Gauss's Lemma

Gauss's Lemma
Theorem: Let  
\[f \in \mathbb{Z}[x]\]
 . Then  
\[f\]
  is irreducible over  
\[\mathbb{Z}[x]\]
  if and only if  
\[f\]
  is irreducible over  
\[\mathbb{Q}[x]\]
 .
(In other words, Let  
\[f(x)\]
  be a polynomial with integer coefficients. If  
\[f(x)\]
  has no factors with integer coefficients, then 
\[f(x)\]
  has no factors with rational coefficients.)
Proof: Let 
\[f(x) = g(x)h(x)\]
  be a factorization of 
\[f\]
  into polynomials with rational coefficients. Then for some rational 
\[a\]
  the polynomial 
\[a g(x)\]
  has integer coefficients with no common factor. Similary we can find a rational 
\[b\]
  so that 
\[b h(x)\]
  has the same properties. (Take the lcm of the denominators of the coefficients in each case, and then divide by any common factors.)
Suppose a prime 
\[p\]
  divides 
\[a b\]
 . Since  
\[ a b f(x) = (a g(x))(b h(x)) \]
 
becomes 
\[ 0 = (a g(x)) (b h(x))\]
  modulo 
\[p\]
 , we see 
\[a g(x)\]
  or 
\[b h(x)\]
  is the zero polynomial modulo 
\[p\]
 . (If not, then let the term of highest degree in 
\[a g(x)\]
  be 
\[m x^r\]
 , and the term of highest degree in 
\[b h(x)\]
  be 
\[n x^s\]
 . Then the product contains the term 
\[m n x^{r+s} \ne 0 \pmod {p}\]
 , a contradiction.)
In other words, 
\[p\]
  divides each coefficient of 
\[a g(x)\]
  or 
\[b h(x)\]
 , a contradiction. Hence 
\[a b = 1\]
  and we have a factorization over the integers.
Example: Let 
\[p\]
  be a prime. Consider the polynomial  
\[ f(x) = 1 + x + ... + x^{p-1} . \]

We cannot yet apply the criterion, so make the variable subsitution 
\[x = y + 1\]
 . Then we have  
\[ g(y) = 1 + (y+1) + ... + (y+1)^{p-1} . \]

Note 
\[f(x)\]
  is irreducible if and only if 
\[g(y)\]
  is irreducible.
The coefficient of 
\[y^k\]
  in 
\[g(y)\]
  is  
\[ \sum_{m=k}^{p-1} {\binom{p-1}{k}} = {\binom{p}{k+1}} . \]

The last equality can be shown via repeated applications of Pascal’s identity:  
\[ {\binom{n+1}{k}} = {\binom{n}{k}} + {\binom{n}{k-1}} . \]
 
Alternatively, use the fact  
\[ g(y) = \frac{(y+1)^p - 1}{(y+1) - 1} \]
 
Thus 
\[p\]
  divides each coefficient except the leading coefficient, and 
\[p^2\]
  does not divide the constant term 
\[p\]
 , hence 
\[f(x)\]
  is irreducible over the rationals.