Show that $\omega^2+1$ is a prime number.

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.