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In particular, if R and S are sets of polynomials over a field, then the sets of points where polynomials of R and S are simultaneously zero are Z(R) and Z(S), respectively.

Then:

Z(R) $\cap$ Z(S) = Z(R $\cup$ S)

And

Z(I) $\cup$ Z(J) = Z(IJ) (where I and J are ideals generated by R and S, resp.)

It seems almost obvious that the intersection or union of two affine algebraic sets would still be an affine algebraic set, however I'm not sure how to go about proving the specific equality. I'm sure it's something basic and I'm just overthinking it.

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  • $\begingroup$ First equation amounts to saying points in the intersection of $Z(R)$ and $Z(S)$ are points vanishing on all polynomials in $R$ and $S$, which is clear. For the second one, start by thinking about what happens when $I$ and $J$ both consists of only one polynomial. $\endgroup$ Mar 19, 2016 at 2:44

1 Answer 1

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Claim: $Z(R)\cap Z(S) = Z(R\cup S)$

Proof: $x\in Z(R)\cap Z(S) \Leftrightarrow f(x)=0 \space$ $\forall f\in R$ and $\forall f\in S \Leftrightarrow x\in Z(R\cup S).$


Claim: $Z(I)\cup Z(J) = Z(IJ)$

Proof: "$\subset$": Let $x \in Z(I)$ (the case $x \in Z(J)$ is symmetric). Then $f(x)=0$ for all $f\in I$. Now, if $g\in IJ$, we can write $g=\sum_{j}f_jh_j$ for $f_j \in I $ and $h_j \in J$. Hence we have

$$g(x) = \sum_{j}f_j(x)h_j(x) = \sum_{j}0*h_j(x) = 0 \implies x\in Z(IJ).$$

"$\supset$": Let $x \in Z(IJ)$. Now, if $x\notin Z(I)$, there is a polynomial $f \in I$ such that $f(x) \neq 0$. But now for all $g \in J$ the polynomial $fg \in IJ$ so we must have

$$f(x)g(x)=0 \implies g(x)=0 \implies x\in Z(J).$$

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    $\begingroup$ If $g\in IJ$, we can write $g=\sum_if_ih_i$ where $f_i\in I$, $h_i\in J$. $\endgroup$
    – Hamou
    Feb 19, 2020 at 13:34
  • $\begingroup$ @Hamou Good catch, luckily the proof can be easily fixed by just adding the sum symbol :) $\endgroup$
    – ploosu2
    Feb 19, 2020 at 13:43

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