MHB Solving Ideals in Q[x,y]: Cox et al Chapter 1 Section 4

  • Thread starter Thread starter Math Amateur
  • Start date Start date
Math Amateur
Gold Member
MHB
Messages
3,920
Reaction score
48
I am reading Chapter 1, Section 4, Cox et al "Ideals, Varieties and Algorithms"

Exercise 3(a) reads as follows:

In $$ \mathbb{Q}[x,y] $$ show the following equality of ideals:

<x + y, x - y > = <x, y>

I would appreciate help with this problem.

====================================================

My 'solution' (of which I am most unsure!) is as follows:

Idea generated by x + y, x - y is the ideal

$$ h_1 ( x + y) + h_2 (x - y)$$ where $$ h_1, h_2 \in \mathbb{Q}[x,y] $$

Ideal generated by x, y is the ideal

$$ h_3 x + h_4 y $$ where $$ h_3, h_4 \in \mathbb{Q}[x,y] $$

So

$$ h_1(x + y) + h_2 (x - y) = h_1x + h_1y + h_2x - h_2y $$

$$ = (h_1 + h_2)x + (h_1 - h_2)y $$

$$ = h_3x + h_4y $$

$$ <x,y> $$

Can someone please either correct this reasoning or confirm that is is correct/adequate.

Peter

[This is also posted on MHF]
 
Physics news on Phys.org
Your equation $$\begin{aligned} h_1(x + y) + h_2 (x - y) &= h_1x + h_1y + h_2x - h_2y \\ &= (h_1 + h_2)x + (h_1 - h_2)y \\ &= h_3x + h_4y \end{aligned}$$ shows that every element $h_1(x + y) + h_2 (x - y)$ in the ideal $\langle x+y,x-y\rangle$ generated by $x+y$ and $x-y$ can be expressed in the form $h_3x + h_4y$, where $h_3 = h_1+h_2$ and $h_4 = h_1-h_2$. This shows that $\langle x+y,x-y\rangle \subseteq \langle x,y\rangle.$

So far, so good. To show the reverse inclusion, you must start with an element $h_3x + h_4y \in \langle x,y\rangle$ and express it in the form $h_1(x + y) + h_2 (x - y)$. Your equation shows that this can be done by taking $h_1 = \frac12(h_3+h_4)$ and $h_2 = \frac12(h_3-h_4)$. What you omitted from your solution was to point out that this works, because the fraction $\frac12$ is in $\mathbb{Q}$.

Notice that if the question had asked you to show the same result in $\mathbb{Z}[x,y]$ then it would no longer have been correct, because the fraction $\frac12$ is not available in $\mathbb{Z}.$
 
Opalg said:
Your equation $$\begin{aligned} h_1(x + y) + h_2 (x - y) &= h_1x + h_1y + h_2x - h_2y \\ &= (h_1 + h_2)x + (h_1 - h_2)y \\ &= h_3x + h_4y \end{aligned}$$ shows that every element $h_1(x + y) + h_2 (x - y)$ in the ideal $\langle x+y,x-y\rangle$ generated by $x+y$ and $x-y$ can be expressed in the form $h_3x + h_4y$, where $h_3 = h_1+h_2$ and $h_4 = h_1-h_2$. This shows that $\langle x+y,x-y\rangle \subseteq \langle x,y\rangle.$

So far, so good. To show the reverse inclusion, you must start with an element $h_3x + h_4y \in \langle x,y\rangle$ and express it in the form $h_1(x + y) + h_2 (x - y)$. Your equation shows that this can be done by taking $h_1 = \frac12(h_3+h_4)$ and $h_2 = \frac12(h_3-h_4)$. What you omitted from your solution was to point out that this works, because the fraction $\frac12$ is in $\mathbb{Q}$.

Notice that if the question had asked you to show the same result in $\mathbb{Z}[x,y]$ then it would no longer have been correct, because the fraction $\frac12$ is not available in$\mathbb{Z}.$
Thanks OpalgReally appreciate your help
Peter
 
I asked online questions about Proposition 2.1.1: The answer I got is the following: I have some questions about the answer I got. When the person answering says: ##1.## Is the map ##\mathfrak{q}\mapsto \mathfrak{q} A _\mathfrak{p}## from ##A\setminus \mathfrak{p}\to A_\mathfrak{p}##? But I don't understand what the author meant for the rest of the sentence in mathematical notation: ##2.## In the next statement where the author says: How is ##A\to...
The following are taken from the two sources, 1) from this online page and the book An Introduction to Module Theory by: Ibrahim Assem, Flavio U. Coelho. In the Abelian Categories chapter in the module theory text on page 157, right after presenting IV.2.21 Definition, the authors states "Image and coimage may or may not exist, but if they do, then they are unique up to isomorphism (because so are kernels and cokernels). Also in the reference url page above, the authors present two...
When decomposing a representation ##\rho## of a finite group ##G## into irreducible representations, we can find the number of times the representation contains a particular irrep ##\rho_0## through the character inner product $$ \langle \chi, \chi_0\rangle = \frac{1}{|G|} \sum_{g\in G} \chi(g) \chi_0(g)^*$$ where ##\chi## and ##\chi_0## are the characters of ##\rho## and ##\rho_0##, respectively. Since all group elements in the same conjugacy class have the same characters, this may be...
Back
Top