Proof verification affine curve not isomorphic to plane curve I'm trying to prove that the affine curve $X\subset\mathbb{A}^3$ given by
$\alpha:\mathbb{A}^1\to\mathbb{A}^3$, $t\mapsto(t^3,t^4,t^5)$, is not isomorphic to a plane curve.
Here is what I've done: it is enough to show that the dimension of $\mathcal{O}_{X,(0,0,0)}$ is greater than 2. As $\mathcal{O}_{X,(0,0,0)}=k[X]_{\mathfrak{m_0}}$, where $k[X]$ is the coordinate ring $k[x,y,z]/(x^4-y^3,x^5-z^3)$ and $\mathfrak{m_0}=\{f\in k[X]:f(0,0,0)=0\}=(\bar x,\bar y,\bar z)$, and as the prime ideals in $k[X]_{\mathfrak{m_0}}$ are in bijective correspondence with the prime ideals in $k[X]$ which are contained in $\mathfrak{m_0}$, it will suffice to find a chain of length 4 of prime ideals between $(0)$ and $\mathfrak{m_0}$ in $k[X]$. I've thought of $(0)\subset (\bar x)\subset (\bar x,\bar y)\subset \mathfrak{m_0}$.
Could somebody point out some possible flaw? I've never got along with irreducibility issues so my apologizes for every stupid confussion.
Thanks in advance! 
 A: I skip the proof that $(0)\subsetneqq(x^4-y^3)\subsetneqq(x^4-y^3,x^5-z^3)\subsetneqq(x,y,z)$ is a maximal chain of prime ideals in $k[x,y,z]$.
By theory, one knows that $\left(\overline0\right)=\left(\overline{x}^4-\overline{y}^3,\overline{x}^5-\overline{z}^3\right)\subsetneqq\left(\overline{x},\overline{y},\overline{z}\right)$ is a chain of prime ideals in $k[X]$; in particular:


*

*$\dim_{Krull}k[X]\geq1$;

*$\left(\overline{x},\overline{y},\overline{z}\right)=\overline{\mathfrak{m}}$ is a maximal ideal of $k[X]$.


Because $X$ is the image via a continuous and surjective map of $\mathbb{A}^1$, that is of an irreducible space of Krull dimension $1$, then $\dim_{Krull}X\leq1$; by previous statement: $\dim_{Krull}k[X]=1$.
At this point, by previous reasoning, one can easy to prove that $\dim_{Krull}k[X]_{\overline{\mathfrak{m}}}=1$; but $\overline{\mathfrak{m}}$ in $k[X]_{\overline{\mathfrak{m}}}$ has three ($3$) generators, indeed $\overline{x}\notin\left(\overline{y},\overline{z}\right)$ and so on; in other words, $k[X]_{\overline{\mathfrak{m}}}$ is not a regular local ring.
By definition $O=(0,0,0)$ is not a regular point of $X$; the Zariski cotangent space of $X$ at $O$ is
\begin{equation*}
\left(T_OX\right)^{\vee}=\overline{\mathfrak{m}}_{\displaystyle/\overline{\mathfrak{m}}^2}=\left(\overline{x},\overline{y},\overline{z}\right)_{\displaystyle/\left(\overline{x}^2,\overline{xy},\overline{y}^2,\overline{yz},\overline{z}^2,\overline{zx}\right)}\cong\left\{\left[a\overline{x}+b\overline{y}+c\overline{z}\in k[X]_{\mathfrak{m}}\right]\mid a,b,c\in k[X]_{\mathfrak{m}}\setminus\overline{\mathfrak{m}}\right\}
\end{equation*}
and its dimension over $\kappa(O)=k$ is $3$; in the same way, one can compute the Zariski cotangent space of $\mathbb{A}^3$ at $O$.
Considering the canonical projection $\pi:k[x,y,z]\to k[X]$, passing to the localization at $\mathfrak{m}$, again $\varpi=\pi_{\mathfrak{m}}:k[x,y,z]_{\mathfrak{m}}\to k[X]_{\overline{\mathfrak{m}}}$ is a surjective morphism of local rings; so $\varpi(\mathfrak{m})=\overline{\mathfrak{m}}$. In this way, one can consider the $k$-linear surjective morphism
\begin{equation*}
\left(d_Oi\right)^{\vee}:[a]\in\left(T_{i(O)}\mathbb{A}^3\right)^{\vee}\cong k^3\to[\varpi(a)]\in\left(T_OX\right)^{\vee}\cong k^3,
\end{equation*}
where $i$ is the inclusion of $X$ in $\mathbb{A}^3$.
Up to regular isomorphisms, if there exists an inclusion $j$ of $X$ in $\mathbb{A}^2$ then one can find a $k$-linear surjective morphism $\left(d_Oj\right)^{\vee}$ from $k^2$ onto $k^3$: this is a contraddition and so $X$ is not a plane curve.
A: Proof that $X$ is not isomorphic to a plane curve using tangent space:
It is enough to find a point $p\in X$ such that the tangent space $T_p(X)$ has dimension $>2$ (as a vector space).
Writing $X=\{(t_1,t_2,t_3)\in\mathbb{A}^3:t_1^4-t_2^3=t_1^5-t_3^3=0\}$ we can compute the tangent space at $p$ as
$$
T_p(X)=\{v=(t_1,t_2,t_3)\in k^3:L_1(v)=L_2(v)=0\},
$$
where $L_1(v)=\sum_{i=1}^3\frac{\partial(t_1^4-t_2^3)}{\partial t_i}(p)\cdot t_i,$ $L_2(v)=\sum_{i=1}^3\frac{\partial(t_1^5-t_3^3)}{\partial t_i}(p)\cdot t_i$.
Choosing $p=0$ we get that $L_0(v)\equiv L_1(v)\equiv 0$, so $T_0(C)=k^3$. In particular
$$
\text{dim}_kT_0(X)=3>2.
$$
