1. ## Ideals in Q[x,y]

I am reading Chapter 1, Section 4, Cox et al "Ideals, Varieties and Algorithms"

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.

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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

2. ## 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 $\displaystyle \mathbb{Z}[x,y]$ 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>.

3. ## Re: Ideals in Q[x,y]

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 $\displaystyle \mathbb{Z}[x,y]$ 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 $\displaystyle \mathbb{Z}[x,y]$ instead?"

... In my 'proof' I really only showed that $\displaystyle <x + y, x - y> \ \ \subseteq \ \ <x, y>$ since I showed the following

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

$\displaystyle \Longrightarrow \ f = h_3x + h_4y \ \Longrightarrow \ f \in <x. y>$

so $\displaystyle <x + y, x - y> \ \ \subseteq \ \ <x, y>$

It appears that the above reasoning would work for $\displaystyle \mathbb{Z} [x,y]$ ... 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 $\displaystyle \mathbb{Z} [x,y]$ since, as you point out, we need to use rational numbers.

Thanks again for the help!

Peter

4. ## 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.