Affine Algebraic Variety.

**skamoni** · March 2nd 2010, 12:08 PM

Find an ideal $I$ in $K[x,y]$ such that $V(I)$ consists of the point $(1,1)$ and the lines $x=0$ and $y=x+1$

Any help would be greatly appreciated...

**Opalg** · March 2nd 2010, 12:16 PM

See this thread.

**NonCommAlg** · March 2nd 2010, 01:43 PM

Originally Posted by Opalg

See this thread.

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

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

**tonio** · March 2nd 2010, 07:03 PM

Originally Posted by NonCommAlg

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

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

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