Identifying rings of bounded functions

Question

Let $S$ be the subset of $\mathbb{R}^2$ defined by the inequalities $$0\le\sqrt{2}x-y<1.$$ The set of polynomials in the ring $\mathbb{R}[x,y]$ that are bounded on $S$ obviously forms a subring $B$. I suspect that $B=\mathbb{R}[\sqrt{2}x-y]$. Can anyone suggest a proof or a counterexample?

Answer 1

$X \mapsto \sqrt{2}x - y$ and $Y \mapsto y$ defines an isomorphism from $\mathbb{R}[X, Y]$ to $\mathbb{R}[x, y]$. In $\mathbb{R}[X, Y]$ you are asking which polynomials are bounded in the strip $0 \le X < 1$. But if $f = f(X, Y)$ is bounded for $X = 1/2$ say, all the coefficients of $Y$ must vanish, hence $f \in \mathbb{R}[X]$. Your claim now follows by pulling this back into $\mathbb{R}[x, y]$.

Answer 2

The inclusion $\mathbb R[\sqrt 2 x-y]\subseteq B$ is clear. Consider the algebra homomorphism $\phi\colon \mathbb R[x,y]\to\mathbb R[t]$ given by $x\mapsto t$, $y\mapsto \sqrt 2 t$. The kernel of $\phi$ is precisely $\mathbb R[\sqrt 2 x-y]$. Let $f\in B$. We want to show $\phi(f)=0$. Assume $\phi(f)\ne 0$. Then $\phi(f)$ is not bounded and for each $M$ we can pick $\tau_M\in\mathbb R$ with $|\phi(f)(\tau_M)|>M$. But then also $|f(\sqrt 2\tau_M,\tau_M)|=|\phi(f)(\tau_M)|>M$ with $(\sqrt 2\tau_M,\tau_M)\in S$, i.e. $f$ is not bounded on $S$.