### Divisors of ordinals

If $\alpha,\beta,\gamma$ are ordinals with $\alpha=\beta\gamma$, then $\beta$ is called a **left-hand divisor** of $\alpha$, and $\gamma$ is called a **right-hand divisor** of $\alpha$. If $\alpha$ is such that there exists no decomposition of the form $\alpha=\beta\gamma$ where $\beta,\gamma<\alpha$, then we say that $\alpha$ is a **prime factor**.

**Proposition 1.** Every ordinal number $>0$ has finitely many distinct right-hand divisors.

*Proof.* Suppose $\alpha$ has infinitely many right-hand divisors, so that there exists some infinite ascending chain of divisors $\delta_1<\delta_2<\cdots$ such that By Proposition 2 from here , we have that the $\gamma_i$ comprise a strictly descending chain, so that $\gamma_1>\gamma_2>\cdots$. But this would imply that there exists an infinite descending chain of ordinals, which is impossible. Hence, $\alpha$ can only have finitely many distinct right-hand divisors. $\blacksquare$

Now we may begin to classify the ordinals which are prime factors.

**Proposition 2.** If $\rho$ is a prime component , then $\rho+1$ is a prime factor.

*Proof.* Suppose that $\rho$ is a prime component and $\rho+1$ is *not* a prime factor, so that $\rho+1=\alpha\beta$, where $\alpha,\beta<\rho+1$. Since $\rho+1$ is a successor ordinal, we have that $\alpha$ and $\beta$ must both be successor ordinals, so that $\alpha=\alpha'+1$ and $\beta=\beta'+1$ for some $\alpha',\beta'$. Then we have that which implies that However, we have that $\alpha'<\rho$ because $\alpha'+1<\rho+1$, implying that $(\alpha'+1)\beta'<\rho$ as well, contradicting the assumption that $\rho$ was a prime component (since prime components cannot be written as the sum of two lesser ordinals). Hence, we have a contradiction, and $\rho+1$ must be a prime factor. $\blacksquare$

# set-theory

# order-theory

# ordinals

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