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Section 4.4 Polynomials

Definition 4.4.1.

A polynomial \(f(x)\) with integer coefficients is reducible provided there are non-constant polynomials \(h(x)\) and \(g(x)\) with integer coefficients such that

\begin{equation*} f(x) = h(x)g(x). \end{equation*}

Definition 4.4.2.

A polynomial \(f(x)\) with integer coefficients is irreducible if it is not reducible.

Let

\begin{equation*} f(x) = a_0 + a_1x +...+ a_nx^n \end{equation*}

be a non-constant polynomial with integer coefficients and suppose there is a prime \(p\) such that \(p\vert a_i\) for \(i = 0, 1,..., n-1,\) and \(p\nmid a_n,\) and \(p^2 \nmid a_0.\) For the purpose of a contradiction, suppose that \(f(x)\) is reducible, say \(f(x) = g(x)h(x)\) with

\begin{equation*} g(x) = \sum_{i=0}^{s}b_ix^i \hspace{5mm}\mathrm{and}\hspace{5mm} h(x) = \sum_{i=0}^{t}c_ix^i, \end{equation*}

and \(s, t \geq 1.\) Since \(a_0 = b_0c_0\) is divisible by \(p\) but not \(p^2,\) precisely one of \(b_0,\) \(c_0\) is divisible by \(p;\) without loss of generality, say \(p\vert b_0.\) Now \(p\nmid b_s,\) otherwise it would divide \(a_n = b_sc_t.\) Thus there is some \(i \leq s\) such that \(p\) divides \(b_0, b_1,..., b_{i-1}\) but not \(b_i.\) Now

\begin{equation*} a_i = \sum_{k=0}^{i}b_kc_{i-k}, \end{equation*}

with \(p\) dividing both \(a_i\) (since \(i \leq s = n-t \lt n\)) and \(\sum_{k=0}^{i-1}b_kc_{i-k},\) so \(p\vert b_ic_0.\) This means, by a previous theorem, that \(p\vert b_i\) or \(p\vert c_0,\) a contradiction. Thus \(f(x)\) is irreducible.

To show that \(\Phi_p(x)\) is irreducible, we cannot apply Eisenstein's criterion directly; however, it is sufficient to show that the polynomial

\begin{equation*} f(x)=\Phi_p(x+1) \end{equation*}

is irreducible. Now from an earlier lecture we know that

\begin{equation*} \Phi_p(x)=\frac{x^p-1}{x-1}, \end{equation*}

so we have

\begin{equation*} f(x) =\frac{(x+1)^p-1}{x}=\sum_{i=0}^{p-1}\binom{p}{i}x^i=x^{p-1}+\binom{p}{p-1}x^{p-2}+...+\binom{p}{2}x^1+\binom{p}{1}. \end{equation*}

Since \(p\) is a prime \(p\) divides \(\binom{p}{i}\) for \(i=1,2,...,p-1.\) Also, \(p\) does not divide the leading coefficient (\(= 1\)) of \(f(x),\) and \(p^2\) does not divide the constant term, \(\binom{p}{1}=p.\) Thus, by Eisenstein's criterion, \(f(x)\) is irreducible, and hence \(\Phi_p(x)\) is irreducible.

By contrapositive. Suppose that \(m\) is not a power of \(2.\) Then \(m = 2^nq\) for some odd \(q \gt 1.\) Now the polynomial \(f(x) = x^q+1\) has a root at \(x =-1,\) so it is divisible by \(x + 1.\) Since \(q \gt 1, x + 1\) is a proper factor of \(f(x);\) thus, setting \(x = y^{2^{n}}\) we see that the polynomial

\begin{equation*} g(y) = f(y^{2^{n}})=y^m+1, \end{equation*}

which has a proper factor \(y^{2^{n}}+1.\) Setting \(y = 2\) this says that \(2^{2^{n}}+1\) is a proper factor of the integer \(g(2) = 2^m + 1,\) which is, by means of having a proper factor, not prime.