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Factoring operator and fundamental theorem of algebra

  1. Feb 15, 2007 #1
    I haven't taken any abstract algebra course so I do not know if this is the right section to post this question in.

    Anyway, I am learning differential equation right now and my prof. recently showed factorization of differential operator.

    For example, let D be the differential operator, he then said that any thing in the form:
    where I is the identity can be factorized into:
    where multiplication means composition of operator

    I am curious to know what kind of algebraic structure has this property... and if I have a random operator, let's say F, what axioms does F have to satisfy in order for
    to be factorizable like a polynomial?

    more specifically, If I define an operator [tex]\Delta x(n)=x(n+1)-x(n)[/tex]
    how can I show (what axioms does Delta have to satisfy) that polynomials in [tex]\Delta[/tex] can be factorized?
    (I'm reading a book on differences equation... so I would like to know)
    Last edited: Feb 15, 2007
  2. jcsd
  3. Feb 15, 2007 #2


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    Observe that the set of all such differential operators is nothing more than the ring of polynomials in one variable over C. You can factor polynomials, therefore you can factor such differential operators.
  4. Feb 15, 2007 #3


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    commutativity is useful, which is why he stuck to linear combinations of the D operator. those all commute with each other.

    then they behave exactly like polynomials in X. almost.

    they may also satisfy relations that polynomials do niot, but that means they behave even better.
  5. Feb 15, 2007 #4
    thx for the quick response.

    but what do I need to do to show that they are the ring of polynomials?

    if I want to prove something is a group, i show a set of elements and a binary operator. then I prove associativity, identity and the inverse axioms... what about proving something is the "ring of polynomials"?

    for instance, to prove that factoring works for differential operators, what do I have to show?

    Do I have to show D commute with D and Identity and stuffs like:

    ... or something similar?
  6. Feb 15, 2007 #5


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    the polynomial ring k[X] has the nice property that given any algebra R over k, and any element D of that algebra, there is a unique k algebra map from k[X] to R sending X to D.

    Thus if D satisfies no relations, the map is an isomorphism. on the iother hand if you view D as an operator satisfyijng some equation like D^2+1 = 0, then the ring of poloynomials in D acting on the solutiions of that equation is a quotient of k[X] by the ideal generated by X^2+1.
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