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How to describe an element of the Ring $\Bbb Z[x]/(x^2-3,2x+4)$?

Or is it isomorphic to any well known ring?

What is meant by quotienting $\Bbb Z[x]$ by a kernel like that?

rschwieb
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EuReka
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2 Answers2

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Hint $\rm\ \ \ mod\,\ (\color{#0a0}{3\!-\!x^2},\ \color{#c00}{2x\!+\!4})\!: $ $\quad 6\!\!\!\!\! \color{#0a0}{\ \overset{ \overset{\ \ \ \ \Large{3\ \ \equiv\ \ x^2}}{\phantom{M}} }\equiv}\! $ $ 2x^2\!\!\!\color{#c00}{\ \overset{ \overset{\ \Large{2x\ \ \equiv\ \ -4}}{\phantom{M}} }\equiv}\!\! $ $ -4x \!\!\!\!\! \color{#c00}{\ \overset{ \overset{\ \Large{-2x\ \ \equiv\ \ 4\ \ \ \ \ \ }}{\phantom{M}} }\equiv}\!\!\!\!\!\! 8 \,\ \Rightarrow\,\ 2\equiv 0$

So $\rm\ \ 2 \in I = (x^2\!-\!3,\, 2x\!+\!4)\ \Rightarrow\ I = (2,I) = (2,\, I\ mod\ 2) = (2, x^2\!+\!1) = (2,(x\!+\!1)^2)$

So $\rm\ \Bbb Z[x]/I \,\cong\, \Bbb Z[x]/(2,(x\!+\!1)^2)\cong\,\Bbb Z_2[x]/((x+1)^2)\cong\, \Bbb Z_2[t]/(t^2)\cong\,$ dual numbers over $\:\!\Bbb Z_2$

Remark $ $ The elimination process used to deduce $\,2\equiv 0\,$ is not ad-hoc. Rather it is a special case of a normal form algorithm for such ideals, e.g. see the discussion on Hermite and Smith normal forms in Henri Cohen's A Course in Computational Number Theory.

Bill Dubuque
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  • Now I =I mod 2 clear . But why tis = x^2+1 and=(x+1)^2? – EuReka Dec 20 '13 at 15:18
  • @EuReka Because $\rm\ mod\ \color{#c00}2!:\ \ \color{#c00}{2\equiv 0}\ \Rightarrow\ (x+1)^2 \equiv x^2!+\color{#c00}2x+1\equiv x^2+1\ \ $ – Bill Dubuque Dec 20 '13 at 15:21
  • ok ok .....thank you I think I have understood...But to be confident I have to do similar problem... Can you give some similar one... – EuReka Dec 20 '13 at 15:43
  • @Eureka You can generate many similar examples simply by varying the coefficients, for example, change $,\color{#0a0}3,$ and $,\color{#c00}4,$ to some other integers. – Bill Dubuque Dec 20 '13 at 16:35
  • Sorry to bother you for your answer after 10 years. I have a question: Does this method always works on any ideals generated by few elements? @BillDubuque – Lucas Oct 07 '23 at 16:54
  • @Lucas Yes, see the book I cited I on Hermite / Smith normal form algorithms (which work over any Dedekind domain) and related methods. – Bill Dubuque Oct 07 '23 at 17:16
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It is the universal ring containing an element $x$ such that $x^2=3$ and $2x=-4$. It follows

$6=2x^2=-4x$

and on the oher hand $4x=-8$, hence $2=0$. But then the relations simplify to $0=x^2-1=(x+1)^2$. Thus, we get

$\mathbb{F}_2[x]/(x+1)^2$

This cannot be simplified anymore (except for the trivial substitution $y:=x+1$). It is the unique $\mathbb{F}_2$-algebra with $4$ elements with a nontrivial nilpotent element (the other ones being $\mathbb{F}_2 \times \mathbb{F}_2$ and $\mathbb{F}_4=\mathbb{F}_2[x]/(x^2+x+1)$).

