\begin{theorem}[Dirichlet's Theorem on Primes in Arithmetic Progressions \cite{Silverman}]
Let a and m be integers with gcd(a,m)=1. Then there are infinitely many primes that are congruent to a modulo m. That is, there are infinitely many prime numbers p satisfying $$p\equiv a\pmod{m}.$$
\end{theorem}
The traditional proof offered by Dirichlet for this theorem is a complicated one, involving calculus with complex numbers, and it would thus be useful to have a ``Euclidean'' proof as well. \cite{Silverman}Before venturing into if this is possible, however, it will be necessary to extend our study of prime numbers to polynomials.
The traditional proof offered by Dirichlet for this theorem is a complicated one, involving calculus with complex numbers, and it would thus be useful to have a ``Euclidean'' proof as well. \cite{Silverman}As it turns out, this is unfortunately not possible in every case, as there are limits to the scenarios which allow use of the Euclidean argument. Before defining these limits, however, it will be necessary to extend our study of prime numbers to polynomials.
\section{Prime Divisors and Polynomials}
A polynomial's prime divisors, as well some properties involving prime divisors, will be essential in exploring the limits of the ``Euclidean'' proof.
@ -191,9 +191,43 @@
\begin{proof}
Let $f(x)$ and $g(x)$ be any two polynomials with integer coefficients with degrees $N$ and $M$, respectively. By Lemma 3.3, we can assume that $N>M$. \par
The remainder of the proof will be done through complete induction. If either polynomial is linear, then by Corollary 3.2.1, they share an infinite number of prime divisors. Thus, degrees of $N=2$ and $M=1$ provide our base case. \par
Now, assume that the theorem is true for all pairs of polynomials whose highest degree is less that $N$. Again, consider $f(x)$ of degree $N$ and $g(x)$ of degree $M$. If $M=1$, apply Corollary 3.2.1. Then suppose $N>M>1$, and construct $$b^if(x+c)=g(x)h(x)+g_1(x),$$ where $f$, $g$, $g_1$, and $c$ follow the conditions of lemmas 3.2-3.5, including a choice of $c$ so that $g_1(x)$ is not constant. \par
Now, assume that the theorem is true for all pairs of polynomials whose highest degree is less that $N$. Again, consider $f(x)$ of degree $N$ and $g(x)$ of degree $M$. If $M=1$, apply Corollary 3.2.1. Then suppose $N>M>1$, and construct $$b^if(x+c)=g(x)h(x)+g_1(x),$$ where $f$, $b^i$, $g$, $g_1$, and $c$ follow the conditions of lemmas 3.2-3.5, including a choice of $c$ so that $g_1(x)$ is not constant. \par
Let $y=x+c$. Note that $g(x)$ and $g_1(x)$ can be rearranged into functions of $x+c$. Call these new functions $g_2(y)$ and $g_3(y)$, respectively, where $g_3$ has degree at most $M-1$ and both $g_2(y)$ and $g_3(y)$ have coefficients that are polynomials in $c$. Based on our assumption, since both $g_2(x)$ and $g_3(y)$ have degree less than $N$, the theorem holds for these two polynomials. Furthermore, if a prime $p$ is one of the common prime divisors of $g_2$ and $g_3$, then it must also be a prime divisor of $b^if(x+c)=b^if(y_0)$. Thus, because Lemma 3.5 guarantees that there are an infinite number of values of $c$ for which this argument holds, and since $b^i$ has only a finite number of prime divisors, $f$ and $g$ have infinitely many common prime divisors.
\end{proof}
\section{Limits of the Euclidean Argument}
While the results of the results of the preceding section are essential to achieving our primary goal, we will still need to make use of two more results. Their proofs are outside of the scope of this paper and will therefore be omitted.
\begin{theorem}[\cite{MurtyandThain}]
Let $H$ be a subgroup of $(\mathbb{Z}/k\mathbb{Z})$*. Then there is an irreducible polynomial $f$ so that all the prime divisors of $f$, with a finite number of exceptions, belong to the residue class of $H$. Any such prime divisor belonging to any residue class of $H$ divides $f$.
