Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

3 Answers 3

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

$$(a_1,\cdots,a_n)=\{r_1a_1+\cdots+r_na_n:r_1,\cdots,r_n\in R\}.$$

share|improve this answer

Note that an ideal is closed under addition and subtraction, and $3-2=1$.

Since the gcd of two integers is an integer linear combination of the two integers, any non-zero ideal is generated by the gcd of the elements of the ideal. If this gcd is $1$, the ideal is all of $\mathbb{Z}$. If this \gcd is composite, say $ab$ where $a$ and $b$ are greater than $1$, then the ideal is properly contained in the ideal generated by $a$, so is not maximal. If the gcd is the prime $p$, then the ideal is maximal.

share|improve this answer

For one, $\mathbb{Z}$ is a PID, so every ideal can be expressed as one generated by a single element. Conveniently, $\mathbb{Z}$ is also a Euclidean domain. The Euclidean algorithm gives us a convenient way of rewriting ideals generated by two or more elements as ideals generated by a single element.

That is, given two integers $x$ and $y$, then there exists a smallest positive integer $d$ such that $d = mx + ny$. Therefore, just applying the properties of an ideal, it is easy to see that $\langle x, y \rangle = \langle gcd(x, y) \rangle$. [See footnote]. The moral of this story is that we need only consider principal ideals.

Now, suppose that some $m \in \mathbb{Z}$ is not prime, and can therefore be written as $m = xy$ for some other integers $x$ and $y$. Then $\langle m \rangle \subset \langle x \rangle$ and $\langle m \rangle \subset \langle y \rangle$. By definition, $\langle m \rangle$ cannot be maximal.

Now, suppose that some $p \in \mathbb{Z}$ is prime. Then thinking about the properties of ideals, is it possible to have $\langle p \rangle \subset \langle x \rangle$ for some other $x \in \mathbb{Z}$? In particular, is $p \in \langle x \rangle$? Answer these questions correctly, and you can conclude that an ideal in $\mathbb{Z}$ is maximal if and only if it is generated by a single prime number.


Footnote: This is coming from the fact stated in blue's comment, wherein an ideal of a ring $R$ is defined: $\langle x_1, x_2, ..., x_n \rangle = \{r_1x_1 + r_2x_2 + ... + r_nx_n : r_i \in R\}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.