Generalized Euclid Lemma

Question

Could anyone check my proof? Thanks in advance.

Statement: If $p$ is a prime and $p$ divides $a_{1}a_{2}...a_{n}$ prove that p divides $a_{i}$ for some $i$.

Proof: Suppose $p$ is a prime that divides $a_{1}a_{2}...a_{n}$ but not $a_{j}$ for all $j\ne i$. Let the set of those $j$ be $J$. Then there exist integers $s$ and $t$ such that $1=a_{k1}a_{k2}...a_{kr}s+pt$ where $k1,k2,k3,...,kr$ corresponds distinctly to each element in $J$ and $r$ is the number of elements in $J$. Now if we multiply the equation above with $a_{i}$ to get $a_{i}=a_{1}a_{2}...a_{n}s+pa_{i}t$. As $p$ divides the right-hand side, $p$ also divides $a_{i}$. Here we assume that there is only one such number as if there are two or more, these can all be multiplied and reduced to just one number.

Answer 1

In your proof, you mention $i$, but you don't explain where it comes from. It's not quite a proof by contradiction either, and is a bit confusing.

The generalized Euclid lemma follows from the original Euclid lemma by induction, where the base case is $n = 2$ (Euclid's lemma). Informally, the proof looks like this:

Since $p$ is prime and $p \mid (a_1)(a_2 \ldots a_n)$, it follows by Euclid's lemma that $p \mid a_1$ or $p \mid a_2 \ldots a_n$. Likewise, if $p \mid (a_2)(a_3 \ldots a_n)$, it follows by Euclid's lemma that $p \mid a_2$ or $p \mid a_3 \ldots a_n$. Continuing in this manner, we find that $p \mid a_1$ or $p \mid a_2$ or ... or $p \mid a_n$, as desired.

Can you see how to do this induction proof rigorously?