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

Mathematics Asked by 2132123 on August 3, 2020

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.

We have $$8(x^2-x)-2x(4x+2)=-12x=-3(4x+2)+6$$, so $$6in (x^2-x,4x+2)$$ ,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 $$Rsimeq mathbb{F}_2[x]/(x^2-x)timesmathbb{F}_3[x]/(x^2-x,x-bar{1})=mathbb{F}_2[x]/(x^2-x)timesmathbb{F}_3[x]/(x-bar{1})$$, which finally yields $$Rsimeqmathbb{F}_2timesmathbb{F}_2timesmathbb{F}_3.$$

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

Let $$f$$ be the ring morphism $$f:Rto Pinmathbb{Z}[x]mapsto ([P(0)]_2,[P(1)]_2, [P(1)]_3)inmathbb{F}_2timesmathbb{F}_2times mathbb{F}_3$$.

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 $$a,b,cinmathbb{Z}$$, we need to find $$Pinmathbb{Z}[x]$$ 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 $$(b-a)+2k+aequiv c [3]$$, 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 $$Pinmathbb{Z}[x]$$ such that $$f(P)$$ is trivial. We want to prove that $$Pin (x^2-x,4x+2)$$. 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.

Correct answer by GreginGre on August 3, 2020

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 $$mathbb Z[x]/(x^2-x,4x+2)simeqmathbb Z[x]/(2,x)timesmathbb Z[x]/(6,x-1)simeqmathbb Z/2mathbb Ztimesmathbb Z/6mathbb Z.$$

Answered by user26857 on August 3, 2020