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Find a set of polynomials $\{P_1, \dots, P_n\}$, all of whose coefficients are real numbers, whose common zero set is the given set.

I know what a zero set is, but I think my confusion comes from what a common zero set is. Do I interpret that as the intersection of their zero sets? I'm pretty sure I have the first two parts of the question, but I'm going to include them anyway in case I"m off.

1) $\{(3, y) : y \in \mathbb{R}\}$ in $\mathbb{R}^2$

My answer: Choose $P_1 = x - 3$, so the zero set gives $x = 3$, and $y$ can be anything.

2) $\{(1, 2)\}$ in $\mathbb{R}^2$

My answer: Choose $P_1 = x - 1$ and $P_2 = y - 2$, so the common zero set is when $x = 1$ and $y = 2$, as desired.

3) $\{(1, 2), (0, 5)\}$ in $\mathbb{R}^2$

I was thinking about choosing $P_1 = x - 1$, $P_2 = y - 2$, $P_3 = x$, $P_4 = y - 5$, but then the intersection of all of the zero sets would be empty. My other idea was something like $P_1 = (x-1)x$, and $P_2 = (y-2)(y-5)$, but then the intersection of the zero sets includes the points $(1, 2), (0, 5), (1, 5),$ and $(0, 2)$ which is more than I want.

4) Generalize the method from Part (3) to any finite set of points in $\mathbb{R}^2$.

I imagine I can figure this out once I have part 3 done, but I wanted to include the whole problem for context.

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    $\begingroup$ It's generally easier to use square over $\mathbb R.$ For (2) you can choose $P_1=(x-1)^2+(y-2)^2.$ For (2), you can take a product of such formulae: $$P_1=\left((x-1)^2+(y-2)^2\right)\left(x^2+(y-5)^2\right)$$ In both cases, you only need one polynomial. $\endgroup$ – Thomas Andrews Apr 11 at 19:52
  • $\begingroup$ This generalizes to finite sets quite easily. $\endgroup$ – Thomas Andrews Apr 11 at 19:55

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