\(\textbf{Lemma 1. }\) Every monic polynomial over \(F\) with integer degree \(n\) where \(n > 1\), is divisible by an irreducible monic polynomial over \(F\).
\(\textbf{Proof. }\)Let \(p(x)\) to be any monic polynomial over \(F\). Consider the set of divisors of \(p(x)\) whose degrees are greater than or equal \(1\) and less than \(n\). It is either empty or nonempty. If it is empty, then \(p(x)\) is an irreducible polynomial by definition and it can be considered as a divisor of itself. If it is nonempty, then the least-integer principle says that some polynomials of its divisors have a smallest degree, and we call one of them as \(d(x)\). Now we have to show that \(d(x)\) is an irreducible monic polynomial by contradcition. If \(d(x)\) had a divisor whose degree is larger than \(1\) and less than \(\text{deg}(d(x))\), then \(d(x)\) is not irreducible and so would \(n\). H However, this is impossible because \(d(x)\) was the monic polynomial with the smallest degree in the set of divisors of \(p(x)\). Thus \(d\) is an irreducible monic polynomial over \(F\). \[ \tag*{$\square$} \]
\(\textbf{Lemma 2. }\) Every monic polynomial over \(F\) \(p(x)\) with degree \(n,n \geq 1\), can be written as a product of irreducible monic polynomials over \(F\).
\(\textbf{Proof}\). From \(\textbf{Lemma 1}\) \(\textbf{Lemma 1. }\)Every monic polynomial over \(F\) with integer degree \(n\) where \(n > 1\), is divisible by an irreducible monic polynomial over \(F\). , we know that there is an irreducible monic polynomial \(p_1(x)\) such that \(p_1(x)\;|\; p(x)\). Hence there exists a polynomial \(n_1(x)\) where its degree \(0\leq n_1< n\) such that \(p(x) = p_1(x)n_1(x)\). If \(n_1(x)\) is a constant number, then we can know that \(n_1(x) = 1\), or \(p(x)\) will not be a monic polynomial since \(p_1(x)\) is monic as well. Then \(p(x) =p_1(x)n_1(x) = p_1(x)\) is a irreducible monic polynomial over \(F\). If \(n_1(x)\) has degree larger or equal to \(1\), then from \(\textbf{Lemma 1}\) Every monic polynomial over \(F\) with integer degree \(n\) where \(n > 1\), is divisible by an irreducible monic polynomial over \(F\). . again, there is an irreducible monic polynomial that divides \(n_1(x)\). That is, \(n_1(x) = p_2(x)n_2(x)\), where \(p_2(x)\) is an irreducible monic polynomial and its degree \(0\leq n_2 \lt n_1\). If \(n_2 =1\), again we are done: \(p(x) = p_1(x)p_(x)\) is written as a product of irreducible monic polynomials over \(F\). But if the degree of \(n_2(x)\) is larger than or equal to one, then we apply \(\textbf{Lemma 1}\) Every monic polynomial over \(F\) with integer degree \(n\) where \(n > 1\), is divisible by an irreducible monic polynomial over \(F\). . once again and get that \(n_2(x) = p_3(x)n_3(x)'\) with \(p_3(x)\) an irreducible monic polynomial with degree \(0\leq \text{deg}(n_3(x)) \lt \text{deg}(n_2(x))\). If \(n_3=1\), we are done. If not, we continue the same process. We will sooner or later come to \(n_i = 1\) since \(\text{deg}(p(x)) > \text{deg}(n_1(x)) > \text{deg}(n_2(x)) >\dots\) and each \(n_i(x)\) has a positive degree; such a sequence cannot continue forever. For some \(k\), we will have \(n_k(x) = 1\), in which case \(p(x) = p_1(x)p_2(x)\cdots p_k(x)\) is the desired expression of \(p(x)\) as a product of irreducible monic polynomials over \(F\). \[ \tag*{$\square$} \]
\(\textbf{Note.}\) each \(p_i(x)\) may not be in unique.