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Kernel is a principle ideal?

  1. Mar 31, 2007 #1
    1. The problem statement, all variables and given/known data
    Prove that the kernel of the homomorphism Z[x]->R sending x to 1+sqrt(2) is a principle ideal, and find a generator for this ideal.

    Z is the integers
    R is the real numbers

    3. The attempt at a solution
    I assume sending x to sqrt(2) is an example. We should first find the actual map shoudn't we? Otherwise we can't find the kernel of this map. I have tried like

    1. Sending the coefficients of x to C+sqrt(x) or C(1+sqrt(2)) where C is the coefficient of x.
    2. Sending the constant of the polynomial to C+1+sqrt(2) or C(1+sqrt(2)) where C is the consant of the polynomial.
    3. Sending the coefficient of the greatest power in the polynomial to C+sqrt(2) or C(1+sqrt(2)) where C is the coefficient of the greatest power of the polynomial.

    All 3 example maps do not form a homomorphism.
     
  2. jcsd
  3. Mar 31, 2007 #2
    Or does this homomorphism only sends polynomials in Z[x] with a single elements to 1+sqrt(2) guided by x->1+sqrt(2) so x^2->(1+sqrt(2))^2 and x^0->(1+sqrt(2))^0. 2x->2(1+sqrt(2)).
     
  4. Mar 31, 2007 #3

    HallsofIvy

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    Z[x] is the ring of all polynomials with integer coefficients. This homomorphism essentially "evaluates" the polynomial at [itex]x= 1+ \sqrt{2}[/itex].

    For example the polynomial [itex]ax^2+ bx+ c[/itex] is mapped into [itex]a(1+ \sqrt{2})^2+ b(1+ \sqrt{2})+ c[/itex]. Yes, that is a homomorphism and its kernel is the set of polynomials having [itex]1+ \sqrt{2}[/itex] as a zero. Obviously, if P(x) has [itex]1+ \sqrt{2}[/itex] as zero, then any product P(x)Q(x) has also. It should also be obvious that if a polynomial with integer coefficients has [itex]1+ \sqrt{2}[/itex] as a zero, then it must also have [itex]1- \sqrt{2}[/itex] and so has [itex](x-1-\sqrt{2})(x-1+\sqrt{2})= (x-1)^2- 2= x^2- 2x- 1[/itex] as a factor.
     
  5. Mar 31, 2007 #4
    I see, that makes more sense. It means evaluate the polynomial at 1+sqrt(2). I think they should have been more clear with it as it took me 2 hours and still couldn't figure it out what the question meant until HallsofIvy came along.

    Although with your last point, there are many integer coefficient polynomials with 1+sqrt2 as a zero but 1-sqrt2 not being a zero. i.e g(x)(x-(1+sqrt2)) contains many polynomials with 1+sqrt2 being a zero but very few with 1-sqrt2 also as a zero. i.e let g(x)=x-1 and many other g(x). However if the polynomial was in C[x] (which this question does not involve) then zeros do come in complex conjugates.

    Looks like the hard bit of the question was working out what it's asking and working out what map it describes.
     
    Last edited: Mar 31, 2007
  6. Mar 31, 2007 #5
    No, I made a mistake in my last post. g(x)(x-(1+sqrt2)) is not in Z[x]. Now I understand why you said x^2-2x-1 must be a factor of a polynomial in order for it to lie in the kernel.
     
  7. Apr 1, 2007 #6

    HallsofIvy

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    Sorry, but I thought the problem was very clear. What was NOT clear to me was why you were looking at things like "Sending the coefficients of x to C+sqrt(x) or C(1+sqrt(2)) where C is the coefficient of x" or "Sending the coefficient of the greatest power in the polynomial to C+sqrt(2) or C(1+sqrt(2)) ". Those couldn't possibly be homomorphisms- they don't use all of the "information" in the polynomial.
     
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