Edit: The whole point of my answer is that you don't have to calculate with ideals and residue classes ... just compute in the quotient ring, then everything is simple.

Martin Brandenburg
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  • I am newcomer in Ring Theory. By ideal(x,y) of ring(comm with identity) R i know {rx+sy: r,s in R} By Z[x]/(p) i mean the elements of the form f(x)+(p)..then i will divide f(x) by p to get the simplest form. I can not follow the rest of the part after 0=2. can you help me Sharing a link to get more info on that... Or explain further. – EuReka Dec 20 '13 at 14:44
  • @EuReka I gave more explicit details of the inferences in my answer, see the justifications over the congruence signs. For example, in the first one, we deduce that $6 \equiv 2x^2$ by multiplying by $,2,$ the congruence shown above it, viz. $3 \equiv x^2$ (which is true because, mod $I!:$ $\ \ 3-x^2\equiv 0\ $ $\Rightarrow$ $\ 3\equiv x^2)$. – Bill Dubuque Dec 20 '13 at 15:13
  • Dear @EuReka : Even more explicit details appear at the solution at the duplicate question. – rschwieb Dec 20 '13 at 15:58
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    @rschwieb I don't agree, e.g. the linked dupe does not give any clue whatsoever how it derives the key step, that $,2 = 2(X^2-3)-(2X+4)(X-2),$ so $,2\in I.,$ Instead, this equation is simply pulled out of a hat - like magic! – Bill Dubuque Dec 20 '13 at 16:34
  • @Bill Dubuque I dont think so. if (x,y) is the ideal I then rx+sy must be in I for r,s is in R (comm with id). Now 2=r and s= (X-2) both in Z[x]. But your method is from Number theory? It is also interesting. I did not know it before. I will check and read about it more. – EuReka Dec 20 '13 at 16:46
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    @EuReka I didn't mean that the inference "so $,2 \in I"$is magic; rather the magic is in the discovery of the key equation $,2 = 2(X^2-3)-(2X+4)(X-2),$ implying the ideal membership. The process I use is essentially a modular form of a generalized Euclidean algorithm. Just as for the normal Euclidean algorithm, we can run it in "extended" form, keeping track of the ideal membership relations. Doing that would yield this equation (or equivalent). It can be considered as a "witness" for the ideal membership claim. Pedagogically, I think it is better to derive it vs. pull it out of a hat. – Bill Dubuque Dec 20 '13 at 17:00
  • ok I agree with you Bill Dubuque. – EuReka Dec 20 '13 at 17:09
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    Dear @BillDubuque : Sorry, I did not mean to interrupt the self-commercial. I just thought the learner should have access to as many details as possible. If you disagree with that then I don't have any response. It was not (as you seem to have assumed) meant to be a statement on the quality of your answer. Regards. – rschwieb Dec 20 '13 at 17:26
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    @rschwieb My primary purpose here is to teach the beautiful methods that I have learned from many masterful expositors: Dedekind, Noether, Krull, etc. (see my location, "Shoulder of Giants). If you insist on viewing that as "commercialization" then perhaps it is more constructive to view it as advertising the many beautiful ideas that have been passed on from these talented teachers. – Bill Dubuque Dec 20 '13 at 17:36
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    Dear @BillDubuque : I thought we already established that these elaborate justifications of yourself are not warranted. I am not (and nobody is, as far as I can see) concerned with the reasons you post what you post. I would much rather you spend this effort instead for new answers. Regards – rschwieb Dec 20 '13 at 17:54
  • @Martin Regarding your edit, I agree that it is simpler to work universally (and that is what I personally do). I was tempted to do so also in my answer, but I thought that the equivalent congruential language might prove easier for beginners. By the way, apologies that the above tangential discussion got attached to your answer. – Bill Dubuque Dec 21 '13 at 00:50