Quotient ring of poly ring Z[x]

  • Thread starter jakelyon
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  • #1
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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.
 

Answers and Replies

  • #2
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What is the cardinality of Z/2Z? Then look at the size of Z[x]/(2x).
 
  • #3
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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.
 
  • #4
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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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