Equalities of ideals in Q[x,y]

Question

I'm trying to prove something about polinomyal ideals. So I have to use this proposition:

Let $I \subset k[x_{1}, \ldots, x_{n}] $ be an ideal, and let $f_{1}, \ldots, f_{s} \in k[x_{1}, \ldots, x_{n}].$ Then these are equivalent:

$(i) f_{1}, \ldots, f_{n} \in I.$

$(ii) \left<f_{1}, \ldots, f_{n}\right> \subseteq I.$

And I have to prove this equality in $\mathbb{Q} [x,y]:$

$\left< x+y, x-y\right> = \left< x,y \right>$.

I tried to do this first $\left<x+y, x-y \right> \subset \left< x,y \right>$ like this:

Let $\left< x,y \right> = I \in \mathbb{Q}[x,y] $ an ideal, and let $x+y, x-y \in \mathbb{Q}[x,y].$ So I want to say that $x+y, x-y \in \left< x,y \right> $ and here is where I don't know if I'm wrong or what:

First, I think, is obvious that $x+y \in \left< x,y \right>$, since $ \left< x,y \right> = h_{1}(x) + h_{2}(y),$ for $ h_{1},h_{2} \in \mathbb{Q}[x,y] $, then I say that $h_{1}=h_{2}=1$ so $x+y \in \left< x,y \right> $ and then I use the same argument for $x-y$ but with $h_{2}=-1.$ And then I use the proposition to say that $\left< x+y, x-y\right> = \left< x,y \right>$.

And I don't know if this is correct, if not, I appreciate if someone could help me proving that equality. Thanks!

Answer 1

To prove that you could do something like this:

Let $I_1=\left<x+y,x-y\right>$ and $I_2=\left<x,y\right>$

Using the proposition, if you prove that $x,y\in I_1$ you get $I_2\subseteq I_1$. If you prove that $x+y,x-y\in I_2$ then you prove $I_1\subseteq I_2$. If you prove both you get $I_1=I_2$.

For example, to prove $x+y\in I_2$. As $x,y\in I_2$ and $I_2$ is an additive subgroup $\mathbb{Q}[x,y]$, we get $x+y\in I_2$. The rest of it can be proved the same way.

So, it's exactly as you say, but you should order it a little. It will help you if you think what is it that you want to prove, and what does the proposition you use mean.

In this case, furthermore, it is not needed the complete concept of an ideal to show the result, it's only needed that it's an additive subgroup ("Linear combinations" are not totally needed, just sums and additive inverses)