\end{theorem}
As an example, consider the subgroup $H=\{1,3\}$ in $(\mathbb{Z}/k\mathbb{Z})$* and the irreducible polynomial $f(x)=4x+3$. Any prime $p$ that divides $f$ is either $1\pmod{4}$ or $3\pmod{4}$ (with only a finite number of exceptions). In addition, if any prime is $1\pmod{4}$ or $3\pmod{4}$, it is guaranteed to divide $f$. \par
There is also one specific case of of this theorem that will be needed in the final proof:
\begin{corollary}
If $\phi_k$ is the $k$-th cyclotomic polynomial, then all of the prime divisors of $\phi_k$ are $\equiv1\pmod{k}$ or divide $k$.
\end{corollary}
An immediate consequence of Corollary 4.1.1 is that there are infinitely many prime divisors of $\phi_k$ that are congruent to $1\pmod{k}$. This is because there are only a finite number of primes that divide $k$, and we have already proven that $\phi_k$ must have infinitely many prime divisors. Combining this logic with that of Theorem 3.6, because any two polynomials have an infinite number of common prime divisors, \textit{any} polynomial with integer coefficients must have infinitely many prime divisors that are congruent to $1\pmod{k}$. With this remarkable result in hand, we finally have the tools necessary to prove to define the limits of ``Euclidean'' proofs:
\begin{theorem}[\cite{MurtyandThain}]
If $l^2\equiv1\pmod{k}$ then there are infinitely many primes $\equiv l\pmod{k}$, provided there is at least one.
\end{theorem}
\begin{proof}
Let $l$ be an integer. If $l\equiv1\pmod{k}$, apply Corollary 4.1.1. Thus we are left with the case in which $l$ is not congruent to $1\pmod{k}$. \par
We will now consider two groups: the group $(\mathbb{Z}/k\mathbb{Z})$*, which is the integers modulo $k$, and one of its subgroups $H=\{1,l\}$. Now, letting $\zeta$ be a primitive $k$-th root of unity, construct the polynomial $h(\zeta)=(u-\zeta)(u-\zeta^l)$. If $m_1,m_2,\dots,m_s$ are the coset representatives of $H$ in $(\mathbb{Z}/k\mathbb{Z})$*, then we want to choose $u$ so that $h(\zeta^{m_i})$ is distinct for $i=1,2,\dots,s$. The case in which they are not distinct is only possible for a finite number of choices of $u$. Then, according to Theorem 4.1 with subgroup $H$ and $\eta=h(\zeta)$, there exists a polynomial $$f(x)^2=\prod_{(a,k)=1}(x-(u-\zeta^a)(u-\zeta^{la})),$$ whose prime divisors, except for a finite number, are congruent to $1$ or $l\pmod{k}$. Additionally, $f(0)=\phi_k(u)$, where $\phi_k(u)$ is the $k$-th cyclotomic polynomial. If we then let $u$ be a non-zero multiple of $k$, then $f(0)=\phi_k(u)\equiv1\pmod{k}$. Thus every prime divisor $p$ of $f(0)$ is such that $p\equiv1\pmod{k}$, because Corollary 4.1.1 guarantees that every prime divisor of $\phi_k$ is either congruent to $1\pmod{k}$ or is a multiple of $k$. \par
We will now consider a prime $p\equiv l\pmod{k}$ that does not divide the discriminant of $f$. By Theorem 5, $p\mid f(b)$ for some $b\in\mathbb{Z}$. If $p^2\mid f(b)$, then $f(b+p)=f(b)+pf'(b)\equiv pf'(b)\pmod{p^2}$. However, $f'(b)\neq0\pmod{p^2}$, because $p\nmid D(f)$ implies that $f$ can have no double roots modulo $p$. Thus, it follows from $f(b)\equiv0\pmod{p^2}$ that $f(b+p)\neq0\pmod{p^2}$. In this case, replace $b$ with $b+p$ so that the choice of $b$ is always such that $p^2\nmid f(b)$. \par
