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I am looking for an example, with a direct proof, of a commutative ring with unity , which is not a Principal Ideal ring and every ideal is generated by at most $2$ elements.

Any example or proof I can think of goes by some non direct theory like coming up with an example of a non PID Dedekind domain and showing every ideal in a Dedekind domain is generated by two elements .. etc.

Can we find an example with a direct and relatively elementary proof ?

Thanks in advance

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    $\begingroup$ It's more straightforward to choose one that is rank 2 over $\Bbb{Z}$. That way any ideal is also rank 2 over $\Bbb{Z}$, and thus generated by at most two elements. Then if it's not a PID, you are done. Therefore, choose a non PID quadratic integer ring. For example, $\Bbb{Z}[\sqrt{-5}]$. $\endgroup$ – jgon Nov 8 '19 at 5:00
  • $\begingroup$ @jgon: yeah but how does one show it has rank $2$ ? $\endgroup$ – user102248 Nov 8 '19 at 5:12
  • $\begingroup$ @jgon: I can't think of a direct, not high-powered, way of showing every ideal of $\mathbb Z[\sqrt {-5}] $ is two generated $\endgroup$ – user102248 Nov 8 '19 at 5:13
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    $\begingroup$ To show it has rank 2, observe that $1,\sqrt{-5}$ is a free basis. Is there a problem with saying submodules of finitely generated free modules over a PID are free of rank at most the rank of the original module? What is the target audience here? $\endgroup$ – jgon Nov 8 '19 at 5:15
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For a very easy example, let $k$ be a field, let $V$ be a $2$-dimensional vector space over $k$, and let $R=k\oplus V$ with multiplication defined by $(a,v)\cdot(b,w)=(ab,bv+aw)$. Thinking of $(a,v)$ as a formal sum $a+v$, this is just the multiplication you get by saying $vw=0$ for $v,w\in V$.

Now suppose $I\subseteq R$ is an ideal. Then $I$ is generated by any subset which spans it as a $k$-vector space. Since $R$ is $3$-dimensional as a $k$-vector space, this automatically means $I$ is generated by at most two elements unless $I=R$, but if $I=R$ then $I$ is generated by the single element $1$.

On the other hand, $\{(0,v):v\in V\}$ is an ideal in $R$ that cannot be generated by one element, since any single element will only generate its span in $V$ which is not the whole ideal.

(This ring can also be described as the quotient $k[x,y]/(x^2,xy,y^2)$, where $V$ is spanned by $x$ and $y$.)

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Expanding on my comment

Example: $R=\Bbb{Z}[\sqrt{-5}]$

Proof: $R$ is free of rank 2 as a $\Bbb{Z}$-module, generated by $1$ and $\sqrt{-5}$. Thus any $\Bbb{Z}$-submodule also has rank at most 2. Hence any ideal of $R$ has rank at most 2 as a $\Bbb{Z}$-submodule, and is thus generated by at most 2 elements.

It just remains to prove that there is an ideal generated by exactly two elements. Note that $$2\cdot 3 =6= (1+\sqrt{-5})(1-\sqrt{-5}).$$ This suggests that the full factorization of 6 ought to be something like $$(2,1+\sqrt{-5})(2,1-\sqrt{-5})(3,1+\sqrt{-5})(3,1-\sqrt{-5}).$$

It suffices to prove that $(2,1+\sqrt{-5})$ is not principal. Suppose $(2,1+\sqrt{-5})=(\alpha)$. Then $\alpha \mid 2$ and $\alpha \mid 1+\sqrt{-5}$, so $N(\alpha)\mid N(2)=4$ and $N(\alpha)\mid N(1+\sqrt{-5})=6$. Thus $N(\alpha)\mid 2$. So $\alpha = a + b\sqrt{-5}$, and $N(\alpha) = a^2 + 5b^2 \le 2$. Hence $b=0$ and $a=\pm 1$. Thus $\alpha = \pm 1$. So the only way it is possible for $(2,1+\sqrt{-5})$ to be principal is if it is the unit ideal. However, it is not, since we have $$\Bbb{Z}[\sqrt{-5}]/(2,1+\sqrt{-5}) \cong \Bbb{Z}/(2) \ne 0.$$ The isomorphism is defined by sending $\sqrt{-5}$ to $1$.

Note

I will edit if you update with a clearer description of the audience, but I think this should be an appropriate explanation for most undergrads with a first course in algebra that covers rings, ideals and the fundamental theorem of finitely generated abelian groups. (Certainly not all first courses in algebra cover these topics, so that's not a given). Let me know how elementary you require, and I'll see what I can do. I have a hard time imagining making it more elementary from here though.

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  • $\begingroup$ Yeah, I was hoping not to use that submodules of free module over a PID of finite rank has rank at most that of the module ... but I guess I really can't avoid that .. $\endgroup$ – user102248 Nov 8 '19 at 5:25
  • $\begingroup$ @user102248 You can make it slightly more elementary if you say free abelian group, since these are $\Bbb{Z}$-modules, but yeah I really am not sure how to avoid something like that. Not without essentially reproving it in a special case. $\endgroup$ – jgon Nov 8 '19 at 5:27
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    $\begingroup$ A proof that submodules of free $\mathbb{Z}$-modules are free and of rank at most the ambient module is found in this answer: math.stackexchange.com/questions/548552/… $\endgroup$ – Ben Blum-Smith Nov 8 '19 at 12:14

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