I cannot understand why maximal ideal in a ring of integers is generated by a single prime number. For example, if we choose 2 and 3 as generators of ideal, then all multiples of 2 and 3 will be in ideal, but those that do not have 2 and 3 as factors will not be included in an ideal, guaranteeing that it is a proper ideal, not trivial whole ring ideal or zero ideal. So what am I getting wrong here?
[Math] why is maximal ideal of a ring of integers generated by a single prime number
idealsring-theory
Best Answer
Maximal ideals in a number ring (the ring of integers of a number field) need not be principal, contrary to what you say. But I'm guessing you don't realize "ring of integers" has a more general meaning in algebraic number theory, and you're just using the phrase to refer to $\Bbb Z$. In $\Bbb Z$, you say that $(2,3)$ will include multiples of $2$ and $3$, but will not include any numbers that don't have $2$ or $3$ as a factor. But $3-2=1$ is in $(2,3)$, hence any multiple of $1$ is in this ideal, hence this ideal is the whole ring since all integers are multiples of $1$. The problem seems to be that you don't know what an ideal is, so I'd recommend going back to the definition and reviewing. Ideals are additive subgroups closed under ambient multiplication. In a ring $R$, the ideal $(a_1,\cdots,a_n)$ is defined to be the smallest ideal containing the elements $a_1,\cdots,a_n$. In particular, it must contain all $R$-linear combinations of them.
That is, in any commutative ring $R$, we have the equality
$$(a_1,\cdots,a_n)=\{r_1a_1+\cdots+r_na_n:r_1,\cdots,r_n\in R\}.$$