# Ideals in Q[x,y]

#### Peter

##### Well-known member
MHB Site Helper
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

[This is also posted on MHF]

#### Opalg

##### MHB Oldtimer
Staff member
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
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