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I have encountered this while going over the wikipedia page "irreducible polynomial".

A polynomial that is irreducible over any field containing the coefficients is absolutely irreducible. By the fundamental theorem of algebra, a univariate polynomial is absolutely irreducible if and only if its degree is one. On the other hand, with several indeterminates, there are absolutely irreducible polynomials of any degree, such as $x^{2}+y^{n}-1$, for any positive integer $n$.

How can I show that $x^2+y^n-1$ is absolutely irreducible for any positive integer $n$? I'm new so I don't have enough tools to tackle this. Any help is appreciated.

テレビ スクリーン
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    See here. It's Eisenstein, smart Eisenstein. – Sarvesh Ravichandran Iyer Jan 27 '21 at 09:51
  • Its perfectly clear now. Thanks a lot :) – テレビ スクリーン Jan 27 '21 at 10:06
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    You are welcome! Remember that if you are in a first course then for absolute irreducibility you won't require more than this. Advanced tools like the Newton polygon are available for harder polynomials. – Sarvesh Ravichandran Iyer Jan 27 '21 at 10:07
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    The argument of user Prism in @Teresa Lisbon's link works for $\mathbb C$ but not over a field of positive characteristic $p$. Prism's answer fails over such a field at lines 5 and 6: " ...and $y-1$ divides $y^{n}-1$ to the first power, that is, $y-1 \mid y^{n}-1$ but $(y-1)^2 \not\mid y^{n}-1$". This is false for all $n$ divisible by $p$. – Georges Elencwajg Jan 27 '21 at 10:35
  • @GeorgesElencwajg I understand, but I thought we were discussing absolute irreducibility here, which is over the complex numbers. I see that the polynomial may indeed have any base field now, but this was not specified. You answer anyway covers the entire situation, I have up voted you in this regard. – Sarvesh Ravichandran Iyer Jan 27 '21 at 10:37
  • Dear @Teresa Lisbon, I agree that it is much more usual for absolute irreducibility to be defined over subfields of $\mathbb C$, and my mentioning characteristic $p$ is a symptom of the professional deformation of an algebraic geometer! Your link to Prism's argument is quite useful and relevant: I did not mean to criticize it but only wanted to analyze a more general context. The more important point is the sense of fair-play that you display in your comment and this is more valuable in my eyes than any mathematical result. Kudos, Teresa! – Georges Elencwajg Jan 27 '21 at 11:04
  • @GeorgesElencwajg Your professional deformation is completely understandable to me, since I do the same when the context is that of probability. Having a more general context is also useful to OP. Thanks once again. – Sarvesh Ravichandran Iyer Jan 27 '21 at 11:07
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    @Teresa Lisbon Дорогоя Тереза, вы меня удивлаете: на вашем описании себя вы используете, думаю, тамильский и хиндий алфавиты. Но ваше имя и фамилия европейские . Могу ли вас спросить об этой загадке, если это не секрет? – Georges Elencwajg Jan 27 '21 at 13:06
  • @GeorgesElencwajg Oh, of course you may! I am an Indian national, my primary languages are Tamil , Hindi and Kannada with a smattering of some North Indian dialect thrown in, my English is not the best around. I can read and write in Cyrillic, and understand about $30$% of what you wrote because I've studied some Russian (I understood secret, European etc.). I'm fascinated about the Balkans and about Russia, which I had the pleasure of visiting once as well. My role model is Teresa Lisbon, as you can probably see as well. I would guess you are French? Belgian? – Sarvesh Ravichandran Iyer Jan 27 '21 at 13:10
  • @GeorgesElencwajg Also, you seem to know Ravi Vakil : please tell him I was a fan of his book titled "A mathematical mosaic" : that book brought me into mathematics, and part shapes why I write my answers the way I do. – Sarvesh Ravichandran Iyer Jan 27 '21 at 13:17
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    Dear @Teresa, thank you for your answer. I did not know the fictional heroin Teresa Lisbon and wrote that I was intrigued by your having that European name (I thought it was yours!), while writing in three non-latin alphabets in your self-presentation. I unfortunately don't know Ravi Vakil although I greatly admire him as a researcher and as a pedagogue. We had some mathematical interactions on MathOverflow, for which I am very grateful, but nothing more. I strongly encourage you to send him an e-mail: I'm sure he will be happy to hear from such a nice compatriot (at least culturally) as you. – Georges Elencwajg Jan 27 '21 at 14:14
  • @GeorgesElencwajg Thanks for the compliment, I will keep it in mind. – Sarvesh Ravichandran Iyer Jan 27 '21 at 14:20

1 Answers1

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a) Over a field $k$ of characteristic $2$ and for $n=2s$ even, it is not true that the polynomial is irreducible.
Indeed, since $-1=1$ in characteristic $2$:$$x^2+y^{2s}-1=(x+y^s-1)^2$$ b) In all the other cases the polynomial $x^2+y^n-1$ is irreducible because $1-y^n$ is not a square in $k[y]$.
The reason is that in the decomposition in linear factors of $1-y^n$ over an algebraic closure $\bar k$ of $k$ some factor will appear with odd exponent:
(i) This is clear in characteristic $0$ since $1-y^n$ is separable, i.e. all its linear factors appear with exponent $1$.
(ii) And in characteristic $p\gt2$ you can write $1-y^n=(1-y^r)^{p^e}$ with $1-y^r$ separable (equivalently $p$ does not divide $r$), so that all linear factors appear with exponent $p^e$, an odd number, so that $1-y^n$ is not a square either

Georges Elencwajg
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