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

[Math Amateur]

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:

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 withLet $l \in <f_1, h>$ ...

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

 

[fresh_42]

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/Polyn...9zout.27s_identity_and_extended_GCD_algorithm

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

 

[Math Amateur]

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/Polyn...9zout.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