# If $x^3 = y^2$, why is $y/x$ transcendental?

Let the ring $$A=\mathbb{k}[x,y]/(x^3-y^2)$$, and set $$t = \frac{y}{x}$$. We can form the subring $$\mathbb{k}[t]\subset \operatorname{Frac}(A)$$, the smallest ring containing $$t$$. We have identities like $$t^2=x$$ and $$t^3 = y$$.

Now - why is $$\mathbb{k}[t]$$ isomorphic to $$\mathbb{k}[X]$$ - the ring of polynomials in one variable? In other words, how we know that for any non-zero polynomial $$p\in \mathbb{k}[X]$$, $$p(t)\neq 0$$, i.e. $$t$$ is transcendental?

Motivation:

1. The question is motivated by another question, about the normalization of A, in which was determined that the normalization of $$\widetilde{A}$$ is indeed equal to $$\mathbb{k}[t]$$. In there, the author of the question explicitly states in the comments that $$t$$ is not algebraic over $$\mathbb{k}$$, but with no proof. So the proof should trivial, but still I don't see it.
2. I think I have a proof of this fact, but it is unnecessarily complicated. I'm looking for a one-sentence proof. Nevertheless, I would be really thankful for a proof verification. Let $$w\in \mathbb{k}[T]$$ be a polynomial such that $$w(t) = 0$$ in $$\mathbb{k}[t]$$. It is of the form $$w(T) = a_n T^n + \cdots+ a_1 T + a_0$$ Using the definition of $$t$$, I make a polynomial $$w' \in \mathbb{k}[X,Y]$$ $$w'(X,Y) = a_n Y^n + a_{n-1} Y^{n-1} X + \cdots +a_1 Y X^{n-1} + a_0 X^n$$ so that $$w'(x,y) = x^n w(t)$$ in $$A$$. Therefore $$w'(x,y) = 0$$. It means that $$w' \in \ker p$$, where $$p$$ is a natural projection $$p: \mathbb{k}[X,Y] \to \mathbb{k}[x,y]/(x^3-y^2)$$, so we have $$w'(X,Y) = (X^3-Y^2)v(X,Y)$$ for $$v \in \mathbb{k}[X,Y]$$. So we have a factorization $$a_n Y^n + a_{n-1} Y^{n-1} X + \cdots + a_1 Y X^{n-1} + a_0 X^n = (X^3 - Y^2)v(X,Y)$$ Assuming $$a_n \neq 0$$ we see that $$-a_0 Y^{n-2}$$ should be among addends of $$v$$ (by comparing coefficients). Then we would have $$-a_0 Y^{n-2} X^3$$ in the product $$(X^3-Y^2)v(X,Y)$$. But the coefficient in $$w'$$ before the $$Y^{n-2} X^3$$ term is $$0$$, so to cancel it, we need either $$-a_0 Y^{n-4} X^3$$ or $$a_0 Y^{n-2}$$ term in $$v(X,Y)$$. But it cannot be the latter - we already determined the coefficient before $$Y^{n-2}$$ to be $$-a_0$$. So it must be $$-a_0 Y^{n-4} X^3$$. But then, analogously, we get $$-a_0 Y^{n-4} X^6$$ that need to be canceled in the product. Continuing like this, after $$\lceil{\frac{n}{2}}\rceil$$ steps we no longer could form the term for canceling, because the exponent would need to be negative. That leads to contradiction.
• Your proof is "unnecessarily complicated" indeed. The approach is natural, but why don't try the following substitution $X\to T^2$ and $Y\to T^3$ in $$a_n Y^n + a_{n-1} Y^{n-1} X + \cdots + a_1 Y X^{n-1} + a_0 X^n = (X^3 - Y^2)v(X,Y)$$ This gives you immediately $a_i=0$ for all $i$. Dec 31, 2018 at 22:30
• @user26857 Thank you, that technique will certainly be useful to me in the future.
– mz71
Dec 31, 2018 at 22:35

$$t^2 = x$$ and $$x$$ is transcendental; if $$t$$ were algebraic over $$\mathbb{k}$$, so would $$x$$.
$$x$$ is transcendental over $$\mathbb{k}$$ in $$A$$ because $$X$$ is transcendental in $$\mathbb{k}[X]$$ and there's a map $$A\to \mathbb{k}[X], x\mapsto X^2, y\mapsto X^3$$.
• How we know that $x$ is transcendental in A?
• @mzg147: Every multiple of $x^3-y^2$ must have positive degree in $y$ and thus no polynomial involving just $x$ vanishes in $A$. Dec 30, 2018 at 20:26
• @mzg147 : here's another slick proof : consider $\mathbb{k}[x,y]\to \mathbb{k}[X], x\mapsto X^2, y\mapsto X^3$. Then this factors through $A\to \mathbb{k}[X]$, and $x\mapsto X^2$, $X^2$ being transcendental over $\mathbb{k}$, so is $x$ over $\mathbb{k}$ in $A$ Dec 30, 2018 at 20:28