Show that $\omega^2+1$ is a prime number. 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.
 A: 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.
A: Here is a proof of the lemma you don’t know how to prove:

Lemma:$$(\omega+1)\omega=\omega^2$$

Taking my definition of multiplication from wikipedia, we consider a grid of points that is laid out as $\omega+1$ from left to right and as $\omega$ from top to bottom. We assign this grid a lexicographical order such that if two points are in different rows, the lower point is larger and if two points are in the same row, the right-most point is larger.
Since each row is arranged as $\omega+1$, there is a unique last element in every row. Call this element $k_\alpha$ when it appears in row $\alpha$.
Now rearrange the grid by taking $k_\alpha$ and moving it from being the last element of row $\alpha$ and instead make it the first element of row $s(\alpha)$. We claim that this doesn’t change the order type of the grid because $k_\alpha$ remains greater than every other element in any row below row $s(\alpha)$ and less than any other element in any row above row $\alpha$.
However, this new arrangement is precisely an $\omega\times\omega$ grid, so it has order type $\omega^2$. Thanks JDH for helping me see this.
A: Well, let's just look at the possible products of the form $(\omega i+j)(\omega k+l)$ with finite $i,j,k,l$. It is obvious that we only can get something involving $\omega^2$ is $i\ne 0$ and $k\ne 0$, therefore in the following, both are assumed.
First we have distributivity of the left factor over the right one, therefore
$$(\omega i+j)(\omega k+l) = (\omega i + j)\omega k + (\omega i + j)l$$
Next, we have
$$(\omega i + j)l = \begin{cases}
\omega i + j + \omega i + j + \ldots + \omega i + j = \omega il + j & l\ne 0\\
0 & l = 0
\end{cases}$$
or using the Kronecker delta
$$(\omega i + j)l = \omega il + j(1-\delta_{l0})$$
Further we have
$$(\omega i + j)\omega = \sup_{n<\omega} (\omega i + j) n = \sup_{n<\omega} (\omega i n + j) = \omega^2$$
and therefore
$$(\omega i + j)\omega k = \omega^2 k$$
Putting it all together, we therefore get
$$(\omega i+j)(\omega k+l) = \omega^2k + \omega i l + j = \omega^2 k + \omega i l + j$$
Now we want
$$(\omega i+j)(\omega k+l) = \omega^2 k + \omega i l + j(1-\delta_{l0}) \stackrel!= \omega^2+1$$
Then we obviously need
$$\begin{aligned}
k &= 1\\
i l &= 0\\
j(1-\delta_{l0}) &= 1
\end{aligned}$$
Since $i\ne 0$, the second equation reduces to $l=0$. However this means the third equation reduces to $0=1$ which is of course a contradiction.
Therefore $\omega^2+1$ cannot be written as product.
