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Is there an easy way to show that $x^4+8$ is irreducible over $\mathbb Z$ without trying to write it as a product of polynomials of lower degrees?

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Is this for a class? If so, what theorems have you covered? –  Brian Fitzpatrick Feb 9 at 21:35
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You may try Eisenstein's criteria. Try to substitute (x+1) or (x-1). –  user60887 Feb 9 at 21:36
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You can also simplify the product-of-polynomials approach a good bit: since you know it has no roots in $\mathbb{Z}$ (no roots in $\mathbb{R}$, in fact!) then it must be a product of quadratics if anything; write it as $(x^2+ax+b)(x^2+cx+d)$. Now compare coefficients; for instance, looking at the coefficients of $x^3$ implies $a=-c$. $bd=8$ similarly diminishes the possibilities there hugely, etc. –  Steven Stadnicki Feb 9 at 21:39
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(And of course, if you know the roots then you can find the quadratic polynomials that it does factor into and just show that they're not in $\mathbb{Z}[x]$.) –  Steven Stadnicki Feb 9 at 21:40
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@ user60887 That didn't help me here since the only prime that divides 8 is 2, and 2^2 | 8. And substituting x+1 and x-1 gives constant terms 9 and 7 respectively, where 7 does not divivde any other coefficient. Or did you have anything else in mind? –  Improve Feb 9 at 21:44
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4 Answers 4

Since $\ 9^4\!+8\ $ is prime, it is irreducible by Cohn's irreducibility test. Or, with smaller prime, consider $\,f(2x) = 8(2x^4\!+1).\,$ Then $2x^4\!+1$ is irreducible by Cohn, since $\,2\cdot 3^4+1 = 163\,$ is prime (in fact, a very famous prime, the largest Heegner number, which explains why Euler's famous prime producing polynomial $\,n^2-n+41\,$ yields distinct primes for $\,n = 1,2,\ldots,40).$

Remark $\ $ Cohn's criterion can be viewed as an example of the general idea that the factorizations of a polynomial are constrained by the factorizations of the values that it takes. If one pushes this idea to the hilt one obtains a simple algorithm for polynomial factorization using factorization of its integer values and (Lagrange) interpolation. The ideas behind this algorithm are due in part to Bernoulli, Schubert, Kronecker. The algorithm is of more theoretical value than practical, since nowadays much more efficient algorithms are known.

There are also other closely related results. In $1918$ Stackel published the following simple:

Theorem If $\rm\, f(x)\,$ is a composite integer coefficient polynomial then $\rm\, f(n)\, $ is composite for all $\rm\,|n| > B,\, $ for some bound $\rm\,B.\,$ In fact $\rm\, f(n)\, $ has at most $\rm\, 2d\, $ prime values, where $\rm\, d = {\rm deg}(f)$.

The simple proof can be found online in Mott & Rose [3], p. 8. I highly recommend this delightful and stimulating $27$ page paper which discusses prime-producing polynomials and related topics.

Contrapositively, $\rm\, f(x)\, $ is prime (irreducible) if it assumes a prime value for large enough $\rm\, |x|\, $. As an example, Polya-Szego popularized A. Cohn's irreduciblity test, which states that $\rm\, f(x) \in \mathbb Z[x]\,$ is prime if $\rm\, f(b)\, $ yields a prime in radix $\rm\,b\,$ representation (so necessarily $\rm\,0 \le f_i < b).$

For example $\rm\,f(x) = x^4 + 6\, x^2 + 1 \pmod p\,$ factors for all primes $\rm\,p,\,$ yet $\rm\,f(x)\,$ is prime since $\rm\,f(8) = 10601\rm$ octal $= 4481$ is prime. Cohn's test fails if, in radix $\rm\,b,\,$ negative digits are allowed, e.g. $\rm\,f(x)\, =\, x^3 - 9 x^2 + x-9\, =\, (x-9)\,(x^2 + 1)\,$ but $\rm\,f(10) = 101\,$ is prime.

Conversely Bouniakowski conjectured $(1857)$ that prime $\rm\, f(x)\, $ assume infinitely many prime values (excluding cases where all the values of $\rm\,f\,$ have fixed common divisors, e.g. $\rm\, 2\: |\: x(x+1)+2\, ).$ However, except for linear polynomials (Dirichlet's theorem), this conjecture has never been proved for any polynomial of degree $> 1.$

Note that a result yielding the existence of one prime value extends to existence of infinitely many prime values, for any class of polynomials closed under shifts, viz. if $\rm\:f(n_1)\:$ is prime, then $\rm\:g(x) = f(x+ n_1\!+1)\:$ is prime for some $\rm\:x = n_2\in\Bbb N,\:$ etc.

