# Irreducible but not prime element

I am looking for a ring element which is irreducible but not prime.

So necessarily the ring can't be a PID. My idea was to consider $R=K[x,y]$ and $x+y\in R$.

This is irreducible because in any product $x+y=fg$ only one factor, say f, can have a $x$ in it (otherwise we get $x^2$ in the product). And actually then there can be no $y$ in $g$ either because $x+y$ has no mixed terms. Thus $g$ is just an element from $K$, i.e. a unit.

I got stuck at proving that $x+y$ is not prime. First off, is this even true? If so, how can I see it?

This is impossible: any polynomial ring over a field is a U.F.D. In such domains, irreducible elements are prime.

The simplest example is the ring of quadratic integers $\;\mathbf Z[i\sqrt 5]$, which is not a U.F.D.. In this ring, we have $$2 \cdot 3=(1+i\sqrt 5)(1-i\sqrt 5),$$ so that $2$ divides the product $\;(1+i\sqrt 5)(1-i\sqrt 5)$, but doesn't divide any of the factors, since it would imply the norm $N(2)=4$ divides $N(1\pm i\sqrt 5)=6$. $2\;$ is irreducible for similar reasons: if $a+ib\sqrt 5$ is a strict divisor of $2$ and a non-unit, its norm $a^2+5b^2$ is a non-trivial divisor of $4$, i.e. $\;a^2+5b^2=2$. Unfortunately, this diophantine equation has no solution.

Thus, $2$ is a non-prime irreducible element. The same is true for all elements in these factorisations of $6$.

Another example, with polynomial rings:

Consider the ring of polynomial functions on the cusp cubic $$R=\mathbf C[X,Y]/(X^2-Y^3).$$ This is an integral domain, as the curve is irreducible. Actually, we have a homomorphism: \begin{align*} \mathbf C[X,Y]&\longrightarrow\mathbf C[T^2,T^3]\\ X&\longmapsto T^3,\\ Y&\longmapsto T^2. \end{align*} This homomorphism is surjective, and its kernel is the ideal $(X^2-Y^3)$, so that it induces an isomorphism $R\simeq \mathbf C[T^2,T^3]$.

If we denote $x$ and $y$ the congruence classes of $X$ and $Y$ respectively, we have $x^2=y^3$. The element $y$ is irreducible, for degree reasons, but it is not prime, since it divides $x^2$ but doesn't divide $x$.

• There are simpler examples, e.g. see my answer. – Bill Dubuque Jul 26 '16 at 14:06
• @Bill Dubuque: Simpler to prove, but exotic! ;o) – Bernard Jul 26 '16 at 14:32
• Perhaps. But (real) polynomial rings are more familiar than rings of algebraic integers for many students. Rings of that form are often a good sources of counterexamples (so often that ring theorists study them at length). – Bill Dubuque Jul 26 '16 at 14:37
• @user26857: I'm sorry but I don't know the contents of this site by heart. It even happens that I know I've more or less already answered a similar question, but I'm unable to find where. In such a case, it happens I give the answer again. – Bernard Jul 26 '16 at 21:04
• Thanks, @Sebastiano! – Bernard Jul 11 at 12:25

Let $\rm\ R = \mathbb Q + x\:\mathbb R[x],\$ i.e. the ring of real polynomials having rational constant coefficient. Then $\,x\,$ is irreducible but not prime, since $\,x\mid (\sqrt 2 x)^2\,$ but $\,x\nmid \sqrt 2 x,\,$ by $\sqrt 2\not\in \Bbb Q$

• $\mathbb{Z}+x\mathbb{Q}[x]$ will work as well. or $\mathbb{R}$ and $\mathbb{C}$. To abstract this more just let $R_{0}\subset R_{1}$ be any two rings in which $R_{1}$ contains a multiplicative inverse to an element in $R_{0}$. – Mars Apr 5 '17 at 7:45
• @Morph Such rings are a rich source of (counter)examples, e.g. see M. Zafrullah, [Various facets of rings between D[X] and K[X]](lohar.com/researchpdf/axbx.pdf) – Bill Dubuque Apr 5 '17 at 21:46
• @Morph The element $x$ in $\mathbb{Z} + x \mathbb{Q}[x]$ is reducible: $x = 2(x / 2)$. The irreducible elements are $\pm$ primes and polynomials irreducible in $\mathbb{Q}[x]$ with $\pm 1$ as a constant term. The ring $\mathbb{Q} + x\mathbb{R}[x]$ works because $\mathbb{Q}$ is a subfield of $\mathbb{R}$. – Robert D-B Jun 30 at 19:17