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Let's restrict ourselves to commutative rings (not necessarily with unity).

Is there a simpler example of a non-principal ideal than $\langle a,x\rangle$ in $R[x]$, where $a\in R$ is not a unit (and therefore $R$ is not a field)? All other examples that come to mind involve more complicated polynomial rings and seem to be particular cases of the previous example.

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    $\begingroup$ How much more simple can you get?? $\endgroup$ – Matt B Sep 23 '15 at 15:56
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    $\begingroup$ That's what I'm asking. $\endgroup$ – Fibonacci Sep 23 '15 at 16:03
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    $\begingroup$ Let $k$ be a field. In the ring of $k$-polynomials with no linear term, the ideal of elements with no constant term is nonprincipal. $\endgroup$ – Lubin Sep 23 '15 at 16:05
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The classic example for a ring that is not a polynomial ring is the ideal $(2,1+\sqrt{-5})$ in $\mathbb{Z}[\sqrt{-5}]$. I don't think there's going to be anything simpler if you rule out examples from polynomial rings.

But personally I find the conceptually simplest example is the ideal $(x,y)$ in $R[x,y]$, where $R$ can be any non-zero ring at all. It requires the least amount of thought to see that it is non-principal. However it fits into the pattern you've ruled out, since $R[x,y]=R[x][y]$.

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    $\begingroup$ dumb question: does the notation $(a,b)$ for defining an ideal represent the set $S = {ma+nb}$ for integers $m,n$? (in other words, the sum of all multiples of $a$ and $b$) $\endgroup$ – Jason S Sep 30 '17 at 4:59
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    $\begingroup$ @JasonS: Not a dumb question at all. In general, if $R$ is a commutative ring, and $a,b\in R$, then the ideal $(a,b)$ of $R$ is the set $$\{ra+sb:r,s\in R\}$$ (see this section on Wikipedia). However, note that for $\mathbb{Z}[\sqrt{-5}]$, that means $(2,1+\sqrt{-5})$ is the set $$\{(r_1+r_2\sqrt{-5})\cdot 2 + (s_1+s_2\sqrt{-5})\cdot(1+\sqrt{-5}):r_1+r_2\sqrt{-5}, s_1+s_2\sqrt{-5}\in\mathbb{Z}[\sqrt{-5}]\}$$ which is bigger than the set $$\{m\cdot 2 + n\cdot(1+\sqrt{-5}):m,n\in\mathbb{Z}\}$$ $\endgroup$ – Zev Chonoles Sep 30 '17 at 7:37
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    $\begingroup$ oh, so the notation $(a,b,c,...)$ is essentially a linear basis for the ideal, with both basis values and coefficients in the ring $R$? $\endgroup$ – Jason S Sep 30 '17 at 20:17
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    $\begingroup$ @JasonS: That's a shaky comparison to make, because there will be relations between the "basis" elements of the ideal (whose more proper name would just be "generators"), whereas for a basis in a vector space, the whole point is that there is no relation between them. For example, the ideal $(4,6)$ inside $\mathbb{Z}$ is $$\{4m+6n:m,n\in\mathbb{Z}\}$$ which is exactly the set of even numbers, $$\{2x:x\in\mathbb{Z}\}$$ otherwise known as $(2)$. In general, for integers $a_1,\ldots,a_n$, we have $(a_1,\ldots,a_n)=(\gcd(a_1,\ldots,a_n))$. $\endgroup$ – Zev Chonoles Oct 1 '17 at 6:18
  • $\begingroup$ How do prove that $(2,1+\sqrt{-5})$ is non principal in $\mathbb{Z}[\sqrt{-5}]$ ? $\endgroup$ – dafnahaktana Sep 28 '18 at 18:29
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This is an answer to the comment of dafnahaktana above to the answer

How do prove that $(2,1+\sqrt{-5})$ is non principal in $\mathbb{Z}[\sqrt{-5}]$ ?

Please inform him, I have not enough reputation to add a comment.

The ideal $(2,1+\sqrt{-5})$ in the commutative ring $\mathbb{Z}[\sqrt{-5}]$ is \begin{align*} &(2,1+\sqrt{-5})\\ =&\{2r_1+(1+\sqrt{-5})r_2:r_1,r_2\in \mathbb{Z}[\sqrt{-5}] \}\\ =&\{2(r_{11}+r_{12}\sqrt{-5})+(1+\sqrt{-5})(r_{21}+r_{22}\sqrt{-5}):r_{11},r_{12},r_{21},r_{22}\in \mathbb{Z} \}\\ =&\left\{ \left(2r_{11}+r_{21}-5r_{22}\right) +\left(2r_{12}+r_{21}+r_{22}\right)\sqrt{-5} :r_{11},r_{12},r_{21},r_{22}\in \mathbb{Z} \right\} \end{align*} Which elements of $\mathbb{Z}[\sqrt{-5}]$ are in $(2,1+\sqrt{-5})$? Let us solve the system \begin{gather*} \begin{cases}2r_{11}+r_{21}-5r_{22}=n_1\\ 2r_{12}+r_{21}+r_{22}=n_2\end{cases} \end{gather*} If $n_1$ and $n_2$ have the same oddity, we can choose any $r_{22}$ an choose $r_{21}$ in such a way that $r_{21}-5r_{22}$ and $r_{21}+r_{22}$ have also the same oddity, then the values of $r_{11}$ and $r_{12}$ follow; if $n_1$ and $n_2$ have opposite oddity, then there is no solution, thus \begin{gather*} (2,1+\sqrt{-5})=\{n_1+n_2\sqrt{-5}:n_1,n_2\in\mathbb{Z}\text{ and $n_1,n_2$ with equal oddity}\}. \end{gather*}

Thus the lattice/grid of $(n_1,n_2)$ in $\mathbb{Z}^2$ cannot be generated by a single "vector".

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