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I'd like to find a proof (as simple as possible) of Eisenstein's criterion of irreducibility. I found here the proof in the picture below. This is rather elegant, yet I do not understand why $g,h$ must have integer coefficients.

Edit:

I found this paper, stating

If a polynomial with integer coefficients is reducible over Q, then it is reducible over Z.

A proof of this is not given though

enter image description here

Ariel Yael
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  • The proof mentions Theorem 17.2 in which apparently the coefficients being integers is shown. – coffeemath Mar 27 '21 at 12:24
  • @coffeemath I see that, but I don't have the book, so I don't have this theorem nor its proof – Ariel Yael Mar 27 '21 at 13:11
  • That reducibility over $\mathbb{Q}$ for a polynomial with integer coefficients implies reducibility over $\mathbb{Z}$ is a result sometimes called “Gauss’s Lemma”, and is a consequence of the fact that product of primitive polynomials is primitive (see Gauss’s proof here. I’m afraid that if you don’t have it, you’ll add a page or so to the proof. – Arturo Magidin Mar 27 '21 at 17:09
  • @ArturoMagidin thanks! and.. just to be clear, the polynomial $3x+3$ is reducible in $\mathbb{Z}$, right? – Ariel Yael Mar 27 '21 at 17:16
  • $3x+3$ can't be factored as the product of two non constant polynomials , so it should be irreducible over $\mathbb Z$. – Aditya Mar 27 '21 at 17:26
  • @Aditya from looking online I understood the following: since 3 is not invertible in Z, the polynomial is reducible (see this and this) – Ariel Yael Mar 27 '21 at 18:12
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    @Aditya: No, that’s wrong. – Arturo Magidin Mar 27 '21 at 19:35
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    Yes, $3x+3 = 3(x+1)$, which is a product of two irreducibles of $\mathbb{Z}[x]$, so it is reducible in $\mathbb{Z}[x]$. But it is not primitive. A primitive polynomial with integer coefficients is irreducible over $\mathbb{Z}$ if and only if it is irreducible over $\mathbb{Q}$. A nonprimitive polynomial with integer coefficients is never irreducible over $\mathbb{Z}$/ – Arturo Magidin Mar 27 '21 at 19:36
  • @Arturo Magidin, ok. I read the definition of reducibility on Wikipedia and it said it has to be the product of two non constant polynomials, but here $p(x)=3$ is a constant polynomial so I thought it was irreducible. – Aditya Mar 28 '21 at 00:49
  • @Aditya: Reducibility where? The article has a gloss on the top, but the actual definition further down talks about polynomials with coefficients in a field, and explicitly talks about polynomials with integer coefficients further down. (cont) – Arturo Magidin Mar 28 '21 at 01:06
  • @Aditya: it states: “ A polynomial with integer coefficients, or, more generally, with coefficients in a unique factorization domain R, is sometimes said to be irreducible (or irreducible over R) if it is an irreducible element of the polynomial ring, that is, it is not invertible, not zero, and cannot be factored into the product of two non-invertible polynomials with coefficients in R. This definition generalizes the definition given for the case of coefficients in a field, because, over a field, the non-constant polynomials are exactly the polynomials that are non-invertible and non-zero.” – Arturo Magidin Mar 28 '21 at 01:06
  • @Aditya: So you didn’t read far enough. – Arturo Magidin Mar 28 '21 at 01:06
  • @ArturoMagidin, Ahh, I see. You're right. – Aditya Mar 28 '21 at 04:57

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This has to do with Gauss's Lemma for polynomials. It states that a primitive polynomial is irreducible over the integers if and only if it is irreducible over the rational numbers.

So in the given proof, they assume that it is reducible over integers to achieve a contradiction in order to prove that it is irreducible over the integers, and hence also over the rationals.

Aditya
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