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Ideals and GCDs in k[x] ... ... Cox et al ...

  • #1
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Homework Statement



I am reading the undergraduate introduction to algebraic geometry entitled "Ideals, Varieties and Algorithms: An introduction to Computational Algebraic Geometry and Commutative Algebra (Third Edition) by David Cox, John Little and Donal O'Shea ... ...

I am currently focused on Chapter 1, Section 5: Polynomials of One Variable ... ... and need help with the proof of Proposition 8, part 3 ...

Proposition 8 of Chapter 1 (including Definition 7 which is relevant) reads as follows:


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In the above text from Cox et al we read the following:


" ... ... To prove part (iii), let ##h \ = \ GCD(f_2, \ ... \ ... \ , f_s)##. We leave it as an exercise to show that

##<f_1, h> \ = \ <f_1, \ ... \ ... \ , f_s>##

... ... "


I need help to show that ##<f_1, h> \ = \ <f_1, \ ... \ ... \ , f_s>## ... ...


Homework Equations



These are all introduced in context, in 3 below ... ...



The Attempt at a Solution




Work so far ...

We need to show that ##<f_1, h> \ \subset \ <f_1, \ ... \ ... \ , f_s>## ... and also that

##<f_1, \ ... \ ... \ , f_s> \ \subset \ <f_1, h>##


So to show ##<f_1, h> \ \subset \ <f_1, \ ... \ ... \ , f_s>## we start with


Let ##l \in <f_1, h>## ...

Then, by definition of ##<f_1, h>##, we have that ##l = f_1 t_1 + h t_2 where t_1, t_2 \in k[x]## ...


Now we have that ##h = GCD(f_2, \ ... \ ... \ , f_s)## ... BUT ... how do we use this in the proof?


Note that we also have

##(1) \ h = GCD(f_2, \ ... \ ... \ , f_s) \ \Longrightarrow \ h \text{ divides } f_2, \ ... \ ... \ , f_s##

## \ \Longrightarrow \ h = f_2 u_2, h \ = \ f_3 u_3, \ ... \ ... \ , h \ = \ f_s u_s##

for some ##u_2, \ ... \ ... \ , u_s \in k[x]## ...


(2) ##k[x]## is a PID so that:

##<f_1, h > \ = \ <v>## for some polynomial ##v \in k[x]## ... ...


... but, how do we use (1) and (2) in the required proof ...



Can someone please help me to complete the proof of ##<f_1, h> \ = \ <f_1, \ ... \ ... \ , f_s>## ...

Help will be appreciated ...

Peter
 

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Answers and Replies

  • #2
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The easy part is ##I := <f_1, \dots , f_s> \;⊆ \;<h>\; ⊆\; <h,f_1>\;## because every ##f_i## can be written ##f_i = g_i \cdot h## by definition of ##h##. Thus any ##f = \sum a_i f_i ∈ I## gets ##f = \sum a_i g_i h ∈ \; <h>.##
The other way, your first, is essentially Bézout's Lemma. I'm too lazy to type it from Wikipedia. Maybe you get along with what is written there:
https://en.wikipedia.org/wiki/Bézout's_identity

It might as well be the quoted proposition 6 or corollary 4. It means that you can write ##h_2 = p_1 f_1 + p_2 f_2## if ##h_2## is the greatest common divisor of ##f_1## and ##f_2## and then by induction ##h = p_1 f_1 + \dots + p_s f_s## which is what we need for ##h \in I.##

It applies to PID, so you may substitute "integers" by "polynomials". There is a written proof and a link to the polynomial version (without proof as far as I could see): https://en.wikipedia.org/wiki/Polynomial_greatest_common_divisor#B.C3.A9zout.27s_identity_and_extended_GCD_algorithm

If you have further questions to it or about the Wiki entry, please let me know.
 
  • #3
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The easy part is ##I := <f_1, \dots , f_s> \;⊆ \;<h>\; ⊆\; <h,f_1>\;## because every ##f_i## can be written ##f_i = g_i \cdot h## by definition of ##h##. Thus any ##f = \sum a_i f_i ∈ I## gets ##f = \sum a_i g_i h ∈ \; <h>.##
The other way, your first, is essentially Bézout's Lemma. I'm too lazy to type it from Wikipedia. Maybe you get along with what is written there:
https://en.wikipedia.org/wiki/Bézout's_identity

It might as well be the quoted proposition 6 or corollary 4. It means that you can write ##h_2 = p_1 f_1 + p_2 f_2## if ##h_2## is the greatest common divisor of ##f_1## and ##f_2## and then by induction ##h = p_1 f_1 + \dots + p_s f_s## which is what we need for ##h \in I.##

It applies to PID, so you may substitute "integers" by "polynomials". There is a written proof and a link to the polynomial version (without proof as far as I could see): https://en.wikipedia.org/wiki/Polynomial_greatest_common_divisor#B.C3.A9zout.27s_identity_and_extended_GCD_algorithm

If you have further questions to it or about the Wiki entry, please let me know.

Thanks for the help fresh_42 ...

Still reflecting on your post ... but basically followed that ...

Thanks again ... most helpful

Peter
 

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