The twisted cubic is an affine variety.

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.

Solution 1:

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.

Solution 2:

There is an obvious isomorphism $Y \cong \mathbb{A}^1$. This proves everything else.

Solution 3:

Expand a little on Martin's answer.

Define $f:X \to Y$ as $$f(x) = (x, x^2, x^3)$$ Define $g:Y \to X$ as $$g(x,y,z) = x$$

Then $f^{-1} = g$, $f$ is an isomorphism between $A^1$ and Y.

Now by p.g.29 of Shafarecich's Basic algebraic Geometry 1, there exists an isomorphism between coordinate rings of $A^1$ and $ Y$, which means $$k[x] \cong k[x,y,z]/((x^2-y),x^3-z)$$

Since k[x] is an integral domain, $k[x,y,z]/((x^2-y),x^3-z))$ is an integral domain, which means that $((x^2-y),x^3-z))$ is irreducible.