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Apr 19 at 14:43 comment added Jérémy Blanc @IosifPinelis: yes, I had seen it before writing my comment. I completely agree with him. My comment gives a negative answer to the question asked here. It is certainly not what the person asking had in mind, so it shows that the question needs to be clarified.
Apr 18 at 12:20 comment added Iosif Pinelis @JérémyBlanc : Please see the comment by YCor.
Apr 18 at 6:01 comment added Jérémy Blanc If $Q(x,y)=x^2+y^2+1$, then $Q$ is irreducible and defines a conic, that has no real point. Then, $P(x,y)=x^4+y^2+6$ is a polynomial that is zero on the real points of the conic but is not a multiple of $P$.
Apr 12 at 20:34 comment added user526214 Now, in Hilbert's Nullstellensatz it is assumed that the field $\mathbb{K}$ is algebraically closed. How is this reconciled with the fact that in the question the polynomial $P(x,y)$ has real coefficients and the variables $x, y$ are also real?
Apr 12 at 20:29 comment added user526214 I am referring to the class of polynomials with real coefficients defined in $\mathbb{R}^2$ that cancel out on an irreducible conic in $\mathbb{R}^2$
Apr 12 at 20:18 comment added YCor Set of zeros in what? In $\mathbf{R}^2$, the set of zero of many unrelated polynomials is empty.
Apr 12 at 20:09 answer added Iosif Pinelis timeline score: 2
Apr 12 at 19:21 comment added user526214 What if we assume that the conic $Q(x,y)$ is irreducible?
Apr 12 at 19:00 comment added Iosif Pinelis What about $P(x,y)=x-y$ and $Q(x,y)=(x-y)^2$?
Apr 12 at 18:09 comment added Stanley Yao Xiao 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.
S Apr 12 at 18:00 review First questions
Apr 12 at 20:25
S Apr 12 at 18:00 history asked user526214 CC BY-SA 4.0