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Are there infinitely many irreducibles in the ring of integers of any algebraic number field ?

I tried to follow the same argument as we usually do for integers. Suppose there are finitely many irreducibles, say $p_1,\ldots ,p_n$ and let $\alpha :=1+ p_1\cdots p_n$. Now if $\alpha $ is not a unit then it must have an irreducible $p$ such that $p|\alpha$ but then $p$ can not be any of the $p_i$'s and we have a contradiction. $\textit{What if $\alpha$ is an unit ? Is it possible for $\alpha$ to be an unit ?}$

Of course, one can replace $p_1\cdots p_n$ by $p_1^{k_1}\cdots p_n^{k_n}$ for any $k_1,\ldots ,k_n\in\mathbb{N}$ and the same argument would go through.

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Can the same irreducible element divide two different rational primes? Offhand, I think the answer is no, and then the problem is solved. But my gut feeling may be wrong. – Daniel Fischer Jun 27 '13 at 16:03
One note that might be helpful - if there are finitely many units, then the argument goes through by your last sentence. You should probably also assume there is at least one irreducible as well. I'm thinking in terms of general rings, so maybe some of this is covered by the assumption that it's the ring of integers of an algebraic number field (I'm not a number theorist, so I don't really have a good feel for those). – Matthew Pressland Jun 27 '13 at 16:16
@DanielFischer If $a,b$ are rational integers which are relatively prime in $\mathbb Z$, they can't have a non-unit factor in common since $ax+by=1$ has a solution... – Thomas Andrews Jun 27 '13 at 16:16
To avoid $\alpha$ being a unit, use $\beta=|N(p_1\dots p_n)| + 1$, where $N$ is the norm of the field. – Thomas Andrews Jun 27 '13 at 16:19
@DanielFischer: I am sorry, you are correct. Could you write your previous comment as an answer? – pritam Jun 29 '13 at 12:04
up vote 4 down vote accepted

$\alpha$ can be a unit as the element $1+\sqrt{2}$ in the ring of integers $\mathbb{Z}[\sqrt{2}]$ shows. Here's a slight variation of your argument: Let $p_1,\ldots,p_n$ be all the irreducibles. Since $x=p_1\cdot\ldots\cdot p_n$ is integral over $\mathbb{Z}$ it satisfies: $$ 0\neq-a_0=x^n+\ldots+a_1x=x(x^{n-1}+a_{n-1}x^{n-2}+\ldots+a_1)=x\cdot y\in\mathbb{Z}. $$ Taking the negative of $y$ if necessary wlog we assume $xy>0$. Then $1+xy$ is not a unit of $\mathbb{Z}$ (and so also not a unit of the ring of integers $R$). Now you can conclude with your argument.

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Consider the mapping

$$\iota \colon \mathbb{P} \to \mathfrak{P}(R);\quad \iota(p) = \lbrace a \in R \colon a \mid p \land a \text{ is irreducible}\rbrace,$$

where $R$ is the ring under consideration, and $\mathbb{P}$ is the set of (positive) rational primes.

Since the norm of the field (w.r.t. $\mathbb{Q}$) is integer-valued on $R$, and multiplicative, each division chain from a nonzero element must stop (at an irreducible element), and hence $\iota(p) \neq \varnothing$ for all $p$.

Let $z \in \iota(p) \cap \iota(q)$. Then $z \mid \gcd_{\mathbb{Z}}(p,\,q)$. If $p \neq q$, then $\gcd_{\mathbb{Z}}(p,\,q) = 1$, hence $\iota(p) \cap \iota(q)$ contains only units - but $\iota(p)$ contains no units by definition. Thus $p\neq q \Rightarrow \iota(p) \cap \iota(q) = \varnothing$. $\iota(p)$ is closed under $x \sim y \iff \bigl(\exists \varepsilon \in R^\ast\bigr)(y = \varepsilon\cdot x)$.

Hence for any choice function $c$,

$$c \circ \iota \colon \mathbb{P} \to R$$

is injective, and $\lbrace c(\iota(p)) \colon p \in \mathbb{P} \rbrace$ is an infinite family of pairwise non-associated irreducible elements of $R$.

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Following Thomas Andrews hint, consider $$\beta:=|N(p_1\cdots p_n)|+1=|N(p_1)|\cdots |N(p_n)|+1$$ Then $\beta\in\mathbb{Z}\subseteq \mathfrak{O}_K$ and $\beta\geq 2$, where $K$ is the number field of degree $d$. Now $\beta $ is a non-unit since $N(\beta )=\beta^d\ne 1$; then $\beta$ has an irreducible factor say $p$ and $\beta =p\gamma$. Then we have $$p\gamma =|N(p_1)|\cdots |N(p_n)|+1$$ Now if $p=p_i$ for some $i$, then as every ideal divides it's norm, we have$$\langle p\rangle|\langle N(\langle p_i\rangle)\rangle\Rightarrow\langle p\rangle \supseteq\langle |N(p_i)|\rangle\Rightarrow |N(p_i)|=p\lambda$$ where $\lambda\in\mathfrak{O}_K$. Hence we have $p||N(p_i)|$ and we have a contradiction.

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You have to know that $N(p_i)$ is divisible by $p_i$, but otherwise this looks fine. Michalis's answer is very similar, in that his $|a_0|$ can be made to be equal to $|N(p_1\dots p_n)|$... – Thomas Andrews Jun 27 '13 at 17:08
@Thomas Andrews: I have shown at the end that $N(p_i)=p\lambda =p_i\lambda$ – pritam Jun 27 '13 at 17:13
I don't see you proving that $p_i|N(p_i)$, only asserting it. "Now if $p=p_i$, then $\left<p\right>|\left<N(p_i)\right>...$. – Thomas Andrews Jun 27 '13 at 17:17
@Thomas Andrews: I have used a theorem which says every ideal divides it's norm. – pritam Jun 27 '13 at 17:22
@pritam: For "$\beta$ has an irreducible factor", you need the Noetherianness of this ring (which Dedekind domains in fact have), am I right? – Tomas Jun 28 '13 at 16:40

If the extension over $\mathbb Q$ is galois, then we can show that there are in fact infinitely many irreducible integers in $\mathbb Z$. Of course this does not answer the question, hence is in community wiki.

If $L/\mathbb Q$ is galois with galois group $G$, then, from the Chebotarev density theorem, we know that the set of prime integers which are unramified and with Frobenius=identity has density $\not=0$, so there are infinitely many such. Suppose $p\in \mathbb Z$ and $\text{Frob}_L(p)=\text{identity}_{L}$, then $(p)$ remains a prime ideal in $L$, that is to say, $p$ is an irreducible element in the Dedekind domain $L$. So there are infinitely many irreducible elements.

Unfortunately, when the extension if not galois, the above argument can only produce infinitely many prime ideals, not necessarily principal, since we have no density theorem this time. Further, when the prime ideal is not principal, then it cannot contain any irreducible, otherwise the prime ideal would be generated by that irreducible element. Hence this argument can offer nothing concerning the irreducibles.
Per chance there are results that can solve this problem?
P.S. That $(\text{Frob}_L(p)=\text{identity}_L)\implies ((p) \text{ is prime in }L)$ is a consequence of the fact that the order of the decomposition group of $(p)$ is equal to the order of $\text{Frob}_L(p)$.

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