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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.
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    $\begingroup$ 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 $i$. $\endgroup$
    – user26857
    Dec 31, 2018 at 22:30
  • $\begingroup$ @user26857 Thank you, that technique will certainly be useful to me in the future. $\endgroup$
    – mz71
    Dec 31, 2018 at 22:35

1 Answer 1

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$t^2 = x$ 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 $A\to \mathbb{k}[X], x\mapsto X^2, y\mapsto X^3$.

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  • $\begingroup$ How we know that $x$ is transcendental in A? $\endgroup$
    – mz71
    Dec 30, 2018 at 20:25
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    $\begingroup$ @mzg147: Every multiple of $x^3-y^2$ must have positive degree in $y$ and thus no polynomial involving just $x$ vanishes in $A$. $\endgroup$ Dec 30, 2018 at 20:26
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    $\begingroup$ @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 $A$ $\endgroup$ Dec 30, 2018 at 20:28
  • $\begingroup$ Thank you. Please add this to your answer, so I can mark it as answered. $\endgroup$
    – mz71
    Dec 30, 2018 at 21:05

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