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Gcd as generator of ideal

  1. Nov 23, 2007 #1
    Hi.

    Given a field F and the space of polynomials over it, F[x], I was wondering why it is that the monic generator of the ideal generated by any two polynomial f, g from F[x] is exactly the gcd of f and g.

    Thank you!

    Kriti
     
  2. jcsd
  3. Nov 23, 2007 #2

    morphism

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    What are your thoughts on the matter? This pretty much follows from the definitions.
     
  4. Nov 23, 2007 #3
    Yeah, that's the impression I got, but I think I might just be missing out some basic connection. So far, I have the following.

    Let I be the ideal generated by f and g.
    Since d = gcd(f, g), we know that d|f and d|g so d|(af+bg) for some a,b in F. And therefore d is in I.

    I cannot think of how to show this is the monic generator, however. Maybe I need to know what is required to show that a given polynomial is the monic generator of an ideal.

    Also, I'm sorry for the plaintext. I have not yet figured out the formatting features yet!

    Thank you!
     
  5. Nov 23, 2007 #4

    morphism

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    Do you know that every ideal in F[x] is generated by single element (i.e. that F[x] is a principal ideal domain)? In fact, the monic generator of any ideal I is the one of smallest degree in I. (You can prove this using the division algorithm: Let h be the monic polynomial of smallest degree in I. Use the division algorithm to write f=hq+r [where deg(r) < deg(h)] for any f in I. Conclude that r=0.)

    Now let I be the ideal generated by f and g. Let d=af+bg be the monic generator of I. It now follows that any divisor of f and g divides d.
     
  6. Nov 23, 2007 #5
    I follow the first 2 statements and the last statement. But I'm not sure where to apply the division algorithm.

    Let me ask this...is the following the right way to go about it?
    Begin with the idea of what a monic polynomial generator is. Call it d.
    By the definition of I, d is in I, and so is of the form d = af+bg.
    By this statement, any divisor of f and g also divides d.
    This is the definition of gcd and so d = gcd(f, g) = monic generator of I.

    Or do we approach in the opposite direction?
    Begin with the idea what gcd is, and then show that this gcd is the monic generator of I.

    I'm sorry if I'm coming across a little dense. I get the main idea but the details seem to be evading me.

    Thank you so much!
     
  7. Nov 24, 2007 #6

    morphism

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    I was outlining a proof of the fact that every ideal of F[x] has a monic generator (this is always the case in polynomial rings over fields, but not necessarily in general polynomial rings). If you already know this fact, you can disregard what I wrote between the parentheses.

    Now given that we can find a monic generator d for the ideal (f,g), then what you said here is correct:
     
  8. Nov 26, 2007 #7

    mathwonk

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    This statement :
    "Begin with the idea what gcd is, and then show that this gcd is the monic generator of I",

    is not doable, because there is no unique gcd. any unit multiple of one gcd is another one.

    i.e. a given gcd need not be monic, assuming you have the correct definition of gcd as a common divisor which is itself divisible by every other common divisor.

    maybe this is causing your difficulties.
     
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