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I have found this in a book:

but I don't know why the curve $x^m-y^n=0$ is irreducible when n, m are coprimes.

Can someone help me?

RghtHndSd
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Rebecca Ghaff.
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1 Answers1

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Here's one way to see it. There may be cleaner ones, but this has the advantage of being a brute-force approach that doesn't require any cleverness or special machinery.

Consider the ring $R := k[x, y] / (x^m - y^n) = k[x, \sqrt[n]{x^m}]$. We want to show that $R$ is an integral domain, because this is equivalent to $(x^m - y^n)$ being prime, which is equivalent to the curve $x^m - y^n = 0$ being irreducible.

A general element of $R$ looks like

$$p_0(x) + p_1(x) \sqrt[n]{x^m} + p_2(x) \left(\sqrt[n]{x^m}\right)^2 + \cdots + p_{n-1}(x) \left(\sqrt[n]{x^m}\right)^{n-1}$$

where the $p_i$ are polynomials, and moreover this representation is unique. (For uniqueness, we're using that $m, n$ are relatively prime).

So, take another element, say

$$q_0(x) + q_1(x) \sqrt[n]{x^m} + q_2(x) \left(\sqrt[n]{x^m}\right)^2 + \cdots + q_{n-1}(x) \left(\sqrt[n]{x^m}\right)^{n-1},$$

multiply them together, and show that if you get zero from this then either all the $p_i$ were zero or all the $q_i$ were zero.

Daniel McLaury
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