# Ideals in Q[x,y]

• Jul 26th 2013, 07:58 PM
Bernhard
Ideals in Q[x,y]
I am reading Chapter 1, Section 4, Cox et al "Ideals, Varieties and Algorithms"

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

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

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

$= h_3x + h_4y$

$$

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

Peter
• Jul 30th 2013, 04:00 PM
Deveno
Re: Ideals in Q[x,y]
There is "something missing" from your proof....can you see what it is? I'll give you a hint: what would happen if you tried to use $\mathbb{Z}[x,y]$ instead?

The most common way of proving two sets are equal in mathematics is to show they contain each other. Clearly, <x+y, x-y> is contained in the ideal generated by x and y, as both x+y and x-y are in <x,y> and <x+y,x-y> is the MINIMAL ideal containing x+y and x-y.

So the "hard" part is showing that <x,y> is contained in <x+y, x-y>, which amounts to showing that both x and y are elements of <x+y,x-y>. And:

x = (1/2)(x+y) + (1/2)(x-y) (note that we need to use rational numbers, integers wouldn't do!)

y = (1/2)(x+y) - (1/2)(x-y), both of which are clearly elements of <x+y, x-y>.
• Jul 31st 2013, 07:24 AM
Bernhard
Re: Ideals in Q[x,y]
Quote:

Originally Posted by Deveno
There is "something missing" from your proof....can you see what it is? I'll give you a hint: what would happen if you tried to use $\mathbb{Z}[x,y]$ instead?

The most common way of proving two sets are equal in mathematics is to show they contain each other. Clearly, <x+y, x-y> is contained in the ideal generated by x and y, as both x+y and x-y are in <x,y> and <x+y,x-y> is the MINIMAL ideal containing x+y and x-y.

So the "hard" part is showing that <x,y> is contained in <x+y, x-y>, which amounts to showing that both x and y are elements of <x+y,x-y>. And:

x = (1/2)(x+y) + (1/2)(x-y) (note that we need to use rational numbers, integers wouldn't do!)

y = (1/2)(x+y) - (1/2)(x-y), both of which are clearly elements of <x+y, x-y>.

Having to think a bit about your statement "... <x,y> is contained in <x+y, x-y>, which amounts to showing that both x and y are elements of <x+y,x-y> ... but I will work on this ...

Now, to address your question: " ... There is "something missing" from your proof....can you see what it is? I'll give you a hint: what would happen if you tried to use $\mathbb{Z}[x,y]$ instead?"

... In my 'proof' I really only showed that $ \ \ \subseteq \ \ $ since I showed the following

$f \in \ \Longrightarrow \ f = h_1 (x + y) + h_2 (x - y)$ where $h_1, h_2 \in \mathbb{Q}$

$\Longrightarrow \ f = h_3x + h_4y \ \Longrightarrow \ f \in $

so $ \ \ \subseteq \ \ $

It appears that the above reasoning would work for $\mathbb{Z} [x,y]$ ... is that correct?'

Following your proof for the second part (showing that <x,y> is contained in <x + y, x - y> ) this would not work for $\mathbb{Z} [x,y]$ since, as you point out, we need to use rational numbers.

Thanks again for the help!

Peter
• Jul 31st 2013, 09:06 AM
Deveno
Re: Ideals in Q[x,y]
Try proving a more general statement: if an ideal I of a ring R contains the set S, then I contains <S>, the ideal generated by S.