Show that in $\mathbb{Z}[x,y]$, $\langle x+y,x-y\rangle\subsetneq\langle x,y\rangle$

Question

I have shown that in $\mathbb{Q}[x,y]$, $$\langle x,y\rangle=\langle x+y,x-y\rangle,$$ by stating that $$\{x,y\}\subset\langle x+y,x-y\rangle$$ because both $x$ and $y$ can be written as elements of the latter; $$x=\frac{1}{2}(x+y)+\frac{1}{2}(x-y), \quad y=\frac{1}{2}(x+y)-\frac{1}{2}(x-y).$$ (The other direction is pretty trivial).
I think that simply saying $1/2\notin \mathbb{Z}$ is not good enough of an argument to show that $\langle x+y,x-y\rangle\subsetneq\langle x,y\rangle \text{in } \mathbb{Z}[x,y]$, because what if there is another way to write $x$ as an element of $\langle x+y,x-y\rangle$ ?
I have tried proving that assuming $x \in \langle x+y,x-y\rangle$ leads to a contradiction (mainly by taking the degrees of the polys), but I haven't had any success. Help?

Answer 1

Consider the homomorphism $\phi:\Bbb Z[x,y]\to\Bbb Z/2\Bbb Z$ given by $\phi(x)=\phi(y)=1$. Then $\langle x+y,x-y\rangle\subseteq\ker\phi$ but $\langle x,y\rangle\not\subseteq\ker\phi$.

Answer 2

I tried to show it using 'mundane' methods as follows.

Suppose there exist $f(x,y), g(x,y) \in \mathbb{Z}[x,y]$ such that $x = (x+y)f + (x-y)g$. Observe that $f \neq g$, otherwise you factor $x = (2x)f$ and no way you can get rid of the $2$. Then this can be rewritten as $x = x(f+g) + y(f-g)$, and then we relabel this as $x = x \cdot h(x,y) + y \cdot k(x,y)$ for $h, k \in \mathbb{Z}[x,y]$, with $h = f+g, k = f-g$. Note that $k(x,y) = f - g \neq 0$.

Now, it is clear that the constant term of $h(x,y)$ must be 1, because there is no other way for $x$ term to appear on the RHS. But recall that $h(x,y) = f(x,y) +g(x,y)$. Thus, the constant term of $f + g$ is 1. But then the constant term of $f-g$ cannot be zero! (because of constant terms of f and g sum to one, and their difference is zero, you can solve to show that the constant terms are both equal 0.5). Then, if we look back at the equation $x = x \cdot (f+g) + y \cdot (f-g)$, we see that this must be a contradiction, because on the RHS there is a term $y \cdot (\text{const term of }f-g)$ which cannot be cancelled out by anything.