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I just have a quick question consider the map: $$\psi : \mathbb{C}[x,y] / (y - x^2) \rightarrow \mathbb{C}[t]$$ $$x \mapsto t$$ $$y \mapsto t^2$$

I want to check that this map is injective. In particular, I showed that it has inverse map. I interested to know why is the kernel of the map above trivial? The way I did that was I showed it is the pull-back of a map between varieties which is injective. Is it possible to do that more directly ?

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The map $\psi$ is already an isomorphism of $\mathbb{C}$-algebras:

Let $\varphi \colon \mathbb{C}[t] \to \mathbb{C}[x,y]/(y - x^2)$ be the unique homomorphism of $\mathbb{C}$-algebras with $\varphi(t) = \overline{x}$. Then $$ \psi(\varphi(t)) = \psi(\overline{x}) = t $$ and thus $\psi \varphi = \operatorname{id}$, and similarly $$ \varphi(\psi(\overline{x})) = \varphi(t) = \overline{x} $$ and $$ \varphi(\psi(\overline{y})) = \varphi(t^2) = \overline{x^2} = \overline{y} $$ and thus $\varphi \psi = \operatorname{id}$.

Note that this corresponds to the fact that the varieties $V_1 = \{ (x,y) \in \mathbb{C}^2 \mid y = x^2 \}$ and $V_2 = \mathbb{C}$ are isomorphic via the polynomials maps $$ V_1 \to V_2, \quad (x,y) \mapsto x \qquad\text{and}\qquad V_2 \to V_1, \quad x \mapsto (x,x^2). $$

Jendrik Stelzner
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  • Is it possible to prove that kernel is zero without exhibiting an inverse map ? –  Jun 12 '18 at 22:51
  • @Newbie It's probably a bit over-powered, but you could use a Groebner basis computation to calculate the kernel of the map $\mathbb{C}[x,y] \to \mathbb{C}[t]$ with $x \mapsto t, y \mapsto t^2$ and thus show that kernel is generated by $y-x^2$. – Daniel Schepler Jun 12 '18 at 23:22