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I need help to identify $\mathbb{Z}[x]/(x^2-3,2x+4)$.

I've been solving such problems in an approach like:

$$ 2(x^2-3)=2x^2-6, x(2x+4)=2x^2+4x \\ (2x^2+4x)-(2x^2-6)=4x+6, 2(2x+4)=4x+8 \\ (4x+8)-(4x+6)=2 $$

What shall I do next, please? Thank you.

Simon

Mon
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  • Hi Simon, maybe add a few of your thoughts to this (I think good!) problem. This will help it be well received. – operatorerror Jan 17 '18 at 04:10
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    If you don't want a question marked as a duplicate, then you need to ask a question that can't be answered simply by copying the answers from the other post. For example, "I don't understand this specific aspect of (insert link to an answer you want to understand), can you explain it in more detail?" –  Jan 17 '18 at 04:31
  • good Simon. We have $2\in (x^2-3,2x+4)$ and $(x^2-3,2x+4)=(x^2-3,2x+4,2)=(x^2-3,2(x+2),2)=(x^2-3,2)$ – Mustafa Jan 17 '18 at 05:05

2 Answers2

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we have $2=2(x^2-3)+(2-x)(2x+4)$ and then $(x^2-3,2x+4,2)=(2,x^2-3)$

$\mathbb Z[x]/(x^2-3,2x+4) \cong\mathbb Z[x]/(2,x^2-3)\cong \mathbb Z_2[x]/(x^2-1) \cong \mathbb Z_2[x]/(x-1)^2 \cong \mathbb Z_2[x]/(x)^2 $

Mustafa
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First divide $x^2-3$ by $x+2$, dividing by a linear polynomial is the same as evaluating at the root of the linear term, so $x^2-3=4-3=1$. Thus $x^2-3=q(x)(x+2)+1$ for some $q(x)$. Multiplying by 2, and rearranging, we get $2(x^2-3)-q(x)(2x+4)=2$. Hence $2\in (x^2-3,2x+4)$, so since $2\mid 2x+4$, we have $$(x^2-3,2x+4)=(x^2-3,2x+4,2)=(x^2-3,2).$$

Then $\newcommand{\ZZ}{\mathbb{Z}}\ZZ[x]/(x^2-3,2)\cong \ZZ_2[x]/(x^2-3),$ since the kernel of the natural map $\ZZ[x]\to \ZZ_2[x]/(x^2-3)$ is $(2,x^2-3)$. Then $x^2-3\equiv x^2+1=x^2+1^2\equiv (x+1)^2\pmod{2}$, so $$\ZZ_2[x]/(x^2-3)\cong \ZZ_2[x]/((x+1)^2).$$ Then by a change of variables, $$\ZZ_2[x]/((x+1)^2)\cong \ZZ_2[x]/(x^2).$$

I'm pretty sure it's impossible to simplify this further. The ring has four elements, $0,1,x,x+1$, only 0 and 1 are idempotents, since $x$ is nilpotent, so $x+1$ is a unit. Thus the ring is not a product of smaller rings.

jgon
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  • Hi, I wonder how do you get $0, 1, x, x+1$. And why does $(x^2-3, 2x+4) = (x^2-3,2)$ instead of $(x^2-3, 2x+4) = (x+2, 2)$ since $x^2-3 = (x+2)(x-2) + 1$ – col Feb 01 '23 at 23:51
  • @col Well to get the elements you just list them out, starting with the degree 0 things, $0$ and $1$, the only two elements of $\mathbb{Z}_2:=\mathbb{Z}/2\mathbb{Z}$, and then proceed to the degree 1 things, $x$ and $x+1$, and then you might proceed to degree 2 things, $x^2$, $x^2+1$, $x^2+x$ and $x^2+x+1$, but $x^2\equiv 0$ in this ring, so these are all equivalent to one of the elements listed earlier, and the same goes for any higher degree polynomial. – jgon Feb 02 '23 at 01:06
  • @col As for why you don't have $(x^2-3,2x+4)=(x+2,2)$, it's because the right hand side doesn't contain $x+2$. If it did, it would contain $1$, so the quotient would be 0. – jgon Feb 02 '23 at 01:13
  • Thanks. Just to clarify, do you mean $(x^2-3) = (x+2, 1)$ for your second response? – col Feb 02 '23 at 05:26