TITLE: Show that $\omega^2+1$ is a prime number.
QUESTION [5 upvotes]: Show that $\omega^2+1$ is a prime number.
Is there easy way to show it? I tried, as a warm up, to show that $(\omega+1)\omega\neq \omega^2+1$ and I failed. I am pretty sure I miss something trivial here.
$\gamma$ is a prime number iff for any $\alpha,\beta<\gamma$ we have $\alpha\beta\neq\gamma$
$\omega$ is first infinite ordinal number.

REPLY [8 votes]: To see that ordinal $\omega^2+1$ is prime, suppose that we could factor it as $\omega^2+1=\alpha\beta$, with $\alpha$ and $\beta$ both less than $\omega^2+1$. It is fairly easy to see that $\alpha$ and $\beta$ must also be less than $\omega^2$, and so bounded by $\omega\cdot n$ for some $n$.
If $\beta$ is finite, then since $\alpha<\omega\cdot n$ for some finite $n$, we'd have $\alpha\beta<\omega^2$, so this won't work. 
So $\beta$ must be infinite. If $\alpha$ is finite, then $\alpha\beta<\alpha\omega\cdot n=\omega\cdot n<\omega^2$, since $\alpha\omega=\omega$ for finite $\alpha$. 
So they are both infinite. Note that neither of them can be a limit ordinal, since if one of them was a limit ordinal, then $\alpha\beta$ would also be a limit ordinal, which it isn't. 
So they are both at least $\omega+1$. Consequently, $\alpha\beta$ is at least $(\omega+1)(\omega+1)$, which is equal to $\omega^2+\omega+1$, which is larger than $\omega^2+1$. 
So $\omega^2+1$ is prime.