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Is there a polynomial $f: \mathbb{N}^{2} \longrightarrow \mathbb{N}$ injective except for the $_2$ action?

This polynomial must be invariant under $S_2$ action, i.e $f(x,y)=f(y,x)$. However, if $(x,y)\not =(c,d)$ and $(x,y)\not =(d,c)$ then $f(x,y)\not = f(c,d)$.

I have tried with $f(x,y)= xy+y+x$, however this function is not injective:

$f(x,y)=x+y+xy=f(0,x+y+xy)$.

Tio Miserias
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1 Answers1

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The usual polynomial bijection $\mathbb N^2\to\mathbb N$ is $$p(u,v)=\frac{(u+v+1)(u+v)}{2}+u.$$ Taking $f(x,y)=p(xy,x+y)$ then works.

Substituting, and using $xy+x+y +1=(x+1)(y+1),$ we get:

$$f(x,y)=\frac{(x+1)^2(y+1)^2-(x+1)(y+1)}2+xy.$$

You could also take $f(x,y)=p(x+y,xy).$

If you need $f$ an integer polynomial, you could just double the above.

Any symmetric polynomial in two variables is of the form $h(xy,x+y).$ But you do not need to require $h$ to be injective on all of $\mathbb N^2.$ It is enough to have that $h$ is injective on $\left\{(u,v)\in\mathbb N^2\mid v^2-4u\text{ is a perfect square}\right\}.$ I’d be interested if there are examples $h$ which are not injective on all $\mathbb N^2.$

Thomas Andrews
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  • Just for curiosity, how do you build an injective polynomial $f:\mathbb{N}^{n} /S^{n}\longrightarrow \mathbb{N}$? – Tio Miserias Aug 02 '22 at 22:47
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    The same way. For example, for $3,$ $f(x,y,z)=p_3(xyz,xy+xz+yz,x+y+z),$ where $p_3(u,v,w)=p(u,p(v,w)),$ where $p$ is above. $p_3$ is again a bijection. For $n,$ you need the the “elementary symmetric polynomials.” @TioMiserias – Thomas Andrews Aug 02 '22 at 23:57