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

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!

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 $$f\notin J$$?
• @Masha You say that $$"\text{ I know that } \mathbf{V}(J) = x \in K^n."$$ Here $n=2$, and I write $(a,b)\in K^2$ in stead of $x\in K^2$ to avoid double use of the symbol $x$, which you already use for an indeterminate in the ring $k[x,y]$. – Servaes Apr 25 at 16:56
• @Masha No, because for $f=x^2-1$ you have $f(0,1)=1$. You should also make sure that $f\notin J$. – Servaes Apr 25 at 17:10
• You have misunderstood what $J$ is. It is the ideal generated by $x^2+y^2-1$ and $y-1$. That is to say, it is the set $$J=\langle x^2+y^2-1,y-1\rangle=\{(x^2+y^2-1)\cdot F+(y-1)\cdot G:\ F,G\in k[x,y]\}.$$ In particular $x^3-y^3+1\in J$ because $$x^3-y^3+1=(x^2+y^2-1)\cdot(x-y+1)+(y-1)\cdot(x^2-xy-x-y-2).$$ The lower the degree of a polynomial, the easier it is to check whether it is contained in $J$. So I would suggest looking at linear polynomials, or if those don't work, quadratic polynomials. – Servaes Apr 25 at 17:28
• @Masha And rightly so; it is in general not easy to check whether an element is contained in an ideal, given some generators. In this case $x+y-1\notin J$ but not for the reason you give; the degree is not lower than that of $y-1$, for example. Perhaps you want to ask a new question about determining whether a given element is contained in this ideal. – Servaes Apr 25 at 17:45
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 $$x\notin J$$