That is, $I=(x)$.  (Note that $R/(x)$ is indeed a field, which is isomorphic to $\mathbb{Q}$.)

Now, let $S$ be the ring $\mathbb{Z}_4[X]/(X^2+X+1)$.  (I assume that $\mathbb{Z}_4$ is the ring of integers modulo $4$.)   Denote by $x$ the image of $X$ under the canonical projection $\mathbb{Z}_4[X]\to S$.  Any element of $S$ can be written as $ax+b$ such that $a,b\in\mathbb{Z}_4$.  Let $I$ be an ideal of $S$ such that $I\neq \{0\}$ and $I\neq S$.
If $I$ contain a nonzero constant, then $I=(2)$ (otherwise $1\in I$, and so $I=S$).  If $I$ contains an element $ax+b$ with $a=1$ or $a=3$, then we can assume that $a=1$ since $3=-1$ is invertible in $\mathbb{Z}_4$.  Now, $$(x^2+x+1)-(b^2-b+1)=x^2+x+b(1-b)=(x+1-b)(x+b)\in I.$$
Because $x^2+x+1=0$, we get $b^2-b+1\in I$.  But $b^2-b+1$ is an odd element of $\mathbb{Z}_4$, so it is invertible.  This means $I$ contains $1$, and $I=S$, a contradiction.
Hence, if $I$ contains $ax+b$ with $a\neq 0$, then $a=2$.  Now, suppose that $2x+1\in I$.  Then,
$$2(x^2+x+1)-(x+2)=2x^2+x=x(2x+1)\in I.$$
So, $x+2\in I$, and $1=(2x+1)-2(x+2)\in I$.  This is again a contradiction, as it leads to $I=S$.  So, $2x+1\notin I$.  Similarly, $2x-1$ cannot be in $I$.  
We have now proven that $I$ contains either $2x$ or $2x+2$.  If $2x\in I$, then
$$2x+2=2(x^2+x+1)+(2x+2)=2x^2=x(2x)\in I\,.$$
Similarly, if $2x+2\in I$, then $2x\in I$.  If $2x$ and $2x+2$ are in $I$, then $2=(2x+2)-(2x)\in I$.  That is, $I=\{0,2,2x,2x+2\}=(2)$.
In summary, $S$ has only three ideals: $\{0\}$, $(2)$, and $S$.  The maximal ideal of $S$ is of course $(2)$.  (Note that $S/(2)$ is indeed a field, which is isomorphic to $\mathbb{F}_4=\mathbb{F}_2[X]/(X^2+X+1)$.)
