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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.
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 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.
Here is a way to show that $Y$ is irreducible with little computation: note that since $Y$ is closed (as the set of zeros of two polynomials) we have $Y=Z(I(Y))$, where $I(Y)$ is the ideal of $Y$. The only thing to check is that $I(Y)$, is prime. But it is easy to see that $I(Y)$ is the kernel of a surjective ring homomorphism $\phi:k[x,y,z]\to k[t]$, namely the homomorphism such that $\phi(x)=t,\phi(y)=t^2,\phi(z)=t^3$. Let's check this in detail, even though this is a bit tautological: To say that $p\in k[x,y,z]$ belongs to $\ker\phi$ means that the polynomial $p(t,t^2,t^3)\in k[t]$ is the zero polynomial. This implies, obviously, that for all $t\in k$ we have $p(t,t^2,t^3)=0$, i.e. that $p\in I(Y)$. Conversely, if $p\in I(Y)$ then the polynomial $\phi(p)=p(t,t^2,t^3)\in k[t]$ has any $t\in k$ as a root, which means that $\phi(p)=0$. Hence $I(Y)=\ker\phi$.
It follows that the ring $A(Y)=k[x,y,z]/I(Y)$ is isomorphic to $k[t]$, which is integral, hence $I(Y)$ is prime, and $Y$ is an affine variety isomorphic to the affine line. Q.E.D.
Of course the ideal $I(Y)$ is the radical of $(x^2-y,x^3-z)$, and the above arguments show that in fact $I(Y)=(x^2-y,x^3-z)$, but the computation of $I(Y)$ is not needed.
