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Let $f(x)=x^n+5x^{n-1}+3$ where $n\geq1$ is an integer. Prove that $f(x)$ can't be expressed as the product of two polynomials each of which has all its coefficients integers and degree $\geq1$.

If the condition that each polynomial must have all its coefficients integers was not there, then I needed to show only that $f(x)$ is irreducible over real numbers. But since this condition is given,therefore, we can't exclude the case when $f(x)$ is reducible over real numbers but not with polynomials of integer coefficients. Anyone with idea how to proceed? Thanks in advance!!

user26857
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Aang
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    This is a classic application of Perron's criterion (http://mathoverflow.net/questions/8182/is-a-polynomial-with-1-very-large-coefficient-irreducible), which can be proven using Rouche's theorem in complex analysis. I am not aware of another approach. – Qiaochu Yuan Jun 28 '12 at 05:42

4 Answers4

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The notation $[x^t]P(x)$ denotes the coefficient of $x^t$ in polynomial $P(x)$.

Suppose that $f(x)=g(x)h(x)$, where $g(x),h(x)$ are monic polynomials, $g(x),h(x)\in\Bbb Z[x]$ and $\deg g>0$, $\deg h>0$, $\deg g+\deg h=n$. Let $g(x)=\sum_k\alpha_kx^k$, $h(x)=\sum_k\beta_kx^k$, where $\alpha_k,\beta_k$ is integer whenever $k\ge0$.

First, we have $\alpha_0\beta_0=3$, so $|\alpha_0|=3$ and $|\beta_0|=1$ or $|\alpha_0|=1$ and $|\beta_0|=3$. WLOG, suppose that $|\alpha_0|=3$ and $|\beta_0|=1$. Let $m$ is the smallest positive integer such that $3\nmid\alpha_m$ (such $m$ exists because $g(x)$ is monic). Now we have $$[z^m]f(x)=\alpha_0\beta_m+\alpha_1\beta_{m-1}+\cdots+\alpha_m\beta_0\equiv\alpha_m\beta_0\not\equiv0\pmod3$$ So $m\ge n-1$, and $\deg g\ge n-1$, thus $\deg h\le 1$, hence $\deg h=1$, and $h(x)=x+\beta_0$, where $\beta_0=\pm1$, so $h(\beta_0)=0$, and $f(\beta_0)=0$, but it's obvious that $f(\pm1)\neq0$, Q.E.D.

Yai0Phah
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    +1 This is a very nice argument. It is, of course, an example of using a $3$-adic Newton's polygon, but could also be given as a challenge question after introducing Eisenstein's criterion. – Jyrki Lahtonen Jun 28 '12 at 08:09
  • @JyrkiLahtonen You remind me of Eisenstein's criterion, and I'm just aware that my proof is just a varient of the proof for Eisenstein's criterion. – Yai0Phah Jun 28 '12 at 08:13
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    @Jyrki It's a minor variant of Eisenstein - see my answer. Of course Newton polygon's are the Master Theorem here. – Bill Dubuque Jun 28 '12 at 17:20
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Hint $\ $ This is a minor variant of Eisenstein's criterion. Mod $3$ it factors as $\rm\:x^{n-1}(x+5)\:$ so by uniqueness of factorization if $\rm\:f = gh\:$ then, mod $3,\,$ $\rm\:g = x^j,\, $ $\rm\,h = x^k(x+5),\,$ $\rm\:j\!+\!k = n\!-\!1.\,$ But not $\rm\,j,k > 0\,$ else $\rm\:3\:|\:g(0),h(0)\:$ $\Rightarrow$ $\rm\:9\:|\:g(0)h(0)=f(0).\:$ Hence either $\rm\:j=0,\:$ so $\rm\:f\:$ is irreducible, or $\rm\:k=0,\:$ and $\rm\:f\:$ has a linear factor $\rm\:g,\:$ which is easily ruled out. $\ \ $ QED

Bill Dubuque
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    (+1) But shouldn't that read Proof instead of Hint? – Pedro Jun 28 '12 at 17:20
  • +1 Yes, another nice natural route to bound the degree of potential factors. That seems to be the theme here. – Jyrki Lahtonen Jun 28 '12 at 20:41
  • I have a doubt yesterday. For example, if $\Bbb Z_p={0,1,\ldots,p-1}$, can we conclude that $\Bbb Z_p[x]$ is an Euclidean domain? The uniqueness of factorization might work in such case, but I'm not sure because I'm just a high-school student without good education of algebra. – Yai0Phah Jun 29 '12 at 01:50
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    @Frank Yes, a polynomial ring $\rm:F[x]:$ over a field $\rm:F:$ is a Euclidean domain (via the high-school long division algorithm), so it is a unique factorization domain. This is proved in most every course in university algebra. – Bill Dubuque Jun 29 '12 at 02:27
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You could write $f(x)=x^n+5x^{n-1}+3$ as $$f(x)=x^n+5x^{n-1}+0\cdot x^{n-2}+0\cdot x^{n-3}+...+0\cdot x^{2}+0\cdot x+3$$

Hence by the Eisenstein's Irreducibility Criterion Theorem

Lets take the prime $p$ as $p=3$

Hence $3|0$, $9\nmid3$, $3\nmid5$

Therefore $f(x)$ has a irreducibility factor of at least $n-1$

Now by the factor theorem we can put $x=\pm1,\pm3$ in $f(x)$ but none of the value is a zero of the polynomial

Q.E.D.

Hardik Deshmukh
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Another solution is to use Perron's criterion, as follows:

Since $|5| > 1+|0|+|0|+\cdots+|0|+|3| \Leftrightarrow 5>4$, by Perron's criterion, $f(x)$ is irreducible over $\mathbb{Q}$ and thus cannot be expressed as the product of two nonconstant polynomials with integer coefficients.

Kevin Chen
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