# In $\mathbb{Z}_2[x]$, the ideal $\langle x^2+x+1\rangle$ is maximal

I want to show that in $\mathbb{Z}_2[x]$, the ideal $\langle x^2+x+1\rangle$ is maximal ideal.

I have listed out the table and I can see that the quotient ring is isomorphic to $\mathbb{Z}_2\oplus \mathbb{Z}_2$, but I can't find a ring homomorphism $f:\mathbb{Z}_2[x]\rightarrow \mathbb{Z}_2\oplus \mathbb{Z}_2$ such that the kernel of this homomorphism is exactly the ideal $\langle x^2+x+1\rangle$.

Also, can anyone give me any general suggestion of finding such homomorphisms?

• Joey: is this modified question exactly the question in your mind? – Groups Sep 30 '15 at 5:44
• yes this is exactly the question im asking – Mark Sep 30 '15 at 5:45
• something is misunderstanding here: since the ideal is maximal, quotient would be a field of order $4$; but $\mathbb{Z}_2\oplus \mathbb{Z}_2$ is not a field with usual point-wise multiplication. – Groups Sep 30 '15 at 5:48
• Something is wrong here. Since $(1)^2+1+1=1$ and $0^2+0+1=1$, this polynomial is irreduicible. Thus the ideal is prime, and so the quotient is a domain, but since finite domains are fields, so is the quotient, but $\mathbb{Z}/2^2$ is not, so something about your isomorphism is wrong. – Pax Kivimae Sep 30 '15 at 6:07
• So can we get a homomorphism from ℤ2[x] to some ring such that the kernel is exactly ⟨x2+x+1⟩, I try to think about such a homomorphism, but I can't – Mark Sep 30 '15 at 14:42

This ideal is maximal since any element $ax+b+ < x^2+x+1>$ has an inverse. To find the inverse see that $(ax+b)(cx+d)=(acx^2+(ad+bc)x+bd=(ad+bc+1)x+(bd+1)$ modulo $<x^2+x+1>$. So if $a,b$ both are not zero we can solve for $c,d$ from the equations: $ad+bc+1=0, bd+1=1$.
You got the quotient ring wrong: for example $x^2= x+1$ modulo $x^2+x+1$ but all the elements in $\mathbb{Z}_2\times \mathbb{Z}_2$ are idempotents.
It is a maximal ideal because $x^2+x+1$, a quadratic polynomial with no root in $\mathbf Z_2$, is irreducible in $\mathbf Z_2[x]$ and in PIDs irreducible elements generate maximal ideals.