How does the quotient ring $\Bbb Z[x]/(x^2-x,4x+2)$ look like?

How does the quotient ring $$\Bbb Z[x]/(x^2-x,4x+2)$$ look like?

Normally to solve this you play around with generators until you get something you can work with. I was unsuccessful in reducing it to something neat.

I have proven that $$6$$ is the smallest integer in the ideal. Thus we get $$\Bbb Z[x]/(6,x^2-x,4x+2)$$ I dont see much further simplification. I would want to get rid of $$x^2$$ but i do not see how.

• Define precisely "look like." Commented Jul 24, 2020 at 20:01
• Let $R$ be your ring and $e$ be the image of $x$ inside it, so that $e^2=e$. Then you can show that $R$ is isomorphic to $R/eR \times R/(1-e)R$. This should get you your isomorphism. Commented Jul 24, 2020 at 20:02

We have $$8(x^2-x)-2x(4x+2)=-12x=-3(4x+2)+6$$, so $$6\in (x^2-x,4x+2)$$ ,and $$(x^2-x,4x+2)=(6,x^2-x,4x+2)$$.

Hence, your ring $$R$$ is isomorphic to $$\mathbb{Z}/6 [x]/(x^2-x,-\bar{2}x+\bar{2})$$.

The map given by the Chinese remainder theorem yields an isomorphism $$R\simeq \mathbb{F}_2[x]/(x^2-x)\times\mathbb{F}_3[x]/(x^2-x,x-\bar{1})=\mathbb{F}_2[x]/(x^2-x)\times\mathbb{F}_3[x]/(x-\bar{1})$$, which finally yields $$R\simeq\mathbb{F}_2\times\mathbb{F}_2\times\mathbb{F}_3.$$

If we follow carefully the proof, we get an explicit isomorphism.

Let $$f$$ be the ring morphism $$f:R\to P\in\mathbb{Z}[x]\mapsto ([P(0)]_2,[P(1)]_2, [P(1)]_3)\in\mathbb{F}_2\times\mathbb{F}_2\times \mathbb{F}_3$$.

It is easy to see $$(x^2-x,4x+2)$$ lies in the kernel of $$f$$. The induced map is then the desired isomorphism.

We can also prove this last fact directly, just to double check that everything is correct.

Claim. $$f$$ is surjective.

Given $$a,b,c\in\mathbb{Z}$$, we need to find $$P\in\mathbb{Z}[x]$$ such that $$P(0)\equiv a \ [2], P(1)\equiv b \ [2]$$ and $$P(1) \equiv c \ [3]$$.

We can try $$P=ux+v$$. We can choose $$v=a$$, $$u=(b-a)+2k$$, so the two first equations are satisfied. Now we want $$(b-a)+2k+a\equiv c \ [3]$$, and we take $$k=b-c$$. To sum up $$P=(3b-a-2c)x+a$$ does the job.

Claim. $$\ker(f)=(x^2-x,4x-2)$$.

As we said before, one inclusion is clear, so let $$P\in\mathbb{Z}[x]$$ such that $$f(P)$$ is trivial. We want to prove that $$P\in (x^2-x,4x+2)$$. Dividing by $$x^2-x$$, and replacing $$P$$ by the corresponding remainder,one may assume that $$P=ux+v$$. By assumption, $$v$$ is even, and $$u+v$$ is a multiple of $$2$$ and $$3$$, so $$v=2m$$ and $$u+v=6n$$, that is $$u=6n-v=6n-2m$$. Hence $$P=-2mx+ 6nx+2m=m(2-2x)-n 6x$$. Now $$6x$$ lies in our ideal (since $$6$$ does), and $$2-2x=4x+2-6x$$ also lies in our ideal, so we are done.

Now apply the first isomorphism theorem.

• Could you explain how you used CRT here? I thought it only applies to intersection of ideals and $(x^2-x,4x+2)$ is a sum. Commented Jul 25, 2020 at 16:50
• Use CRT on $\mathbb{Z}/6$ + the canonical isomorphism $(R\times S)[x]\simeq R[x]\times S[x]$. ANyway, if you don't like this argument, I gave a direct proof (Starting from "Let $f$ be the ring morphism....") Commented Jul 25, 2020 at 19:31

We have $$(x^2-x,4x+2)=(x,4x+2)(x-1,4x+2)=(2,x)(6,x-1).$$

The ideals $$(2,x)$$ and $$(6,x-1)$$ are comaximal, and by CRT $$\mathbb Z[x]/(x^2-x,4x+2)\simeq\mathbb Z[x]/(2,x)\times\mathbb Z[x]/(6,x-1)\simeq\mathbb Z/2\mathbb Z\times\mathbb Z/6\mathbb Z.$$