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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!

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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 $f\notin J$?

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    $\begingroup$ @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 $k[x,y]$. $\endgroup$ – Servaes Apr 25 at 16:56
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    $\begingroup$ @Masha No, because for $f=x^2-1$ you have $f(0,1)=1$. You should also make sure that $f\notin J$. $\endgroup$ – Servaes Apr 25 at 17:10
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    $\begingroup$ 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. $\endgroup$ – Servaes Apr 25 at 17:28
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    $\begingroup$ @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. $\endgroup$ – Servaes Apr 25 at 17:45
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    $\begingroup$ @Masha Also, if your question has been answered by one of the given answers, consider accepting an answer so it does not stay in the unanswered queue. This also goes for your other questions. $\endgroup$ – Servaes Apr 25 at 17:46
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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 $x\notin J$

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