# Thread: Algebraic Varieties and Ideals

1. ## Algebraic Varieties and Ideals

In Chapter 1, Section 4 of Cox et al "Ideals, Varieties and Algorithms, Exercise 3(c) reads as follows:

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

$< 2x^2 + 3y^2 -11, x^2 - y^2 - 3> \ = \ $

Any help with this problem would be appreciated.

Peter

2. ## Re: Algebraic Varieties and Ideals

I will give you one half:

$x^2 - 4 = \dfrac{1}{5}(2x^2 + 3y^2 - 11) + \dfrac{3}{5}(x^2 - y^2 - 3)$

$y^2 - 1 = \dfrac{1}{5}(2x^2 + 3y^2 - 11) - \dfrac{2}{5}(x^2 - y^2 - 3)$

3. ## Re: Algebraic Varieties and Ideals

Originally Posted by Deveno
I will give you one half:

$x^2 - 4 = \dfrac{1}{5}(2x^2 + 3y^2 - 11) + \dfrac{3}{5}(x^2 - y^2 - 3)$

$y^2 - 1 = \dfrac{1}{5}(2x^2 + 3y^2 - 11) - \dfrac{2}{5}(x^2 - y^2 - 3)$

Thanks for the guidance and help Deveno

OK so we want to show that in $\mathbb{Q}[x,y]$ we have $<2x^2 + 3y^2 - 11, x^2 - y^2 - 3> \ = \ $

First show that $ \ \subseteq \ <2x^2 + 3y^2 - 11, x^2 - y^2 - 3>$

From Deveno's equations above it follows that $x^2 - 4, y^2 - 1 \in \ <2x^2 + 3y^2 - 11, x^2 - y^2 - 3>$

Thus, $ \ \subseteq \ <2x^2 + 3y^2 - 11, x^2 - y^2 - 3>$ ... ... ... (1)

Now show $<2x^2 + 3y^2 - 11, x^2 - y^2 - 3> \ \subseteq \ $

We can write $2x^2 + 3y^2 - 11 \ = \ 2(x^2 - 4) + 3(y^2 - 1)$

and $x^2 - y^2 - 3 \ = \ (x^2 - 4) - (y^2 - 1)$

Thus $2x^2 + 3y^2 - 11, x^2 - y^2 - 3 \in $

Thus $<2x^2 + 3y^2 - 11, x^2 - y^2 - 3> \ \subseteq \ $ ... ... ... (2)

Equations (1) (2) $\ \Longrightarrow \ \mathbb{Q}[x,y]$ we have $<2x^2 + 3y^2 - 11, x^2 - y^2 - 3> \ = \ $

4. ## Re: Algebraic Varieties and Ideals

Note this shows the generators of ideals in Q[x,y] aren't even unique up to units. Also, this is just high-school algebra in fancier clothes.