# 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.

• Please use the word "prove" appropriately. In what you call "the easy part", you didn't prove anything; you applied a trivial definition. – Fly by Night Nov 15 '13 at 20:03
• Show that the quotient is an integral domain by exhibiting an explicit isomorphism with the polynomial ring in one variable. – Zhen Lin Nov 15 '13 at 20:35
• @ZhenLin Do you mean $k[x,y,z]/(x^2-y,x^3-z)\cong k[x,x^2,x^3]\cong k[t]$? – user75086 Nov 15 '13 at 21:23
• How do we know $Y = Z(x^2-y, x^3-z)$? – Al Jebr Aug 24 '20 at 17:52

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.

• There is no need to assume that $k$ is algebraically closed (and the OP didn't do it): in the last polynomial relation plug in $y=x^2$ and $z=x^3$ to find $f_3=0$ or $g_3=0$. – user89712 Nov 23 '13 at 9:44
• Element calculations make everything complicated ... $I$ is prime because obviously $k[x,y,z]/(I) = k[x]$. – Martin Brandenburg Nov 23 '13 at 16:40
• Thank you for you answer, I have a question, why can you use division algorithm? – user75086 Dec 3 '13 at 11:04
• Not necessarily. It is enough that $k[x,y]$ is a commutative ring, and that leading coefficient of $x^3-z$ is an unit in that ring. The leading coefficient is $-1$ (in variable $z$ of course). Proof is the same as in the case of fields. – Rafael Mrđen Dec 4 '13 at 9:01
• No, no calculation is needed. It is trivial to check that both sides represent the same functor, hence they are isomorphic (Yoneda). – Martin Brandenburg Dec 5 '13 at 0:53

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

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.

New contributor
LVMIMAOQ is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.