Is there possibly a largest prime number?

Below are a couple noteworthy variations on Euclid's classic proof that there are infinitely many primes. The first is a simplification and the second is a generalization to rings with few units.

Theorem $\rm\ \ N (N+1) \;$ has a larger set of prime factors than does $\:\rm N > 0$.

Proof $\ $ $\rm N+1 > 1\,$ so it has a prime factor $\rm P$ (e.g. its least factor $> 1)$. $\,\rm N$ is coprime to $\rm N+1\,$ so $\rm P$ can't divide $\rm N$ (else $\rm\: P$ divides $\rm N+1\:$ and $\rm N$ so also their difference $\rm N+1 - N = 1)$. So the prime factors of $\rm\: N(N+1)$ include all those of $\rm N$ and at least one prime $\rm P$ not dividing $\rm N$.

Corollary $\ $ There are infinitely many primes.

Proof $\, $ Iterating $\rm\, N\to N (N+1)\, $ yields integers with an unbounded number of prime factors.

Below, generalizing Euclid's classic argument, is a simple proof that an infinite ring has infinitely many maximal (so prime) ideals if it has fewer units than elements (i.e. smaller cardinality). The key idea is that Euclid's construction of a new prime generalizes from elements to ideals, i.e. given some maximal ideals $\rm P_1,\ldots,P_k$ then a simple pigeonhole argument employing $\rm CRT$ implies that $\rm 1 + P_1\cdots P_k$ contains a nonunit, which lies in some maximal ideal $\rm P$ which, by construction, is comaximal (so distinct) from the prior max ideals $\rm P_i.\:$ Below is the full proof, excerpted from from some of my old sci.math/AAA/AoPS posts.

Theorem $\ $ An infinite ring $\rm R$ has infinitely many maximal ideals if it has fewer units $\rm U = U(R)$ than it has elements, i.e. $\rm\:|U| < |R|$.

Proof $\rm\ \ R$ has a max ideal $\rm P_1,\:$ since the nonunit $\rm\: 0\:$ lies in some max ideal.
Inductively, suppose $\rm P_1,\ldots,P_k$ are maximal ideals in $\rm R$, with product $\,\rm J.$

$\rm Case\ 1\!: \; 1 + J \not\subset U.\:$ Thus $\rm 1 + J$ contains a nonunit $\rm p,\,$ lying in a max ideal $\rm P.$
It's new: $\rm\: P \neq P_i\:$ since $\rm\: P + P_i = 1\:$ via $\rm\: p \in P,\ 1 - p \in J \subset P_i$

$\rm Case\ 2\!: \; 1 + J \subset U$ is impossible by the following pigeonhole argument. $\rm R/J = R_1 \times \cdots \times R_k,\ R_i = R/P_i\:$ by the Chinese Remainder Theorem.
We deduce that $\rm\ |U(R/J)| \leq |U|\ $ because $\rm\ uv \in 1 + J \subset U \Rightarrow u \in U.$
Thus $\rm|U(R_i)| \leq |U(R/J)| \leq |U|\:$ via the injection $\rm u \mapsto (1,1,\ldots,u,\ldots,1,1).$
$\rm R_i$ field $\rm\: \Rightarrow\ |R| > 1 + |U| \geq |R_i|,\,$ and $\,\rm|J| \leq |U| < |R|\,$ via $\,\rm 1 + J \subset U.$
Therefore $\rm|R| = |R/J|\ |J| = |R_1|\ \cdots |R_k|\ |J|\:$ yields the contradiction that
the infinite $\rm|R|$ is a finite product of smaller cardinals.

I recall the pleasure of discovering this "fewunit" generalization of Euclid's proof and other related theorems while reading Kaplansky's classic textbook Commutative Rings as an MIT undergrad. There Kaplansky presents a simpler integral domain version as exercise $8$ in Section $1$-$1,\:$ namely

(This exercise is offered as a modernization of Euclid's theorem on the infinitude of primes.) Prove that an infinite integral domain with with a finite number of units has an infinite number of maximal ideals.

I highly recommend Kap's classic textbook to anyone seeking to master commutative ring theory. In fact I highly recommend everything by Kaplansky - it is almost always very insightful and elegant. Learn from the masters! For more about Kaplansky see this interesting NAMS paper which includes quotes from many eminent mathematicians (Bass, Eisenbud, Kadison, Lam, Rotman, Swan, etc).

I liked the algebraic way of looking at things. I'm additionally fascinated when the algebraic method is applied to infinite objects. $\ $--Irving Kaplansky

Note $ $ Readers familiar with the Jacobson radical may note that it may be employed to describe the relationship between the units in $\rm R$ and $\rm R/J\:$ used in the above proof. Namely

Theorem $\ $ TFAE in ring $\rm\:R\:$ with units $\rm\:U,\:$ ideal $\rm\:J,\:$ and Jacobson radical $\rm\:Jac(R).$

$\rm(1)\quad J \subseteq Jac(R),\quad $ i.e. $\rm\:J\:$ lies in every max ideal $\rm\:M\:$ of $\rm\:R$

$\rm(2)\quad 1+J \subseteq U,\quad\ \ $ i.e. $\rm\: 1 + j\:$ is a unit for every $\rm\: j \in J$

$\rm(3)\quad I\neq 1\ \Rightarrow\ I+J \neq 1,\qquad\ $ i.e. proper ideals survive in $\rm\:R/J$

$\rm(4)\quad M\:$ max $\rm\:\Rightarrow M+J \ne 1,\quad $ i.e. max ideals survive in $\rm\:R/J$

Proof $\: $ (sketch) $\ $ With $\rm\:i \in I,\ j \in J,\:$ and max ideal $\rm\:M,$

$\rm(1\Rightarrow 2)\quad j \in all\ M\ \Rightarrow\ 1+j \in no\ M\ \Rightarrow\ 1+j\:$ unit.

$\rm(2\Rightarrow 3)\quad i+j = 1\ \Rightarrow\ 1-j = i\:$ unit $\rm\:\Rightarrow I = 1$

$\rm(3\Rightarrow 4)\ \ \ $ Let $\rm\:I = M\:$ max.

$\rm(4\Rightarrow 1)\quad M+J \ne 1 \Rightarrow\ J \subseteq M\:$ by $\rm\:M\:$ max.

Euclid's famous proof is as follows: Suppose there is a finite number of primes. Let $x$ be the product of all of these primes. Then look at $x+1$. It is clear that $x$ is coprime to $x+1$. Therefore, no nontrivial factor of $x$ is a factor of $x+1$, but every prime is a factor of $x$. By the fundamental theorem of arithmetic, $x+1$ admits a prime factorization, and by the above remark, none of these prime factors can be a factor of $x$, but $x$ is the product of all primes. This is a contradiction.

No there is not, here is a collection of proofs;

http://math.mit.edu/~ssam/writings/primes.pdf