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Goal is to prove $y^2-x^3$ is irreducible in $\mathbb C$.

Answers like https://math.stackexchange.com/a/1208947/342943 below explains it but cannot understand fundamentally.

It is in $C(y)[x]$ and has no root in $C(y)$.

Note that: $C(y)$ is a field and since it is cubic, we just need to show it has no root.

So in what form of Gauss or Eisenstein we use it here? And how exactly, I am confused.

Jale'de jaled
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  • If it was reducible, you could write it as a product in $\mathbb{C}[x,y]$ of non-units. This would also give a factorisation in $(\mathbb{C}(y))[x]$. Now, $\mathbb{C}(y)$ is a field, so we can apply the criterion that a polynomial in $(\mathbb{C}(y))[x]$ of degree $2$ or $3$ is irreducible if and only if it has no root (in $\mathbb{C}(y)$). – ThePuix Nov 19 '21 at 01:19
  • Hint $ $ As here, $\ y^2 -x^3\ $ is irreducible in $,\Bbb C[x,y],$ since $,x^3,$ is not a square in the UFD $,\Bbb C[x],,$ since the prime $,x,$ occurs to odd power. – Bill Dubuque Nov 19 '21 at 02:16

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If $f(X)=y^2-X^3$ has a root in $\mathbb C(y)$ as $\frac {g(y)}{h(y)}$, then $g(y)^3=y^2h(y)^3$. Then, $g, h$ must be non-zero.

The degree of left side is $0 \bmod3$, and the degree of right side is $2 \bmod 3$. So it contradicts. $f(X)$ is irreducible polynomial over $\mathbb C(y)$.

user26857
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MH.Lee
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