Question about the ring $A=k[X,Y]/(Y^2-X^3)$, Miles Reid section 4.4 example (iii)

Question

I am studying commutative algebra from the book of Miles Reid’s, and I stumbled upon this example, page 62 example (iii), which is have a very hard time understanding. The example:

Consider the ring $A=k[X,Y]/(Y^2-X^3)$, and write $x,y \in A$ for the classes $X,Y$; the $A$ is not normal: it is not hard to see that $Frac A = k(t)$, where $t=y/x$, and $x=t^2$ and $y=t^3$, either of these relations show that $t$ is integral over $A$, but obviously $t$ not in $A$. Also $k[t]$ is normal since it is a UFD, so it is the integral closure of $A$ in $k(t)$.

Can anyone help me understand what is going on, I know all the terminology but I am really confused about this example. I cannot find the logical continuation between most of the arguments. Thank you very much.

EDIT: My main questions are:

1. What he means by : we write $x,y \in A$ for the classes $X,Y$;
2. Why $Frac A = k(t)$ with $t=y/x$?
3. How do we see that $t$ is integral over A?

Answer 1

Given a commutative ring $A$ and an ideal $I$ of $A$, an element of the quotient ring $A/I$ is a coset of $I$, that is, a subset of $A$ of the form $a + I = \{a + b : b \in I\}$ for some $a \in A$. Specializing to this example, an element of $A = k[X, Y]/(Y^2 - X^3)$ is a coset of the ideal $(Y^2 - X^3)$. When Reid says to write $x, y \in A$ for the classes $X, Y$, this means to denote $$\begin{aligned} x &= X + (Y^2 - X^3) = \{X + (Y^2 - X^3) \cdot P : P \in k[X, Y]\}, \\ y &= Y + (Y^2 - X^3) = \{Y + (Y^2 - X^3) \cdot P : P \in k[X, Y]\}. \end{aligned}$$ Algebraically, this means that $x$ and $y$ are indeterminates that also satisfy the relation $y^2 - x^3 = 0$ (and only this relation and those that logically follow from it), and the ring $A$ is generated over $k$ by $x$ and $y$.

Since the polynomial $Y^2 - X^3$ is irreducible and $k[X, Y]$ is a UFD, the ideal $(Y^2 - X^3)$ is prime, which means $A$ is an integral domain. Furthermore, $x \neq 0$ because $X \notin (Y^2 - X^3)$, which means we can consider the fraction $t = y/x$ as an element of $\operatorname{Frac}(A)$.

Note that $k(t) \subseteq \operatorname{Frac}(A)$ is the smallest field extension of $k$ containing $t$. Since $y^2 = x^3$, we have the following algebraic relations: $$\begin{aligned} t^2 &= y^2/x^3 = x^3/x^2 = x, \\ t^3 &= y^3/x^3 = y^3/y^2 = y. \end{aligned}$$ So $x, y \in k(t)$, and therefore $A \subseteq k(t)$. Since $k(t)$ is a field containing $A$ and is itself contained in $\operatorname{Frac}(A)$, we must have $k(t) = \operatorname{Frac}(A)$.

Finally, for $t$ to be integral over $A$ means that $t$ is the root of a monic polynomial with coefficients in $A$. And indeed, we have $t^2 - x = 0$ and $t^3 - y = 0$, and either of these shows $t$ is integral over $A$.