# Non-existence of an irreducible fraction representation

If $R$ is a unique factorization domain and $F$ its field of fractions then any $z\in F$ can be represented as an irreducible fraction (i.e. $z=a/b$ with $a,b\in R$ so that $a$ and $b$ have no common non-unit factors in $R$).

What is an example of a domain $R$ where this fails, i.e., a domain $R$ and an element $z$ in the field of fractions of $R$ so that $z$ has no representation as an irreducible fraction (meaning $z=a/b$ with $a,b\in R$ always implies that $a$ and $b$ have a common non-unit factor in $R$)?

• Which non-UFD domains have you tried? Nov 13, 2015 at 13:14
• e.g $\mathbb{Z}[\sqrt{-n}]$? Nov 13, 2015 at 13:20
• @NeerajKumar Thanks for the tip! I will look into that later today. Nov 13, 2015 at 13:54
• @NeerajKumar: If we have $z=a/b$ with no representation as an irreducible fraction then I think we can construct a strictly ascending infinite sequence of (principal) ideals $(a)=(a_0)\subsetneq(a_1)\subsetneq(a_2)\subsetneq\cdots$ in $R$. Hence the domains $\mathbb{Z}[\sqrt{d}]$ don't work since they are Noetherian (unless I'm mistaken). Nov 16, 2015 at 1:46

Let $R=\mathbb Z+(X,Y)\mathbb Q[X,Y]$, and $z=\dfrac XY$.
It's obvious that every non-zero integer $n$ divides $X$ and $Y$: we have $X=n\dfrac Xn$ and $Y=n\dfrac Yn$.
Now write $z=\dfrac ab$ with $a,b\in R$, $b\ne0$. It's easily seen that $a(0,0)=0$, and $b(0,0)=0$. This shows that $a,b\in(X,Y)\mathbb Q[X,Y]$, and once again every non-zero integer $n$ divides $a$ and $b$. (In fact, $a=Xf$ and $b=Yf$ for some $f\in\mathbb Q[X,Y]$.)