Showing $k[X] \cong k[X,Y,Z]\big/{(Y-X^2,Z-X^3)}$

Let $k$ be an algebraically closed field. I want to show that $(Y-X^2,Z-X^3)$ is a prime ideal of $k[X,Y,Z]$.

I know that an ideal $\frak a \subseteq$ $R$ of a (commutative ring with $1$) is prime if any only if $R\big/{\frak a}$ is an integral domain.

Thus I understand that if we can show that $k[X,Y,Z]\big/{(Y-X^2,Z-X^3)}$ is an integral domain, we're done.

Now this exercise's solution I am trying to understand is stating that the homomorphism \begin{align*} k[X]&\to k[X,Y,Z]\big/{(Y-X^2,Z-X^3)}\\ X&\mapsto X\end{align*}

has an inverse via

\begin{align*} k[X,Y,Z]\big/{(Y-X^2,Z-X^3)} &\to k[X] \\ X&\mapsto X \\ Y&\mapsto X^2 \\ Z&\mapsto X^3\end{align*}

and thus they claim $k[X,Y,Z]\big/{(Y-X^2,Z-X^3)}$ to be an integral domain.

I don't fully understand the proposed solution and have some questions.

Question 1: How to come up with these maps in the first place?

I am aware of the natural inclusion $k[X]\subset k[X,Y,Z]$, so the first map is kind of an obvious choice, maybe. But how to come up with the inverse map? What role does the quotient ring play in terms of "finding the inverse map" or rather "seeing the obvious choice"?

In other words: Why are these maps "obvious"?

Question 2: I suppose the key observation in having this inverse map is that this shows that $$k[X] \cong k[X,Y,Z]\big/{(Y-X^2,Z-X^3)}$$ and since $k[X]$ is an integral domain, the quotient ring $k[X,Y,Z]\big/{(Y-X^2,Z-X^3)}$ is an integral domain, too. Is that correct?

Thanks for any help!

Solution 1:

The fact that $Y-X^2$ and $Z-X^3$ are quotiented out means that in the quotient you have $Y=X^2$ and $Z=X^3$. Thus it makes sense to see if the map $$\begin{align*} k[X,Y,Z]\big/{(Y-X^2,Z-X^3)} &\to k[X] \\ X&\mapsto X \\ Y&\mapsto X^2 \\ Z&\mapsto X^3\end{align*},$$ is an isomorphism, or at least injective, as then it follows that the quotient is an integral domain.