Proof. Suppose \(f(x) = h(x)\, g(x)\), where both \(h(x)\, , g(x)\in F(x)\) and have deg less than that of \(f(x)\). Then \[\deg (f(x)) = \deg (h(x)) + \deg (g(x))\hspace {0.3cm}\text {and}\hspace {0.3cm} \deg (f(x))\] at least one of \(h(x)\) and \(g(x)\) has degree 1. Say \(g(x) = ax + b\). Thus clearly \(-a^{-1}b\) is a zero for \(f(x)\).
Conversely, suppose \(f(a) = 0\) for some \(a\in F\), then by the factor theorem, we know that \((x - a)\) is a factor of \(f(x)\) then \(f(x)\) is reducible over \(F\).

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Proof. Let \(f(x)\) and \(g(x)\) be primitive. Suppose that the product \(f(x)\,g(x)\) is not primitive. Then the content of \(f(x)\, g(x) \neq 1\). Let \(p\) be a prime divisor of the content of \(f(x)\, g(x)\). Let \(\overline {f(x)}\, , \, \overline {g(x)}\) and \(\overline {f(x)\, g(x)}\) be polynomials obtained from \(f(x)\, , \, g(x)\) and \(f(x)\, g(x)\) mod\(_p\). Then \(\overline {f(x)}\) and \(\overline {g(x)}\) belong to the integral domain \(\mathbb {Z}_p[x]\).
Now \(\, \overline {f(x)}\, \,\overline {g(x)} = \overline {f(x)\, g(x)} = 0\,\) in \(\mathbb {Z}_p[x]\). Thus \(\overline {f(x)} = 0\) or \(\overline {g(x)} = 0\). This means that either \(p\) divides every coefficient of \(f(x)\) or \(p\) divides every coefficient of \(g(x)\). Hence either \(f(x)\) is not primitive or \(g(x)\) is not primitive. This is a contradiction.

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Proof. Suppose \(\, f(x) = g(x)\, h(x)\,\) where \(g(x)\, , \, h(x)\in \mathbb {Q}[x]\). Clearly, we assume \(f(x)\) is primitive since we can divide both \(f(x)\) and \(g(x)\, h(x)\) by the content of \(f(x)\). Let \(a\) be the \(L\subset M\) of the denominators of the coefficients of \(g(x)\) and \(b\) the \(LCM\) of the denominators of \(h(x)\). Then \[a\, b\, f(x) = a\, b \cdot g(x)\, h(x) = a\, g(x) \cdot b\, h(x)\] where \(a\, g(x)\) and \(b\, h(x)\) are in \(\mathbb {Z}[x]\). Let \(c_1\), be the content of \(a\, g(x)\) and \(c_2\) be the content of \(b\, h(x)\). Then \(a\, g(x) = c_1\, g_1(x)\) and \(b\, h(x) = c_2\, h_1(x)\) where both \(g_1(x)\) and \(h_1(x)\) are primitive and \(a b f(x) = c_1 c_2\, g_1(x)\, h_1(x)\). Since \(f(x)\) is primitive then content of \(ab\, f(x)\), it follows that the content of \(c_1 c_2\, g_1(x)\, h_1(x)\) is \(c_1 c_2\). Thus \(a b = c_1 c_2\) and \(f(x) = g_1(x)\, h_1(x)\) where \(g_1(x)\) and \(h_1(x)\in \mathbb {Z}[x]\) and \(\deg g_1(x) = \deg g(x)\). \(\, \deg h_1(x) = \deg h(x)\).

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Proof. If \(f(x)\) is reducible, then \(f(x) = g(x)\, h(x)\) where \(g(x)\, , \, h(x)\in \mathbb {Z}[x]\) and are of degree less than \(f(x)\).
Now, \(\deg \overline {f(x)} = \deg f(x)\). Thus, we have \(\deg \overline {g(x)} \leq \deg g(x) < \deg \overline {f(x)}\) and \(\deg \overline {h(x)} \leq \deg h(x) < \deg \overline {f(x)}\). But \(\overline {f(x)} = \overline {g(x)}\, \overline {h(x)}\) and this is a contradiction to our assumption that \(\overline {f(x)}\) is irreducible over \(\mathbb {Z}_p\).

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Proof. \[f(x) = \Phi _p(x + 1) = \dfrac {(x + 1)^p - 1}{(x + 1) - 1} = (x + 1)^{p - 1} + (x + 1)^{p -2} + \cdots \cdots \cdots + 1 = x^{p-1} + p\, x^{p-2} + \cdots \cdots \cdots + 1\hspace {0.3cm} (*)\] Then, since every coefficient of \(x^{p - 1}\) is divisible by \(p\) by Eisenstein’s Criterion, \(f(x)\) is irreducible over \(\mathbb {Q}\). So if \(\Phi _p(x) = g(x)\, h(x)\) was some non-trivial factorization of \(\Phi _p(x)\), then
\(f(x) = \Phi (x + 1) = g(x + 1)\, h(x + 1)\) would be a nontrivial factorization of \(f(x)\) over \(\mathbb {Q}\). Now since this is impossible, we conclude that \(\Phi _p(x)\) is irreducible.

