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Let $R$ -- be an irreducible plane real algebraic curve (without isolated points).

Suppose that $(x,y)\in R\Leftrightarrow (x,-y)\in R.$

Question: could one find a polynomial $f(x,y)$ with zero set $R$ such that

$$\forall (x,y)\in{\mathbb R^2}\quad f(x,y)=f(x,-y)$$

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    $\begingroup$ The title asks for something not reflection-invariant, but the text inside asks for something invariant. $\endgroup$
    – P Vanchinathan
    Dec 8, 2015 at 4:59

1 Answer 1

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Let $f$ be any polynomial whose zero set is $R$. Then $F(x,y)=f(x,y)f(x,-y)$ works.

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