Let J be an ideal. Find a function in I(V(J)) such that the function f is not in J

Question

Let $J$ be the ideal $\langle x^2+y^2-1,y-1\rangle$. Find $f \in \textbf{I}(\textbf{V}(J))$ such that $f \not \in J$.

I'm confused on a number of aspects here. Firstly, how do I find $\mathbf{V}(J)$ and then following that, how do I find $\textbf{I}(\textbf{V}(J))$.

I know that $\mathbf{V}(J) = x \in K^n$ (where K is an affine space) such that $f(x) = 0$ and $ f\in J$. I also understand that $\textbf{I}(\textbf{V}(J)) = f(x)$ such that $f(x) = 0$ if $x \in V$

So does this mean that $\mathbf{V}(J)$ is all $x's$ where $x^2+y^2-1 =0$ and $y-1 = 0$? Because that would mean that $y = 1$ and then $x = 0$. Then, if that is correct, we would have to find $\mathbf{I}(0)$ so that would be where $f(x) = 0$ and $f(x) \in V$. But wouldn't that mean that $f(x)$ is any polynomial with no constants?

Please let me know where I am mistaken and offer any hints/suggestions/solutions. Thank you!

Answer 1

You have understood correctly what $V(J)$ is, but not what $I(V(J))$ is, probably because your double use of the symbol $x$ is a bit confusing. To clarify, set $$f_1:=x^2+y^1-1\in K[x,y]\qquad\text{ and }\qquad f_2:=y-1\in K[x,y].$$ Here $n=2$, and you have correctly found that $$V(J)=\{(a,b)\in K^2:\ f_1(a,b)=f_2(a,b)=0\}=\{(0,1)\}.$$ It then follows that \begin{eqnarray*} I(V(J))&=&\{f\in K[x,y]:\ (\forall (a,b)\in V(J))(f(a,b)=0)\}\\ &=&\{f\in K[x,y]:\ f(0,1)=0\}. \end{eqnarray*} Can you now find $f\in I(V(J))$ such that $f\notin J$?

Answer 2

Nullstellensatz tells you that $I(V(J))=rad(J)$. Note that $x^2=(x^2+y^2-1)-(y-1)(y+1)$, so $x^2\in J$ which implies that $x\in rad(J)$. Note $x\in I(V(J))$ but $x\notin J$