Square is not an algebraic set.

Question

I am trying to show that square with vertices at $(\pm1,0)$ and $(0,\pm1)$ is not a zero set of a polynomial in $\mathbb{R}[x,y]$. Clearly, square is the zero set of a function $$ f: \mathbb{R}^2\to \mathbb{R}; $$ $$ (x,y)\mapsto |x|+|y|-1, $$ which is not a polynomial. But how to show that there is no polynomial which has square as the zero set?

Answer 1

If the square were the zero-set of a (finite degree) polynomial $f$ in $x$ and $y$, then the line from $-1$ to $1$ on the $x$-axis would be the zero set of the polynomial $g(x) = f(x, 0)$. That's not possible (by the fundamental theorem of algebra).