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Is $1+x+x^2+x^3+x^4$ irreducible over $\mathbb{Z}$?

I understand that if asked for $\mathbb{Q}$, the answer is yes, because $$f(x) = x^{5-1}+ x^{5-2} + x^{5-3} +x^{5-4} +1.$$

Since $5$ is prime, therefore it is irreducible over $\mathbb{Q}$.

But can we say that, since it is irreducible over $\mathbb{Q}$, it is also irreducible over $\mathbb{Z}$? I am not clear about it.

ə̷̶̸͇̘̜́̍͗̂̄︣͟
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Muskaan Madan
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    Any factorization over $\mathbb Z$ is also a factorization over $\mathbb Q$, so irreducible over $\mathbb Q$ implies irreducible over $\mathbb Z$. – lisyarus Aug 18 '20 at 06:30
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    If it was reducible over $\Bbb{Z}$, then wouldn't it be reducible over $\Bbb{Q}$ as well since $\Bbb{Z} \subset \Bbb{Q}$. – Anurag A Aug 18 '20 at 06:30
  • But shouldn't it be opposite? – Muskaan Madan Aug 18 '20 at 06:31
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    Is a polynomial in $\Bbb Z$ also a polynomial in $\Bbb Q$? – pancini Aug 18 '20 at 06:31
  • Yes. A polynomial in $\mathbb{Z}$ is polynomial in $\mathbb{Q}$ – Muskaan Madan Aug 18 '20 at 06:34
  • Is this it? Gauss's lemma, is that a primitive polynomial is irreducible over the integers if and only if it is irreducible over the rational numbers. – Muskaan Madan Aug 18 '20 at 06:38
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    I completely forgot that invertible elements of $\mathbb Q[X]$ are not the same as $\mathbb Z[X]$. @JCAA was right, this is the statement of Gauss's lemma. – lisyarus Aug 18 '20 at 06:44
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    Since $\mathbb Z\subset\mathbb Q$, the polynomial is irreducible in $\mathbb Z$ either – TravorLZH Aug 18 '20 at 07:16
  • Another possibility is drawing a picture. If the graph doesn't intersect the $x$-axis there are no real zeros and hence there are CERTAINLY no integer zeros... – Vincent Aug 18 '20 at 07:38
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    @Vincent: irreducibilty over $\mathbb{Z}$ is not the same as having no integer zeroes. – Paramanand Singh Aug 18 '20 at 07:55
  • O wait, you are right, that only proves it has no linear factors but doesn't rule out that the thing is a produc two degree two irreducible polynomials. – Vincent Aug 18 '20 at 10:05

2 Answers2

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Yes, of course.

Another way:

Let $x=y+1$.

Thus, $$x^4+x^3+x^2+x+1=y^4+5y^3+10y^2+10y+5$$ and use Eisenstein: https://en.wikipedia.org/wiki/Eisenstein%27s_criterion

Michael Rozenberg
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  • We'll only be able to prove it till rationals. We then have to use Gauss lemma as mentioned above, in the end. – Muskaan Madan Aug 18 '20 at 07:09
  • @Muskaan Madan But a polynomial is irreducible over $\mathbb Q$ it's equivalent that it's irreducible over $\mathbb Z$, which gives an answer on your question. I just posted another way to understand that $x^4+x^3+x^2+x+1$ is irreducible over $\mathbb Z.$ – Michael Rozenberg Aug 18 '20 at 07:13
  • That's what I said – Muskaan Madan Aug 18 '20 at 07:17
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    +1 Well this is actually one of the ways we prove the irreducibilty of $f(x) =(x^p-1)/(x-1)$ over $\mathbb {Z} $, given $p$ a prime. The irreducibilty of $f$ in $\mathbb{Q} $ then follows from Gauss lemma. – Paramanand Singh Aug 18 '20 at 07:53
  • @MichaelRozenberg $2x$ is irreducible over $\mathbb Q$ but not over $\mathbb Z$. – lisyarus Aug 18 '20 at 12:42
  • @lisyarus But the degree of $2$ is zero. You example is not correct. See please better a definition of a polynomial irreducibility. – Michael Rozenberg Aug 18 '20 at 15:38
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    Ah, it never occurred to me that irreducible polynomial and irreducible element of the polynomial ring are different things. You are right. – lisyarus Aug 18 '20 at 15:43
  • Using different definitions for irreducible polynomial and irreducible element of a polynomial ring is weird. Just introduce the notion of a primitive polynomial to bridge the discrepancy and make Gauss lemma work. – Ennar Sep 26 '20 at 13:08
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$$f(x) = 1+x+x^2+x^3+x^4$$

$$f(x) = x^{5-1}+ x^{5-2} + x^{5-3} +x^{5-4} +1.$$

Since $5$ is prime, therefore it is irreducible over $\mathbb{Q}$.

By Gauss's lemma,

A primitive polynomial is irreducible over the integers if and only if it is irreducible over the rational numbers.

Using the lemma above,

Since $1+x+x^2+x^3+x^4$ irreducible over $\mathbb{Q}$, therefore, $1+x+x^2+x^3+x^4$ irreducible over $\mathbb{Z}$ too.

Muskaan Madan
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