Let $Y=\{(t,t^2,t^3)\mid t\in k\}$ be the twisted cubic curve. I'm trying to prove this curve is a variety, i.e., it's irreducible and affine algebraic set.
The easier part is to prove the twisted cubic curve is an affine algebraic set $(Y=Z(x^2-y,x^3-z))$.
I don't know how to prove that $Y$ is irreducible, I'm trying to prove that $(x^2-y,x^3-z)$ is prime, I think if I do this I proved what I want, but I found this hard to prove.
I need help to finish this question.
Thanks a lot.
Best Answer
Let's first prove that ideal $I:=(x^2-y, x^3-z)$ is prime. Suppose $f\cdot g \in I$. Using obvious isomorphisms $k[x,y,z] \cong (k[x,y])[z]$, $k[x,y] \cong (k[x])[y]$ and division algorithm, we have $$ f(x,y,z)=(x^3-z)f_1(x,y,z) + (x^2-y)f_2(x,y) + f_3(x), $$ $$ g(x,y,z)=(x^3-z)g_1(x,y,z) + (x^2-y)g_2(x,y) + g_3(x). $$ Now we have $f_3(x) \cdot g_3(x) \in I$, therefore $$f_3(x) \cdot g_3(x)=(x^3-z)h_1(x,y,z) + (x^2-y)h_2(x,y,z).$$ Insert $(x,y,z)=(t,t^2,t^3)$ and get $f_3(t) \cdot g_3(t) =0$ for all $t \in k$. If $k$ is algebraically closed (therefore infinite), we have $f_3 \cdot g_3 = 0$, so $f_3 = 0$ or $g_3= 0$. Then $f \in I$ of $g \in I$, so $I$ is prime (and therefore radical). We have $I(Y)=I(V(I)) = \operatorname{Rad}(I) = I$, which is prime. So $Y$ is irreducible.