Let $V=\mathcal{Z}(xz-y^2,yz-x^3,z^2-x^2y)\subseteq\mathbb{A}^3$.
First, I am asked to prove that the map $\varphi:\mathbb{A}^1\to V$ defined by $\varphi(t)=(t^3,t^4,t^5)$ is a surjective morphism. [For the surjectivity, if $(x,y,z)\neq(0,0,0)$, let $t=y/x$.]
I believe I have done this:
Let $V=\mathcal{Z}(xz-y^2,yz-x^3,z^2-x^2y)\subseteq\mathbb{A}^3$, and consider the map $\varphi:\mathbb{A}^1\to V$ defined by $\varphi(t)=(t^3,t^4,t^5)$. Let $(x,y,z)\in V$ and $(x,y,z)\neq (0,0,0)$. Additionally, let $t=y/x$. Then, we can rewrite the above expressions as follows:$xz-y^2=xz-(xt)^2=0$, $yz-x^3=xtz-x^3=0$, and $z^2-x^2y=z^2-x^3t=0$. Simplifying, we deduce that $x=t^3$, $y=t^4$, and $z=t^5$. On the other hand, if $(x,y,z)=(0,0,0)$, then $t=0$. Therefore, $\varphi(t)=(t^3,t^4,t^5)$ is onto.
However, I am stuck on part b: Describe the corresponding $k$-algebra homomorphism $\tilde{\varphi}:k[V]\to k[\mathbb{A}^1]$ explicitly.
I have a theorem (Theorem 6 on page 663 of Dummit and Foote) which states: Every morphism $\varphi:V\to W$ induces an associated $k$-algebra homomorphism $\tilde{\varphi}:k[W]\to k[V]$ defined by $\tilde{\varphi}(f)=f\circ\varphi$.
However, I am confused about how to 'explicitly' describe this map. I believe that $\tilde{\varphi}(f)=f\circ(t^3,t^4,t^5)$ in this case, but I am unclear on what to do next, and also am having trouble picturing explicitly what $k[W]$ and $k[V]$ look like.
Thank you! (This is in Chapter 15.1 of Dummit and Foote by the way)
Edit: I have another definition that seems pertinent to this question (page 661) which states: If $V\subseteq\mathbb{A}^n$ is an affine algebraic set the quotient ring $k[\mathbb{A}^n]/\mathcal{I}(V)$ is called the coordinate ring of $V$, and is denoted by $k[V]$.
I write here some facts, I hope to be clear:
1) For an affine variety $V$ the ring $k[V]$ can just be thought as the ring of polynomial functions restricted to the variety when embedded in some affine space (and has the algebraic expression that you have written ).
2) The map does explicitly what you wrote above: so you have $f(x,y,z) \in k[V]$ and now you define $g(t) \in k[\mathbb{A}^1]=k[t]$ as $g(t)=f(t^3,t^4,t^5)$. Note that because of the surjectvity of your map , if $g(t)=0$ then $f \equiv = $ on $V$ and so $f=0 $ in $k[V]$ , This tells you that you obtain an embedding of $k$.algebras (that has other interesting properties).
You can think of your variety $V$ (which can be expressed with one parameter $t$) as the affine line stretched in some strange way by your map $\phi$ in the affine space to try to figure what it is happening.