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Proof. Suppose that \(\langle p(x)\rangle \) is maximal in \(\mathbb {F}\). Clearly, \(\langle p(x)\rangle \) is neither the zero nor the 1 in \(\mathbb {F}\). If \(p(x) = g(x)\, h(x)\) is a factorization of \(p(x)\) over \(\mathbb {F}[x]\). Then \(\langle p(x) \rangle \subseteq \langle g(x)\rangle \subseteq \mathbb {F}\). But since \(\langle p(x)\rangle \) is maximal then \(\langle p(x)\rangle = \mathbb {F}\). (Since it implies \(\langle p(x) \rangle = \langle g(x)\rangle )\). Thus deg \(p(x) = \deg g(x)\), hence \(\deg h(x) = 0\). Thus \(p(x)\) cannot be written as a product of polynomials of lower degree in \(\mathbb {F}[x]\).
Conversely, suppose \(p(x)\) is irreducible over \(\mathbb {F}\). Let \(I\) be any ideal of \(\mathbb {F}[x]\) such that \(\langle p(x) \rangle \subseteq I \subseteq \mathbb {F}[x]\). Now, \(\mathbb {F}[x]\) is a principle ideal domain, we have \(I = \langle g(x)\rangle \) for some \(g(x)\in \mathbb {F}[x]\). So, \(p(x)\in \langle g(x)\rangle \). Hence \(p(x) = g(x)\, h(x)\), where \(h(x)\in \mathbb {F}[x]\). Now, since \(p(x)\) is irreducible over \(\mathbb {F}\), it follows that \(g(x)\) or \(h(x)\) is constant. In this case, we have \(I = \mathbb {F}[x]\) or \(\langle p(x) \rangle = \langle g(x)\rangle = I\). Thus \(\langle p(x)\rangle \) is maximal in \(\mathbb {F}[x]\).

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Proof. \(\mathbb {F}[x]\) is a commutative ring with identity, so the quotient by an ideal is a commutative ring with identity. It remains to produce inverses.

Let \(g(x) + \langle p(x)\rangle \) be a non-zero coset, so \(p \nmid g\). Since \(p\) is irreducible, its only divisors are units and associates of itself, so \(\gcd (p,g) = 1\). By the division algorithm in \(\mathbb {F}[x]\) there are \(u,v\) with \[u(x)p(x) + v(x)g(x) = 1 ,\] and reducing modulo \(\langle p(x)\rangle \) gives \(v(x)g(x) \equiv 1\). So \(v(x)+\langle p(x)\rangle \) is the required inverse, and the quotient is a field. □

Proof. Suppose \(p(x)\) is irreducible, then \(\mathbb {F}/\langle p(x)\rangle \) is a field thus it is an integral domain. Let \(\overline {a}(x)\) and \(\overline {b}(x)\) be the images of the natural homomorphism of \(a(x)\) and \(b(x)\). Now, since \(p(x)\, |\, a(x)\, b(x)\), we have \(\overline {a}(x)\, \overline {b}(x) = \overline {0}\), then zero in \(\mathbb {F}/\langle p(x) \rangle \). Thus \(\overline {a}(x) = \overline {0}\) or \(\overline {b}(x) = \overline {0}\) and it follows that \(p(x)\, |\, a(x)\) or \(p(x)\, |\, b(x)\).



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Proof. Argue by induction on \(\deg f\). If \(f\) is irreducible over \(F\) then \(F[x]/\langle f(x)\rangle \) is a field containing a root of \(f\), namely the coset \(x + \langle f(x)\rangle \), and it contains an isomorphic copy of \(F\).

If \(f\) is reducible, write \(f = gh\) with \(\deg g,\deg h < \deg f\). By induction there is an extension of \(F\) in which \(g\) has a zero, and that zero is a zero of \(f\) as well. Either way an extension field containing a zero of \(f\) exists. □

Proof. Since \(p(x)\) is irreducible and \(p(a)=0\), consider the division of \(p(x)\) by \(x-a\) in \(F[x]\): there are \(q(x)\) and a constant \(r\) with \[p(x) = (x-a)q(x) + r .\] Evaluating at \(a\) gives \(r = p(a) = 0\), so \((x-a)\mid p(x)\). As \(p\) is irreducible its only factorisations are trivial, so \(q\) must be a unit and \(p(x) = c(x-a)\) for a constant \(c\neq 0\). Hence \(p\) has degree one. □

Proof. We know that \(\, F(a) \,\approx \,F[x]\big /\langle p(x)\rangle \approx F(b)\), thus \(\, F(a)\approx F(b)\).

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Proof. Suppose \(a\) is a multiple zero for \(f(x)\) in \(E\) then there exist \(g(x)\in E[x]\) such that
\(f(x) = (x - a)^2\, g(x)\). Since \(\, f'(x) = (x - a)^2\, g'(x) + 2(x - a)\, g(x)\). We see that \(f'(a) = 0\). Hence \(x - a\) is a factor of \(f(x)\) and \(f'(x)\). Now, if \(f(x)\) and \(f'(x)\) have a common division of \(+ve\) degree in \(F[x]\), there are polynomials \(h(x)\) and \(k(x)\) such that \(f(x) \, h(x) + f'(x)\, k(x) = 1\). Now, \(f(x)\, h(x) + f'(x)\, k(x)\in E[x]\). Note that this says that \(x-a\) is a factor of \(1\). Which does not make sense. Thus \(f(x)\) and \(f'(x)\) must have a common divisor of \(+ve\) degree.
Conversely, suppose \(f(x)\) and \(f'(x)\) have a common factor of \(+ve\) degree. Let \(a\) be a zero of that common factor. Then \(a\) is a zero of \(f(x)\) and \(f'(x)\). Since \(a\) is a zero of \(f(x)\), then \(f(x) = (x - a)\, q(x)\). Then \(f'(x) = (x-a)\, q'(x) + q(x)\) and \(0 = f'(a) = q(a)\). Thus \(x - a\) is a factor of \(q(x)\) and \(a\) is a multiple zero of \(f(x)\).

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