# Factorization into irreducibles [duplicate]

Let $R$ be a ring with $1$. Let $r$ be an element of $R$, $r\neq 0$, $r$ not a unit. Can I say that $r$ is a product of a finite number of irreducibles or not?

• Maybe you want $R$ to be an integral domain. As I mentioned here, in my answer, there are integral domains without irreducible elements.
– user26857
Apr 28 '13 at 10:01

Let $R$ be the ring of algebraic integers (over $\mathbb{Q}$). There we see the following problem: $$2=(\sqrt2)^2=(\sqrt{\sqrt2})^4=(\root8\of2)^8=(\root{16}\of 2)^{16}=\cdots$$ Obviously you can continute as long as you wish without ever leaving the ring $R$, so you never get to see irreducible elements here.

In a comment Federica asked for a ring that also has irreducible elements but also elements that cannot be written as products a finite number of irreducibles. I think that the following works, but I haven't double-checked everything.

Let $x_k=e^{2\pi i/2^k}$ be a primitive root of unity of order $2^k$. Let $$R=\mathbb{Z}[x_2,x_3,x_4,\ldots].$$ Then the number two still has a never-ending factorization $$2=1-x_2^2=(1+x_2)(1-x_2)=(1+x_2)(1+x_3)(1-x_3)=(1+x_2)(1+x_3)(1+x_4)(1-x_4)=\cdots$$ because we always have $1-x_k=1-x_{k+1}^2=(1+x_{k+1})(1-x_{k+1})$. From the theory of cycltomic field we know that all the factors $1+x_k$ generate the lone prime ideal of $\mathbb{Z}[x_k]$ lying above the prime $2$.

Yet, I think that the two factors of $5=(2+i)(2-i)=(2+x_2)(2-x_2)$ are irreducible. An eventual factor (being a polynomial of a finite number of $x_k$:s) belongs to some ring $\mathbb{Z}[x_k]$. But $5$ generates a subgroup of index two in $\mathbb{Z}_{2^k}^*$, and consequently splits into a product of two prime ideals, namely the ones lying above $(2+i)$ and $(2-i)$, in all the rings $\mathbb{Z}[x_k], k\ge2$. Therefore $2+i$ and $2-i$ are irreducible as elements as well.

• YACP gave a link to a more comprehensive description of this phenomenon! Apr 28 '13 at 10:07
• and if we assume that there exist irreducibles? Is there an example where $R$ have irreducible elements and $r\in R$ cannot be expressed as a product of finitely many irreducibles? Apr 28 '13 at 12:21
• @Federica, is this what you wanted to see? Apr 28 '13 at 13:03

there are rings which have elements which cannot be factors as product of a finite numer of irreducibles.

Let $A=\mathbb Q[x_1,y_1,x_2,y_2,x_3,y_3,\dots]$ be the polynomial ring on two infinite sequences of variables, let $I$ be the ideal of $A$ generated by the elements $x_i-x_{i+1}y_i$, and let $R=A/I$.

In the ring $R$, we have $x_1=x_2y_1=x_3y_2y_1=\cdots=x_{n+1}y_n\cdots y_1$ for all $n\geq1$.

• so $x_1$ can be factorized in infinitely many ways as a product of finitely many factors. How this implies that $R=A/I$ has an element $r$ which cannot be factorized as a product of finitely many irreducibles? Can you do an example of such an $r$ in $A/I$? Apr 28 '13 at 9:36