Abstract algebra: monic gcd of polynomials in a subfield problem

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Homework Help Overview

The discussion revolves around the properties of the greatest common divisor (gcd) of polynomials within the context of fields, specifically examining the conditions under which a monic gcd in a larger field implies the gcd also resides in a smaller subfield.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of the monic condition on the uniqueness of the gcd and its relationship to polynomials in a subfield. Questions arise regarding the necessity of demonstrating that the monic gcd in the larger field equals the gcd in the smaller field.

Discussion Status

Participants are actively engaging with the problem, considering various aspects of the uniqueness of the gcd and how the monic condition influences the relationship between the gcds in the two fields. There is a recognition of the need to connect the properties of the gcd with the linear combinations of the polynomials involved.

Contextual Notes

There is an underlying assumption that the gcd is defined as the unique monic polynomial, and participants are navigating the implications of this definition in the context of polynomial ideals in different fields.

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Homework Statement


Let K \subseteq L be fields. Let f, g \in K[x] and h a gcd of f and g in L[x].

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

The Attempt at a Solution



Assume h is monic.

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

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

Thanks!
 
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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 h \left( x \right) a gcd of f \left( x \right) and g \left( x \right) means that r h \left( x \right) where r \neq 0 \in L is also a greatest common divisor of f \left( x \right) and g \left( x\right). Well, if you let h \left( x \right) be monic and r \in L but r \not \in K, then it is easy to see the r h \left( x \right) \not \in K \left[ x \right].

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

I will premise this by saying that I'm not sure if this is the easiest way to do this problem, but suppose that h \left( x \right) and d \left( x \right) are the unique monic gcd of f \left( x \right) and g \left( x \right) in L \left[ x \right] and K \left[ x \right] respectively. Well, then it can be seen that d \left( x \right) is at least still a common divisor of f \left( x \right) and g \left( x \right) in L \left[ x \right], and d \left( x \right) = p \left( x \right) f \left( x \right) + q \left( x \right) g \left( x \right) where p \left( x \right), q \left( x \right) \in K \left[ x \right] \subset L \left[ x \right]. Can you use this and the fact that h \left( x \right) and d \left( x \right) both monic to show that h \left(x \right) = d \left( x \right)?
 
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?
 
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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