How to prove that if a product is in a prime ideal, then one of the factors must be?

Question

Let $R$ be a commutative ring, and suppose $a_1, \dots , a_n \in R$.

Using induction, prove that if $I \subset R$ is a prime ideal and $a_1\cdot a_2 \cdots a_n \in I$, then there exists some $1 \leq i \leq n$ such that $a_i \in I$.

I started with the base case that when $n=1$. Then $1 \leq i \leq n=1$, then $i=1$ i.e $a_1\in I$ , how do we know that is true for $a_1$. And please what is the following step for $n=k+1$?

Answer 1

Hint: suppose $a_1a_2\cdots a_{k+1}\in I$. Write this as $(a_1a_2)a_3\cdots a_{k+1}\in I$. This is $k$ distinct elements, so what can you say now? Note you'll still have to "prove" the $n=2$ case separately, but this is really just the definition of a prime ideal.

Answer 2

Just to answer the first part of your question specifically. Remember for n=1 your hypothesis is exactly $a_1 \in I$. So actually there is nothing to prove.