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Let $f=xy-z$ and $g=y-z$. Let $I = \langle f,g \rangle$.

a) Show associative variety is the union of two lines $L_1 $ and $L_2$.

b) Show $\mathbb{I}(L_1) \subset I $ and $\mathbb{I} (L_2) \subset I $.

Part (a)

Asking for a proof for arbitrary $f,g$ would be different. Think that it is just an example, so for

$$\begin{aligned} xy-z=0 \\ y-z=0 \end{aligned} $$

just sub in $y=z$ to $xy-z=0 \Rightarrow xy-y=0 \Leftrightarrow (x-1)y=0$.

From this guessing that the two lines are $x=1$ and $y=0$.

Part (b) (This is were I start to break if I did not before)

Def. of associative variety (same as affine variety??? think this is wrong its just I is in it)

$$ V(S) = \{ (x,y,z) \in R^ 3 :f(x,y,z)=0 \wedge g(x,y,z)=0\}. $$

Def. $$ \mathbb{I} (L_1) = \{ f \in R[x,y,z] : f(x,y,z)=0 , \forall (x,y,z)\in L_1 \}.$$

Something is wrong: $\mathbb{I}(L_1)$ is a set of polynomials and $V(S) $ is a set of points.

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    You proved that $V(I)=L_1\cup L_2$. We get $\mathbb I(V(I))=\mathbb I(L_1)\cap \mathbb I(L_2)$. Over an algebraic closed field this means that $\sqrt I=\mathbb I(L_1)\cap \mathbb I(L_2)$. But in your case $I$ is a radical ideal, so $I=\mathbb I(L_1)\cap \mathbb I(L_2)$, and the containments in (b) are in fact reversed. Moreover, some simple manipulations show that $I=(x-1,y-z)\cap(y,z)$ over any field. – 2017-03-09
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    I have no idea what you got wrong in the definitions of $V(S)$ and $\mathbb I(L_1)$. – 2017-03-09
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    @Ugo: It is the first time that I encounter the term "associative variety"; is it a translation of yours from a different language? If so, how does the original sound like? – 2017-03-10

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