Timeline for On zeros of real polynomials in two variables
Current License: CC BY-SA 4.0
12 events
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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 |