Is there a subset of natural numbers that has the same additive and multiplicative structure as the set of ordered pairs of natural numbers under the classical operations of addition and multiplication i.e. is there a (primitive recursive) bijection $f: N \times N \rightarrow U \subset N$ with $f(x,y)+f(u,v)=f(x+u,y+v)$ and $f(x,y) \cdot f(u,v)=f(x \cdot u, y \cdot v)$?


Obviously $f(x,y)=2^x \cdot 3^y$ is not a solution: $2^1 \cdot 3^1 + 2^1 \cdot 3^1 = 2^{1+1} \cdot 3^{1+0}$.



Is there a bijection $f: N \times N \rightarrow U \subset N$ with $f(x,y)+f(u,v)=f(x+u,y+v)$ and $f(x,y) \cdot f(u,v)=f(x \cdot u, y \cdot v)$?
There is no bijection satisfying the addition and multiplication
identities. In fact, even
without the assumption that $f$ is a bijection
there
are only three functions satisfying both identities: for all $x$ and $y$
either 
$f\_1(x,y) = x$, $f\_2(x,y) = y$ or $f\_3(x,y) = 0$.


To see this, use the addition identity repeatedly to deduce that
(i) $f(a,b)$ $=$ $f(a+0,0+b)$ $=$ $f(a,0)+f(0,b) = a\cdot f(1,0)$ $+$ $b\cdot f(0,1)$. In particular,
(ii) $f(1,1) = f(1,0)+f(0,1)$. Now the multiplication identity
yields that $f(1,1)\cdot f(1,1) = f(1,1)$, so $f(1,1)\in \{0,1\}$,
since $0$ and $1$ are the only multiplicative idempotents.
If $f(1,1)=0$, then by (ii) we get $f(1,0)+f(0,1)=0$, forcing $f(1,0)=0=f(0,1)$. Hence by (i) we
get $f(a,b)=0$ for any $a$ and $b$. ($f=f\_3$).


On the other hand, if $f(1,1)=1$, then by (ii) we get either
(x) $f(1,0)=1$ and $f(0,1) = 0$ or (y) $f(1,0)=0$ and $f(0,1) = 1$.
If (x), then (i) yields $f(a,b)=a$ for any $a$ and $b$ ($f=f\_1$),
while if (y),
then (i) yields $f(a,b)=b$ for any $a$ and $b$ ($f=f\_2$).


