Affine Algebraic Sets - Equivalence Statement

Question

I am just learning about affine algebraic sets and am struggling with the following problem.

Consider the algebraic sets; $A=\lbrace (x,y) \mid y=x^2\rbrace$ (parabola) and $B=\lbrace (x,y) \mid y=0\rbrace$ (the $x$-axis). Show that $$\mathcal{I}(A\cap B)=\mbox{rad}(\mathcal{I}(A) + \mathcal{I}(B))$$ Comment: This formula is in fact true in general, for any two algebraic sets.

Basic definitions:

Let $k$ be a field. Let $S$ be a subset of functions in the coordinate ring $k[\mathbb{A}^n]$. Let $A$ be any subset of $\mathbb{A}^n$. Then

• $\mathcal{Z}(S)=\{a_1,a_2,\cdots,a_n\in\mathbb{A}^n|f(a_1,a_2,\cdots,a_n)=0 \hspace{3mm}\forall f\in S\}$
• $\mathcal{I}(A)=\{f\in k[x_1,\cdots,x_n|f(a_1,\cdots,a_n)=0\hspace{3mm}\forall(a_1,\cdots,a_n)\in A\}$
• rad$(I)=\{a\in R|a^k\in I\text{ for some }k\geq 1\}$

Some Known Properties:

• $S\subseteq \mathcal{I}(\mathcal{Z}(S))$

• $\mathcal{I}(A\cup B)=\mathcal{I}(A)\cap\mathcal{I}(B)$

• $A\subseteq B\Rightarrow \mathcal{I}(B)\subseteq\mathcal{I}(A)$

• $\mathcal{Z}(I)\cup\mathcal{Z}(J)=\mathcal{Z}(IJ)$

Note: I have more properties, but I thought these were some that could be helpful for this problem.

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My Attempt:

Assume the hypothesis. By definition, $A=\mathcal{Z}(y-x^2)$ and $B=\mathcal{Z}(y)$. First we will determine $\mathcal{I}(A\cap B)$. Notice \begin{eqnarray*} A\cap B &=& \lbrace (x,y) \mid y=x^2\rbrace \cap \lbrace (x,y) \mid y=0\rbrace\\ &=& \{(x,y)|y=x^2\text{ and }y=0\}\\ &=&\{(x,0)|x^2=0 \}\\ &=&\{(0,0)\} \end{eqnarray*} Therefore by the example shown in class (I can explain the reasoning if not obvious), we know \begin{eqnarray*} \mathcal{I}(A\cap B) &=& \mathcal{I}(\{(0,0)\})\\ &=& \langle x-0,y-0\rangle\\ &=& \langle x,y\rangle \end{eqnarray*}

Notice $\mathcal{I}(A)=\langle y-x^2\rangle $ and $\mathcal{I}(B)=\langle y\rangle$. Then \begin{eqnarray*} \mathcal{I}(A)+\mathcal{I}(B)&=&\langle y-x^2,y\rangle\\ &=&\{f_1(y-x^2)+f_2y|f_i\in k[x,y]\}\\ &=& \{f_1y-f_1x^2+f_2y|f_i\in k[x,y]\}\\ &=& \{-f_1(x^2)+(f_1+f_2)y|f_i\in k[x,y]\}\\ &=& \langle x^2, y\rangle \end{eqnarray*} Recall $k[x,y]/\langle x,y\rangle\cong k$. Because $k$ is a field, we know $k[x,y]/\langle x,y\rangle$ is a field. Therefore $\langle x,y\rangle$ is maximal. Notice $$ \langle x,y\rangle ^2\subseteq \langle x^2,y\rangle \subseteq \langle x,y\rangle $$ Then by proposition, we know rad$(\langle x^2,y\rangle)=\langle x,y\rangle$.

Hence $\mathcal{I}(A\cap B)=\mbox{rad}(\mathcal{I}(A) + \mathcal{I}(B))$.

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Am I on the right track here? I keep second guessing myself. Any input would be much appreciated. Thanks.

Answer 1

we have $I(A)+I(B)=(y-x^2,y)=(x^2,y)$, then $rad (I(A)+I(B))=(x,y)$, therefore $I(A \cap B)=rad({I(A)+I(B)}) $

Answer 2

So you only need to show that $\operatorname{rad} (y-x^2, y) = (x, y)$.

Hint: Show that $\operatorname{rad} I$ is an ideal for all ideals $I$ in any ring $R$. (Maybe you already know that?) Then it suffices to show that:

• $(x, y)$ is a maximal ideal in $k[\mathbb A^2] = k[x,y]$,
• $(y - x^2, y)$ is not all of $k[x,y]$, … so
• $\operatorname{rad} (y - x^2, y)$ is not all of $k[x,y]$, and
• $x, y ∈ \operatorname{rad} (y-x^2, y)$.

Why does it suffice?