Is Z[x]/(2x) Isomorphic to Z/2Z?

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Discussion Overview

The discussion revolves around the question of whether the quotient ring Z[X]/(2x) is isomorphic to Z/2Z. Participants explore the structure of the ideal (2x) and its implications for the elements of the quotient ring, as well as the necessary conditions to achieve an isomorphism with Z/2Z.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant suggests that every element in Z[x] can be expressed as a polynomial with integer coefficients, and questions whether the ideal (2x) corresponds to the set of multiples of 2x.
  • Another participant raises the issue of cardinality, prompting a comparison between Z/2Z and Z[x]/(2x).
  • A participant argues that modding out by (2x) effectively sends x to 0, leading to the conclusion that Z[x]/(2x) is equivalent to Z, not Z/2Z.
  • Another reply clarifies that while 2x = 0 in the quotient, x itself is not zero, and discusses the nature of the elements in Z[x]/(2x) as polynomials with certain coefficients.
  • It is noted that to obtain Z/2Z from a quotient of Z[x], one would need to quotient out by the ideal (2,x), which cannot be generated by a single element.

Areas of Agreement / Disagreement

Participants express differing views on the structure of Z[x]/(2x) and its relationship to Z/2Z. There is no consensus on whether the two are isomorphic, and the discussion remains unresolved.

Contextual Notes

Participants highlight the distinction between integral domains and other types of rings, noting that Z[x] is not a principal ideal domain, which may affect the conclusions drawn about the ideals involved.

jakelyon
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Can anyone explain, in detail, why/why not Z[X]/(2x) is isomorphic to Z/2Z? I know that every element in Z[x] can be written as a_0 + a_1 x + a_2 x^2 + ... with a_i in Z and only finitely many a_i's are nonzero. Now, does (2x) = (2, 2x, 2x^2,...)? Also, the quotient is "like" taking 2x=0, or x=0. Thus, I think that all elements of Z[x]/(2x) would look like a_0/2 for some a_0 in Z. But this does not give Z/2Z does it? Thanks.
 
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What is the cardinality of Z/2Z? Then look at the size of Z[x]/(2x).
 
Thanks for replying daveyinaz, but I am not sure I am following. However, I been doing some reading:

(2x) is the ideal consisting of all linear combinations of 2x (with integer coefficients). Now, by moding Z[x] out by (2x) it is "like" sending x to 0. So, if I am correct, then Z[x]/(2x) = Z[0] =
Z, not Z/2Z, right?

Does this make sense? How would I get Z/2Z then? Thanks.
 
You are right that you can think of the quotient as forcing 2x = 0. However, this does not mean that x = 0, it means that all multiples of 2x by elements in Z[x] are equal to zero. So x is not zero since x is not a multiple of 2x in Z[x]. Also, all of Z will be in that quotient since no integer is a multiple of 2x in Z[x]. Things in Z[x]/(2x) will look like polynomials with any constant term but with coefficients on the other terms being 0 or 1.

Maybe part of your trouble is thinking that all rings are integral domains (integral domains are rings such that if ab = 0 then a = 0 or b = 0). This ring is not and there are more familiar ones that are not either such as the ring of all matrices.

If you want to get Z/2Z from a quotient of Z[x] you would have to quotient out by the ideal (2,x). Note that Z[x] is not a principal ideal domain and this ideal cannot be generated by a single element.
 

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