22
$\begingroup$

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?

$\endgroup$
25
$\begingroup$

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 $x$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ There are simpler examples, e.g. see my answer. $\endgroup$ – Bill Dubuque Jul 26 '16 at 14:06
  • $\begingroup$ @Bill Dubuque: Simpler to prove, but exotic! ;o) $\endgroup$ – Bernard Jul 26 '16 at 14:32
  • $\begingroup$ 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). $\endgroup$ – Bill Dubuque Jul 26 '16 at 14:37
  • 1
    $\begingroup$ @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. $\endgroup$ – Bernard Jul 26 '16 at 21:04
  • 1
    $\begingroup$ Thanks, @Sebastiano! $\endgroup$ – Bernard Jul 11 at 12:25
25
$\begingroup$

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 $\sqrt 2\not\in \Bbb Q$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ $\mathbb{Z}+x\mathbb{Q}[x]$ 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 $R_{0}$. $\endgroup$ – Mars Apr 5 '17 at 7:45
  • $\begingroup$ @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) $\endgroup$ – Bill Dubuque Apr 5 '17 at 21:46
  • $\begingroup$ @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 $\mathbb{R}$. $\endgroup$ – Robert D-B Jun 30 at 19:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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