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Polynomial Ring not a Field

  1. Aug 6, 2007 #1

    First off, my terminology in the title may not be correct or what others use because when searching the forums for "polynomial ring" i had a lot of trouble finding anything that matched what i meant when i coined the term....

    By polynomial ring i mean, for some ring R the polynomial ring over R is the ring R[x] = {a_0 + a_1x + .... + a_nx^n + ... ; a_i is in R} under the operations (+, .)

    Now to my Question -

    In the case that R (as above) is a Field, F, the i understand that F[x] is an integral domain, but not a field because of the simple example that my lecturers love to throw about, x in F[x] does not have an inverse (is not a unit)

    Now i understand this argument and i dont dispute it, but i was wondering if any of you could tell me, or at least start me off (as it would probably help my understanding) on how to show / prove that x has no inverse in F[x]....

    I have thought about this....but im not sure the following argument is correct

    suppose there was g(x) st. xg(x)=1

    then xg(x)=x^0

    hence g(x) = x^(0-1) = x^-1 which is not polynomial and hence not in F[x]

    im thinking maybe a better argument would involve how i started but then somehow using the fact F[x] has no zero divisors would be more correct...but this is just guess work really.

    sorry for the long post


  2. jcsd
  3. Aug 6, 2007 #2
    Also, the degree of xg(x) is 1+deg(g) > 0=deg(1). Your argument seems circular to me because it rests on the premise that x itself doesn't have a polynomial inverse in R[x], which is what you're proving.
  4. Aug 6, 2007 #3

    Yeah i know my attempt a a proof was pretty dismal....

    Thankyou for your reply, i think that would be it. Im wondering does the degree of 1 (mult. identity) have to be 1? Maybe that is a stupid question, is it by definition deg 1 = 1....?
  5. Aug 6, 2007 #4
    I think if f is a constant polynomial other than zero, deg(f) is defined to be zero and if f is the zero polynomial, then its degree is undefined or some people say negative infinity or something. I dunno, maybe some people define the degree of any constant polynomial to be zero...
  6. Aug 6, 2007 #5
    Yeah we had deg 0 = - infinity, but nothing defined about constants so maybe it was assumed...

    anyway, thanks a bundle for sorting that out for me, now at least when im in bed i wont be thinking about maths... : )
  7. Aug 6, 2007 #6
    I'm surprised deg is defined for the zero polynomial but not other constant polynomials...perhaps it's embedded within the definition for other polynomials?
  8. Aug 6, 2007 #7
    hmm....ive looked it up in two text books now

    The first one, Linear Algebra (LAY) - This is one was from 1st year - mentions non zero constant polynomials and says they have degree 0, but its talking about them in the context of the vector space P_n, so im not sure if it would apply

    My second text book, A First Course in Abstract Algebra (FRALEIGH) - this one is from 2nd year - does discuss the degree of a polynomial in the context of the polynomial ring, but does not mention the degree of a non zero constant functin.

    I will just ask my lecturer tomorrow, he comes around to everyone in the class individually before and after every lecture to see how we are going with his course so i shouldnt have any trouble getting to talk to him...

    but for now, SLEEP, its 12.44 here and im tired

    THANK YOU again for your prompt and helpful responses

  9. Aug 6, 2007 #8
    Yes, F[x] is an integral domain.

    To show x has no inverse in F[x] we need to show x*f(x) ! = 1.

    If, f(x) is a non-zero polynomial then deg f(x) = n where n>=0. And so deg (x*f(x)) >= n+1 >=1. Which means xf(x)!=1.

    If, f(x)=0 then x*f(x) = 0 !=1. Unless this is a zero ring, i.e. 1=0. But by convention it is not considered a field. And so this is impossible.
  10. Aug 6, 2007 #9


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    the degree of a non zero constant is zero.
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