Show $k[x]\cap(y-x^2,z-x^3)={0}$ in $k[x,y,z]$. $ k$ field.

Question

Let k be a field.

How could I show that $k[x]\cap(y-x^2,z-x^3)={0}$ in $k[x,y,z]$.

I understand that there's a whole algorithm I could go through with Grobner basis, elimination theorem etc. but is there a simple argument?

This problem arises from finding the dimension of a twisted cubic in Algebraic Geometry.

In fact, $k[x,y,z]/(y-x^2,z-x^3)\simeq k[x] $ just by the isomorphism theorem. But to show the isomorphism directly using the obvious map gets me to my original question. So this suggests my original question ought not be too hard yet it seems to me to be non-obvious. Any comments on this point would be appreciated.

Thanks!

Update:

Thanks for your answers which I've upvoted, but my question is really how to solve the problem directly (e.g. perhaps by considering degrees of various variables) instead of appealing to isomorphism results which, as I mentioned above, I am aware of. And the second question is more philosophically why using isomorphism theorems would make things easier.

Answer 1

There is a unique $k$-linear ring homomorphism $\phi:k[x,y,z]\to k[t]$ such that $\phi(x)=t$, $\phi(y)=t^2$ and $\phi(z)=t^3$, and the ideal $I=(y-x^2,z-y^3)$ is conttained in its kernel, as its generators are. If $f=\sum_{i=0}^na_ix^i$ is an elemntt of of $k[x]$ which is in $I$, then $\phi(f)$ is zero. But $\phi(f)$ is just $\sum_{i=0}^na_it^i$, whose vanishing evidently implies the vanishing of $f$ itself.

N.B. You do not need to know that $k[x]\cap I=0$ in order to show that the obvious isomorphism is an isomorphism. Indeed, as my homomorphism $\phi$ above contains $I$, it passes on to the quotient to give a $k$-linear homomorphim of rings $\bar\phi:k[x,y,z]/I\to k[t]$. On the other hand, it is obvious that there is a $k$-linear homomorphism of rings $\psi:k[t]\to k[x,y,z]/I$ which maps $t$ to (the class of) $x$. Now we want to show that $\psi\circ\bar\phi$ and $\bar\phi\circ\psi$ are identities. For the second one, it is enough to notice that $(\bar\phi\circ\psi)(t)=\bar\phi(\psi(t))=\bar\phi(x)=t$.

For the second, let $\pi:k[x,y,z]\to k[x,y,z]/I$ be the quotient map. As $\pi$ is surjective, to show that $\psi\circ\bar\phi$ is the identity map, it is enough to show that for all $u\in k[x,y,z]$ we have $(\psi\circ\bar\phi\circ\pi)(u)=\pi(u)$, and for that it is enough to show it for $u\in\{x,y,z\}$. This is immediate.

Answer 2

I'm not sure this answers your first question, but here's one way to prove $k[x,y,z]/(y - x^2, z - x^3) \cong k[x]$. The intuition is that $$ k[x,y,z]/(y - x^2, z - x^3) \cong k[x,x^2,x^3] = k[x] $$ but I'm guessing you want something more rigorous. First, let's consider the following lemma.

Lemma. Let $R$ be a unital commutative ring and $R[x]$ be the one-variable polynomial ring over $R$. For $\alpha \in R$, let \begin{align*} \varphi = \text{eval}_\alpha: R[x] &\to R\\ x &\mapsto \alpha \end{align*} be the evaluation homomorphism. Then $\ker(\varphi) = (x-\alpha)$. Moreover, the induced map $\overline{\varphi}: R[x]/(x-\alpha) \to R$ is an isomorphism.

We can use this lemma, along with the Third Isomorphism Theorem, to show the above isomorphism. First note that the lemma implies that $$ \frac{k[x,y]}{(y - x^2)} = \frac{k[x][y]}{(y-x^2)} \cong k[x] \, . $$ Now, since $(y - x^2) \subseteq (y-x^2,z-x^3) \subseteq k[x,y,z]$, then \begin{align*} \frac{k[x,y,z]}{(y-x^2,z-x^3)} &\cong \frac{(k[x,y]/(y-x^2))[z]}{(y-x^2,z-x^3)/(y-x^2)} \cong \frac{k[x][z]}{(0,z-x^3)} \cong \frac{k[x][z]}{(z-x^3)} \cong k[x] \end{align*} by the Third Isomorphism Theorem.