Mathematics Asked by IV_ on November 27, 2020

Let $A$ be an arbitrary algebraic function in dependence of the two variables $x$ and $y$.

Let’s consider the equation $A(x,y)=0$ (equation 1) involving $x$ and $y$. For which kinds of equation 1 can we decide if there is an **irreducible** polynomial equation whose solution set contains the solution set of equation (1)?

I already know that we can multiply both sides of the equation by the denominators of both sides to get an equivalent or non-equivalent equation. And the solution set of this new equation does contain the equivalent solution set. At least in the one-variable case, the equivalent solution set yields an equivalent polynomial equation.

I also see: We can first separate the denominators and the irreducible factors.

The following is a simple first case of corresponding radical equations.

Let

$A_0inoverline{mathbb{Q}}[x,y]$ algebraic,

$Pinoverline{mathbb{Q}}[t]setminusoverline{mathbb{Q}}$,

$P_1inoverline{mathbb{Q}}[x,y]setminus(overline{mathbb{Q}}[x]cupoverline{mathbb{Q}}[y])$,

$rinmathbb{Q}setminusmathbb{Z}$

so that $P((P_1(x,y))^r)$ and $A_0(x,y)$ are coprime,

and let $P^{-1}$ denote a suitable partial inverse of $P$.

$$frac{P((P_1(x,y))^r)}{A_0(x,y)}=0$$

$$P((P_1(x,y))^r)=0$$

$$(P_1(x,y))^r=P^{-1}(0)$$

$$P_1(x,y)=(P^{-1}(0))^frac{1}{r}$$

$$P_1(x,y)-(P^{-1}(0))^frac{1}{r}=0$$

The equation is irreducible if $P$ is irreducible and $P_1(x,y)-(P^{-1}(0))^frac{1}{r}$ is irreducible for all suitable partial inverses $P^{-1}$.

Correct answer by IV_ on November 27, 2020

There may not be. Let A(x,y) = x/y.

A(x,y) = 0 iff x = 0 and y /= 0.

No polynomial can have that property.

Answered by William Elliot on November 27, 2020

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