Let $R$ be a Noetherian domain. Suppose $I = (a_1,...,a_n)$ is a maximal ideal. Then does it mean that each $a_i$ is irreducible?
This is of course true if $I$ is principal but does this result generalize to multiple generators under the assumption that $n$ is chosen to be minimal?
No. For instance, let $R=k[x,y]$ for some field $k$ and $I=(x,y+xy)$. This ideal is maximal because it is equal to $(x,y)$ and cannot be generated by less than $2$ elements, but $y+xy=y(1+x)$ is not irreducible.
What is true is that you can always choose a minimal-sized set of generators that are all irreducible. This is true more generally for any prime ideal. Indeed, suppose you have a minimal-sized set of generators $(a_1,\dots,a_n)$ for a prime ideal $I$ in a Noetherian ring. Then each $a_i$ can be written as a product of irreducible elements, and since $I$ is prime, for each $i$ some irreducible factor $b_i$ of $a_i$ must be in $I$. We then have $(b_1,\dots,b_n)=I$ since $a_i$ is a multiple of $b_i$ for each $i$.