# Find a set of polynomials whose common zero set is $\{(1, 2), (0, 5)\}$.

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.

• 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. – Thomas Andrews Apr 11 at 19:52
• This generalizes to finite sets quite easily. – Thomas Andrews Apr 11 at 19:55