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Abstract algebra: monic gcd of polynomials in a subfield problem

  1. Apr 18, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]K \subseteq L[/itex] be fields. Let [itex]f[/itex], [itex]g \in K[x][/itex] and [itex]h[/itex] a gcd of [itex]f[/itex] and [itex]g[/itex] in [itex]L[x][/itex].

    To show: if [itex]h[/itex] is monic then [itex]h \in K[x][/itex].

    3. The attempt at a solution

    Assume [itex]h[/itex] is monic.

    Know that: [itex]h = xf + yg[/itex] for some [itex]x[/itex], [itex]y \in K[x][/itex].
    So the ideal generated by [itex]h[/itex], [itex](h)[/itex] in [itex]L[x][/itex] equals the ideal [itex](f,g)[/itex] in [itex]L[x][/itex]. Also, since [itex]K[x][/itex] is a principal ideal domain, [itex](f,g)=(d)[/itex] in [itex]K[x][/itex] for some [itex]d \in K[x][/itex], so [itex]d = af + bg[/itex] for some [itex]a[/itex], [itex]b \in K[x][/itex]. So [itex]d[/itex] is a gcd of [itex]f[/itex] and [itex]g[/itex] in [itex]K[x][/itex].

    Now I'm not sure where to go... I know that [itex]h[/itex] is monic and therefore the unique monic gcd of [itex]f[/itex] and [itex]g[/itex] in [itex]L[x][/itex], but not sure how this is useful. Do I need to show that [itex]h = d [/itex]? How can I use the monic assumption to show this?

    Thanks!!
     
  2. jcsd
  3. Apr 19, 2012 #2
    Hmm, well, you the monic assumption is usually used to prove the uniqueness of the gcd. In this case, I'd say the monic assumption is useful since if you do not define the gcd to be monic, then [itex]h \left( x \right)[/itex] a gcd of [itex]f \left( x \right)[/itex] and [itex]g \left( x \right)[/itex] means that [itex]r h \left( x \right)[/itex] where [itex]r \neq 0 \in L[/itex] is also a greatest common divisor of [itex]f \left( x \right)[/itex] and [itex]g \left( x\right)[/itex]. Well, if you let [itex]h \left( x \right)[/itex] be monic and [itex]r \in L[/itex] but [itex]r \not \in K[/itex], then it is easy to see the [itex]r h \left( x \right) \not \in K \left[ x \right][/itex].

    I believe you could let [itex]d \left( x \right)[/itex] be monic as well and you'd want to show that [itex]d \left( x \right) = h \left( x \right)[/itex].

    I will premise this by saying that I'm not sure if this is the easiest way to do this problem, but suppose that [itex]h \left( x \right)[/itex] and [itex]d \left( x \right)[/itex] are the unique monic gcd of [itex]f \left( x \right)[/itex] and [itex]g \left( x \right)[/itex] in [itex]L \left[ x \right][/itex] and [itex]K \left[ x \right][/itex] respectively. Well, then it can be seen that [itex]d \left( x \right)[/itex] is at least still a common divisor of [itex]f \left( x \right)[/itex] and [itex]g \left( x \right)[/itex] in [itex]L \left[ x \right][/itex], and [itex]d \left( x \right) = p \left( x \right) f \left( x \right) + q \left( x \right) g \left( x \right)[/itex] where [itex]p \left( x \right), q \left( x \right) \in K \left[ x \right] \subset L \left[ x \right][/itex]. Can you use this and the fact that [itex]h \left( x \right)[/itex] and [itex] d \left( x \right)[/itex] both monic to show that [itex]h \left(x \right) = d \left( x \right)[/itex]?
     
  4. Apr 19, 2012 #3
    Thanks! I think I can,... does it involve using the fact that a gcd of f and g is the smallest degree polynomial that can be written as a linear combination of f and g, and that d, being a common factor of f and g in L[x], must divide h?
     
  5. Apr 19, 2012 #4
    Yea, pretty much. I also believe you may have seen somewhere that a gcd is the only common divisor which can be written as a linear combination. And the monic condition means it is the unique monic gcd.

    Then you've just shown that making the field bigger doesn't change the gcd.
     
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