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Ideals in Q[x,y]

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am reading Chapter 1, Section 4, Cox et al "Ideals, Varieties and Algorithms"

Exercise 3(a) reads as follows:

In \(\displaystyle \mathbb{Q}[x,y] \) 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

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

Ideal generated by x, y is the ideal

\(\displaystyle h_3 x + h_4 y \) where \(\displaystyle h_3, h_4 \in \mathbb{Q}[x,y] \)

So

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

\(\displaystyle = (h_1 + h_2)x + (h_1 - h_2)y \)

\(\displaystyle = h_3x + h_4y \)

\(\displaystyle <x,y> \)

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

Peter

[This is also posted on MHF]
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
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 $\mathbb{Z}.$
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
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 Opalg


Really appreciate your help



Peter