Show that \[ x^{10}+x^9+x^8+\cdots+x+1 \] is irreducible over \(\mathbb Q\). Is \[ x^{11}+x^{10}+\cdots+x+1 \] irreducible over \(\mathbb Q\)?
Solution
Define \[ P(x)=1+x+\cdots+x^{10}. \]
1. One has\[ P(x) = \frac{x^{11}-1}{x-1}. \]
This is the finite geometric-series identity: \[ (x-1)(1+x+\cdots+x^{10})=x^{11}-1. \] Since both sides are polynomial identities, the quotient formula holds in \(\QQ[x]\).
2. The shifted polynomial satisfies\[ P(x+1) = \frac{(x+1)^{11}-1}{x} = \sum_{j=1}^{11} \binom{11}{j}x^{j-1}. \]
Substitute \(x+1\) into step 1: \[ P(x+1) = \frac{(x+1)^{11}-1}{(x+1)-1} = \frac{(x+1)^{11}-1}{x}. \] The binomial theorem gives \[ (x+1)^{11}-1 = \sum_{j=1}^{11}\binom{11}{j}x^j. \] Dividing by \(x\) gives the displayed sum.
3. Every nonleading coefficient of \(P(x+1)\) is divisible by \(11\).
For \[ 1\leq j\leq10, \] the binomial coefficient is \[ \binom{11}{j} = \frac{11!}{j!(11-j)!}. \] Because \(11\) is prime and neither \(j!\) nor \((11-j)!\) is divisible by \(11\), the factor \(11\) in the numerator does not cancel. Hence \[ 11\mid\binom{11}{j}. \] The leading coefficient corresponds to \(j=11\) and equals \(1\).
4. The constant term of \(P(x+1)\) is \(11\), so it is not divisible by\(11^2\).
In step 2, the constant term is the term with \(j=1\): \[ \binom{11}{1}=11, \] and \(121\nmid11\).
5. The polynomial \(P(x+1)\) is irreducible over \(\QQ\).
Steps <1>3–<1>4 show that Eisenstein’s criterion applies to \(P(x+1)\) with the prime \(11\): every nonleading coefficient is divisible by \(11\), the leading coefficient is not, and the constant term is not divisible by \(11^2\). Therefore \(P(x+1)\) is irreducible over \(\QQ\).
6. The polynomial\[ \boxed{ 1+x+\cdots+x^{10} } \]is irreducible over \(\QQ\).
The substitution \[ \tau:\QQ[x]\longrightarrow\QQ[x], \qquad \tau(f)(x)=f(x+1), \] is a ring automorphism, with inverse \(f(x)\mapsto f(x-1)\). Ring automorphisms preserve reducibility and irreducibility. Since \(P(x+1)\) is irreducible by step 5, so is \(P(x)\).
7. Define\[ Q(x)=1+x+\cdots+x^{11}. \]Then\[ Q(-1)=0. \]
There are twelve alternating terms: \[ Q(-1) = 1-1+1-1+\cdots+1-1 = 0. \]
8. The polynomial \(Q\) is reducible over \(\QQ\).
By step 7, the factor theorem gives \[ x+1\mid Q(x). \] Since \(Q\) has degree \(11\), the quotient has degree \(10\), so this is a nontrivial factorization in \(\QQ[x]\).
9. Thus the answer to the second question is\[ \boxed{\text{No.}} \]
Step 8 shows that the second polynomial is reducible.
10. Q.E.D.
Step 6 proves irreducibility of the first polynomial, and step 9 answers the second question.