Problem:
Solution:
For part (a), it suffice the find the set of polynomials that vanish on the parametric curve.
Analogous to the twisted cubic, it is pretty obvious to me that the polynomials are:
$ y - x^3 $ and $ z - x^4 $.
Let's prove that $ V(y - x^3, z - x^4) $ is the parametric curve $ C = \{(t, t^3, t^4)\} $.
Note that any point in the parametric curve must be in $ V $ by construction, so $ C \subset V(y - x^3, z - x^4) $.
For the reverse inclusion, suppose a point $ p $ outside of $ C $, the point $ p $ must necessarily have the form $ (t, t^3 + a, t^4 + b) $ where not both $ a = 0 $ and $ b = 0 $. Putting $ p $ into the equation we get $ y - x^3 = a $ and $ z - x^4 = b $, which mean at least one of the polynomial must not vanish and therefore the point cannot in $ V $. In other words $ v \in V \implies v \in C $, and so $ V(y - x^3, z - x^4) \subset C $, so we proved that the curve is the variety $ V(y - x^3, z - x^4) $.
For part (b), we claim that $ P = \langle y - x^3, z - x^4 \rangle $ is $ I(V) $.
It is obvious that any polynomial $ p \in P $ must vanish in $ V(y - x^3, z - x^4) $.
We will use the same trick as the last problem. We claim that any polynomial $ f(x, y, z) $ can be written as $ f(x, y, z) = g(x, y, z)(y - x^3) + h(x, y, z)(z - x^4) + r(x) $.
Again, it suffice to prove this for monomial, so we have:
$ \begin{eqnarray*} & & x^ay^bz^c \\ &=& x^a((y - x^3) + x^3)^b((z - x^4) + x^4)^c \\ &=& x^a(x^{3b} + p(x, y)(y - x^3))(x^{4c} + q(x, z)(z - x^4)) \\ &=& x^{a + 3b + 4c} + m(x, y, z)(y - x^3) + n(x, y, z)(z - x^4) \end{eqnarray*} $
As we can represent a monomial, we can also represent any polynomial, so the claim is proved.
Next, for any polynomial $ f(x, y, z) \in I(V) $, we write $ f(x, y, z) = g(x, y, z)(y - x^3) + h(x, y, z)(z - x^4) + r(x) $ and then it need to vanish $ (t, t^3, t^4) $, so we substitute in and get:
$ 0 = f(x, y, z) = g(t, t^3, t^4)(t^3 - t^3) + h(t, t^3, t^4)(t^4 - t^4) + r(t) $, so we get $ r(t) $ is the zero polynomial and therefore $ I(V) \subset \langle y - x^3, z - x^4 \rangle $.
Solution:
For part (a), it suffice the find the set of polynomials that vanish on the parametric curve.
Analogous to the twisted cubic, it is pretty obvious to me that the polynomials are:
$ y - x^3 $ and $ z - x^4 $.
Let's prove that $ V(y - x^3, z - x^4) $ is the parametric curve $ C = \{(t, t^3, t^4)\} $.
Note that any point in the parametric curve must be in $ V $ by construction, so $ C \subset V(y - x^3, z - x^4) $.
For the reverse inclusion, suppose a point $ p $ outside of $ C $, the point $ p $ must necessarily have the form $ (t, t^3 + a, t^4 + b) $ where not both $ a = 0 $ and $ b = 0 $. Putting $ p $ into the equation we get $ y - x^3 = a $ and $ z - x^4 = b $, which mean at least one of the polynomial must not vanish and therefore the point cannot in $ V $. In other words $ v \in V \implies v \in C $, and so $ V(y - x^3, z - x^4) \subset C $, so we proved that the curve is the variety $ V(y - x^3, z - x^4) $.
For part (b), we claim that $ P = \langle y - x^3, z - x^4 \rangle $ is $ I(V) $.
It is obvious that any polynomial $ p \in P $ must vanish in $ V(y - x^3, z - x^4) $.
We will use the same trick as the last problem. We claim that any polynomial $ f(x, y, z) $ can be written as $ f(x, y, z) = g(x, y, z)(y - x^3) + h(x, y, z)(z - x^4) + r(x) $.
Again, it suffice to prove this for monomial, so we have:
$ \begin{eqnarray*} & & x^ay^bz^c \\ &=& x^a((y - x^3) + x^3)^b((z - x^4) + x^4)^c \\ &=& x^a(x^{3b} + p(x, y)(y - x^3))(x^{4c} + q(x, z)(z - x^4)) \\ &=& x^{a + 3b + 4c} + m(x, y, z)(y - x^3) + n(x, y, z)(z - x^4) \end{eqnarray*} $
As we can represent a monomial, we can also represent any polynomial, so the claim is proved.
Next, for any polynomial $ f(x, y, z) \in I(V) $, we write $ f(x, y, z) = g(x, y, z)(y - x^3) + h(x, y, z)(z - x^4) + r(x) $ and then it need to vanish $ (t, t^3, t^4) $, so we substitute in and get:
$ 0 = f(x, y, z) = g(t, t^3, t^4)(t^3 - t^3) + h(t, t^3, t^4)(t^4 - t^4) + r(t) $, so we get $ r(t) $ is the zero polynomial and therefore $ I(V) \subset \langle y - x^3, z - x^4 \rangle $.
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