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?

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    The element $x+y$ is prime, because $K[x,y]/(x+y)\cong K[x]$, which is a domain, if you assume that $K$ is a domain. From this follows that $(x+y)$ is a prime ideal, so $x+y$ is a prime element. – Thijs Jul 26 '16 at 13:30
  • See [this prior question.](http://math.stackexchange.com/q/706352/242) – Bill Dubuque Jul 26 '16 at 13:53
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    @user26857 that dupe is not a dupe. I also already thought about reopening the post linked by Bill Dubuque. We can then dupe-close those two. – quid Jul 26 '16 at 20:30
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    @quid Maybe you can find a connection between the present question and the following: http://math.stackexchange.com/questions/483620/irreducible-but-not-prime, http://math.stackexchange.com/questions/744014/possible-irreducible-but-not-prime, http://math.stackexchange.com/questions/1100272/x2-irreducible-but-not-prime, http://math.stackexchange.com/questions/77012/irreducible-but-not-prime-in-mathbbz-sqrt-5, http://math.stackexchange.com/questions/706352/irreducible-and-not-prime, http://math.stackexchange.com/questions/1743216/r-mathbbz-sqrt-41-show-that-3-is-irreducible-but-not-prime-in-r, – user26857 Jul 26 '16 at 20:43
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    http://math.stackexchange.com/questions/1028486/irreducible-but-not-prime-in-kx-y-x2-y3, http://math.stackexchange.com/questions/1538730/showing-that-x2-and-x3-are-irreducible-but-not-prime-in-kx2-x3, http://math.stackexchange.com/questions/106681/find-irreducible-but-not-prime-element-in-mathbbz-sqrt5 – user26857 Jul 26 '16 at 20:43
  • @user26857 yes http://math.stackexchange.com/questions/706352/irreducible-and-not-prime is a arguably a dupe (although there is the specific verification part). As said, unfortunately it is incorrectly closed as a duplicated. The difference is whether it is about one specific example or any example. This question here is somewhat half-way. I really do not care about my answer. I almost decided to delete it on Bernard's comment, but then this could be awkward. The reason I posted it was in order to give another example, which is polynomial and naturally occurring. – quid Jul 26 '16 at 20:50
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    If anything one could decide to eventually close the question asking about one specific example into one that collects examples in general. However, the sole question of that type I am aware of is this and the one mentioned, which is however closed (by you, among others). @user26857 – quid Jul 26 '16 at 20:55

2 Answers2


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

Bill Dubuque
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    $\mathbb{Z}+x\mathbb{Q}[x]$ 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 05 '17 at 07:45
  • @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]](http://www.lohar.com/researchpdf/axbx.pdf) – Bill Dubuque Apr 05 '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 '20 at 19:17

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 $x$.

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  • 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
  • The second example is probably clearer using $\Bbb C[T^2,T^3]$ throughout and never mentioning $R$. – Greg Martin Jul 26 '16 at 20:06
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    @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
  • I think all you need for $n+m>2$, $\mathbb{C}[T^{n},T^{m}]$ to produce such an element just make sure $n$ does not divide $m$. – Mars Apr 05 '17 at 07:47
  • In what the fact that $2\mid (1+i\sqrt 5)(1-i\sqrt 5)$ gives that $2$ is not prime ? For example, in $\mathbb Z$, $2\mid 3\cdot 4$ and $2$ is prime... Also, I have a question : if $a\mid b$, then it always hold that $N(a)\mid N(b)$, but is the converse true ? Because you justify that $2$ doesn't divide $1+i\sqrt 5$ because $N(2)$ doesn't divide $N(1+i\sqrt 5)$ (sorry if my question are obvious, I'm quite novice in algebra). – NewMath Jan 04 '19 at 13:08
  • If $2$ generated a prime ideal, one of the factors would belong to this ideal, i.e. one of the factors woumld be divisible by $2$. For your second objection, I don't use the converse: I prove the assertion by contrapositive, which is different. – Bernard Jan 04 '19 at 13:41
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    Thanks, @Sebastiano! – Bernard Jul 11 '20 at 12:25
  • @Bernard When others are happy, I'm happy too. – Sebastiano Jul 11 '20 at 15:27