Finally, we proceed with a ``Euclidean'' argument: assume there are a finite number of primes congruent to $l$ modulo $k$, $p,p_1,\dots,p_m$. Additionally, let $q_1,q_2,\dots,q_t$ be the prime divisors of $D(f)$ and $$Q=p_2p_3\cdots p_mq_1q_2\cdots q_t.$$ Note that $p^2$ and $kQ$ are relatively prime, and thus by the Chinese Remainder Theorem there exists a $c$ such that $c\equiv b\pmod{p^2}$ and $c\equiv0\pmod{kQ}$. This also implies that $f(c)\equiv f(b)\pmod{p^2}$ and $f(c)\equiv f(0)\pmod{kQ}$. As already mentioned, all but a finite number of the prime divisors of $f$ are congruent to either $1$ or $l$ modulo $k$. This remaining finite set of primes must either divide $k$ or the discriminant of $f$. Because $f(0)$ is only divisible by primes that are congruent to $1$ modulo $k$, so is $f(c)$. Also $p^2\nmid f(c)$, which implies that $f(c)\equiv l\pmod{k}$. But $f(c)\equiv f(0)\equiv1\pmod{k}$, which is a contradiction. Therefore, there must be infinitely many prime numbers congruent to $l$ modulo $k$.
\end{proof}
This theorem has established the cases in which the conditions exist for the creation of a ``Euclidean'' proof for the infinitude of arithmetic progressions of prime numbers. While it is unfortunate that this argument cannot be universally applied, it still provides a relatively simple and elegant method of proof for many progressions. As a final demonstration of this fact, we will return to the proof of Theorem 2.1 ($3\pmod{4}$ primes) and restructure it to use the methods of the above proof. The same basic method can be applied to any progression that is guaranteed to work by Theorem 4.2.
\begin{proof}
We are considering the subgroup $H=\{1,3\}$ of $(\mathbb{Z}/4\mathbb{Z})$*. First, note that $3^2=9\equiv1\pmod{4}$, which is why we can use the methods of Theorem 4.2 to reprove this result. Let $\zeta=i$ and construct the polynomial
\begin{align*}
f(x)&=x-(u-\zeta)(u-\zeta^3) \\
f(x)&=x-(u-i)(u-i^3) \\
f(x)&=x-(u^2-iu+iu+1) \\
f(x)&=x-u^2+1.
\end{align*}
Since $1$ and $2$ are the coset representatives of $H$, we must choose a $u$ such that $f(\zeta^1)=f(i)$ and $f(\zeta^2)=f(-1)$ are distinct. This is clearly the case for $u=4$, so we now have $f(x)=x-17$. \par
Now, notice that $2$ is not a prime divisor of the translation $f(4x+20)=4x+3$, and so all of its prime divisors are either $1\pmod{4}$ or $3\pmod{4}$. Additionally, $f(4x+20)\equiv3\pmod{4}$, so not all of its prime divisors are $1\pmod{4}$. Suppose that $f(4x+20)$ has a finite number of prime divisors congruent to $3\pmod{4}$ and let $Q$ be their product. Then consider $f(4Q+20)=4Q+3$, which is clearly $3\pmod{4}$. But, $Q$ and $4Q+3$ are coprime, so it has no prime divisor that is $3\pmod{4}$. This is a contradiction, and we have thus proven the infinitude of $3\pmod{4}$ primes.
\end{proof}
\section{Conclusions}
Euclid, when proving that there is an infinite number of prime numbers, formulated an argument whose utility remained untapped for many years. Throughout this paper, that utility has been demonstrated by proving the infinitude of different categories of prime numbers, including the abstract case. Though the argument does have its limits, it provides both elegant and relatively simple proofs when compared to alternative methods, and at the same time demonstrates that proof techniques can be have many useful applications beyond their original use.
\begin{thebibliography}{2}
\bibitem{Silverman} Joseph Silverman, \textit{A Friendly Introduction to Number Theory} p.83-140, Pearson, Boston, 4th edition, 2011.