For further detailed discussion of Bouniakowski's conjecture and related results, including heuristic and probabilistic arguments, see Chapter 6 of Ribenboim's The New Book of Prime Number Records.

[1] Bill Dubuque, sci.math 2002-11-12, On prime producing polynomials.

[2] Murty, Ram. Prime numbers and irreducible polynomials.
Amer. Math. Monthly, Vol. 109 (2002), no. 5, 452-458.

[3] Mott, Joe L.; Rose, Kermit. Prime producing cubic polynomials.
Ideal theoretic methods in commutative algebra, 281-317.
Lecture Notes in Pure and Appl. Math., 220, Dekker, New York, 2001.

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(+1) And again I learned something surprising :D. –  Your Ad Here Feb 9 at 21:45
    
Thanks! It was something like this I was hoping to see:) –  Improve Feb 9 at 21:50
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I’m just so happy you’re here again. –  k.stm Feb 9 at 21:59
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@TooOldForMath I added some further remarks that you may find of interest. –  Bill Dubuque Feb 9 at 22:03
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@Bill Dubuque , Oh shoot , how did i miss that? Thanx! –  neofoxmulder Feb 9 at 22:04
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Proffering a different route. The polynomial $$ x^4+8\equiv x^4-2\pmod5. $$ Therefore it suffices to show that $x^4-2$ is irreducible as a polynomial in $\Bbb{Z}_5[x]$. To see that we observe that $2$ is of order four modulo $5$. Therefore the zeros of this polynomial (in some extension field of $\Bbb{Z}_5$) are primitive sixteenth roots of unity. The smallest exponent $m>0$ with the property $16\mid 5^m-1$ is $m=4$. This means that the zeros of $x^4-2$ all belong to a quartic extension of the prime field. The claim follows.

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I appreciate the answer, but I do not immediately see why we can reduce this to irreducibility in Z_5[X]. –  Improve Feb 9 at 22:13
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If the polynomial were reducible in $\Bbb{Z}[x]$, then reducing that factorization modulo five would give a non-trivial factorization in $\Bbb{Z}_5[x]$. Thus the non-existence of the latter factorization implies the non-existence of a factorization over the integers. It is a one-way street, but are going with the flow. –  Jyrki Lahtonen Feb 9 at 22:16
    
Yes, of course. Thanks! –  Improve Feb 9 at 22:17
    
The problem with this solution is that it needs familiarity with basic properties of finite fields. Following Steven Stadnicki's comment: it may be more accessible to simply find the complex zeros and test that no pairs gives a quadratic factor with integer coefficients. –  Jyrki Lahtonen Feb 9 at 22:21
    
@Jyrki But it is a very useful general technique, which I had planned to mention too, but I got too carried away discussing the other ideas (which deserve to be better known). –  Bill Dubuque Feb 9 at 23:05
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There are many other ways of showing irreducibility here. For instance, if our polynomial is $f(x)$, look at $f(2x)=8(2x^4+1)$. The thing in parentheses is irreducible if and only if $x^4+2$ is irreducible.

But my favorite method is to consider the polynomial to be defined over the $2$-adic numbers, $\mathbb Q_2$, look at its Newton Polygon, which consists of the two points $(0,3)$ and $(4,0)$, with the segment joining them, which hits no other integral points. Therefore irreducible.

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As $x^4+8$ is a primitive polynomial; the only elements of $\mathbb Z[x]$ that divide the polynomial are the units, by Gauss's lemma we need only show $x^4+8$ is irreducible in $\mathbb Q[x]$.

Let $\zeta$ be a root of $x^4+2$. By Eisenstein's criterion, $x^4+2$ is irreducible, so we must have $[\mathbb Q(\zeta):\mathbb Q]=4$, because $\{1,\zeta,\zeta^2,\zeta^3\}$ is a basis of $\mathbb Q(\zeta)$ over $\mathbb Q$. Now, $\zeta^3$ is a root of $x^4+8$, but $\mathbb Q(\zeta^3)=\mathbb Q(\zeta)$, as $\zeta^6=-2\zeta^2$ and $\zeta^9=4\zeta$, hence $[\mathbb Q(\zeta^3):\mathbb Q]=4$, which implies $x^4+8$ is irreducible in $\mathbb Q[x]$.

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