# Affine Algebraic Variety.

• Mar 2nd 2010, 12:08 PM
skamoni
Affine Algebraic Variety.
Find an ideal $\displaystyle I$ in $\displaystyle K[x,y]$ such that $\displaystyle V(I)$ consists of the point $\displaystyle (1,1)$ and the lines $\displaystyle x=0$ and $\displaystyle y=x+1$

Any help would be greatly appreciated...
• Mar 2nd 2010, 12:16 PM
Opalg
• Mar 2nd 2010, 01:43 PM
NonCommAlg
Quote:

Originally Posted by Opalg

unfortunately tonio's answer is not correct because in his example we have $\displaystyle V(I)=\{(0,1)\}.$ recall that for any ideal $\displaystyle I$ of $\displaystyle K[x,y]$ we define $\displaystyle V(I)=\{p \in K^2: \ f(p)=0, \ \forall f \in I \}.$

anyway, to answer the question, you can choose $\displaystyle I=\langle x(y-x-1)(x-1), \ x(y-x-1)(y-1) \rangle.$ then it's easy to see that $\displaystyle V(I)$ is the set you're looking for.
• Mar 2nd 2010, 07:03 PM
tonio
Quote:

Originally Posted by NonCommAlg
unfortunately tonio's answer is not correct because in his example we have $\displaystyle V(I)=\{(0,1)\}.$ recall that for any ideal $\displaystyle I$ of $\displaystyle K[x,y]$ we define $\displaystyle V(I)=\{p \in K^2: \ f(p)=0, \ \forall f \in I \}.$

anyway, to answer the question, you can choose $\displaystyle I=\langle x(y-x-1)(x-1), \ x(y-x-1)(y-1) \rangle.$ then it's easy to see that $\displaystyle V(I)$ is the set you're looking for.

Why "unfortunately"? My idea was wrong but, perhaps, it could push the OP in the right direction. Anyway, she/he now has the correct answer and can compare. Nothing lost, all is profit. Thanx for the correct answer.

Tonio