Let $P(x,y)$ be a polynomial with real coefficients in two real variables $x,y$ such that the set of zeros of $P(x,y)$ is the real conic curve $Q(x,y)=0$. Will it be true that there exists a polynomial $R(x,y)$ such that $P(x,y)=Q(x,y)R(x,y)$? How can this be proven?"
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$\begingroup$ I am not sure if your question is formulated in a precise manner, but it seems what you want is a basic application of Bezout's theorem. $\endgroup$– Stanley Yao XiaoCommented Apr 12 at 18:09
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$\begingroup$ What about $P(x,y)=x-y$ and $Q(x,y)=(x-y)^2$? $\endgroup$– Iosif PinelisCommented Apr 12 at 19:00
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$\begingroup$ What if we assume that the conic $Q(x,y)$ is irreducible? $\endgroup$– user526214Commented Apr 12 at 19:21
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$\begingroup$ Set of zeros in what? In $\mathbf{R}^2$, the set of zero of many unrelated polynomials is empty. $\endgroup$– YCorCommented Apr 12 at 20:18
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$\begingroup$ I am referring to the class of polynomials with real coefficients defined in $\mathbb{R}^2$ that cancel out on an irreducible conic in $\mathbb{R}^2$ $\endgroup$– user526214Commented Apr 12 at 20:29
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1 Answer
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If all zeros of $Q$ in $\mathbb C^2$ are zeros of $P$, then, by Hilbert's Nullstellensatz, $Q$ divides $P^r$ for some natural $r$.
If $Q$ is also irreducible, then it follows that $Q$ divides $P$.
If $Q$ is not irreducible, then $Q$ does not have to divide $P$, as noted in my comment.
Now, how to answer the original question, for real (rather than complex) zeros -- assuming $Q$ is irreducible and the set of real zeros of $Q$ is of cardinality $>1$?
Using an affine transformation, without loss of generality $Q(x,y)$ is $x^2+y^2-1$ or $x^2-y^2-1$ or $y-x^2$.
If $Q(x,y)=x^2+y^2-1$, then $p_1(t):=P(\cos t,\sin t)=0$ for all real $t$ and hence for all complex $t$, since $p_1$ is an entire function. So, in this case, all complex zeros of $Q$ are zeros of $P$.
If $Q(x,y)=x^2-y^2-1$, then $p_2(t):=P(\cosh t,\sinh t)=0$ for all real $t$ and hence for all complex $t$, since $p_2$ is an entire function. So, all complex zeros of $Q$ are zeros of $P$ in this case as well.
If finally $Q(x,y)=y-x^2$, then $p_3(t):=P(t,t^2)=0$ for all real $t$ and hence for all complex $t$, since $p_3$ is an entire function. So, all complex zeros of $Q$ are zeros of $P$ in this latter case as well.
So, it remains to use the first two sentences of this proof.
A more general result is now available.
answered Apr 12 at 20:09
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$\begingroup$ In my question I assume that the conic $Q(x,y)$ defined in $\mathbb{R}^2$ is real and irreducible. Now, what if the zeros of $P(x,y)$ and $Q(x,y)$ coincide in $\mathbb{R}^2$ but not in $\mathbb{C}^2$ as stated to apply Hilbert's Nullstellensatz? So is there a counterexample to my question? $\endgroup$ Commented Apr 12 at 20:55
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1$\begingroup$ @user526214 : I have addressed this concern. $\endgroup$ Commented Apr 12 at 21:12
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1$\begingroup$ @user526214 : Do you have a further response? $\endgroup$ Commented Apr 14 at 2:10