# What is the simplest non-principal ideal?

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.

• How much more simple can you get?? – Matt B Sep 23 '15 at 15:56
• That's what I'm asking. – Fibonacci Sep 23 '15 at 16:03
• 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. – Lubin Sep 23 '15 at 16:05

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 $R[x,y]=R[x][y]$.

• 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 $b$) – Jason S Sep 30 '17 at 4:59
• @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 $$\{m\cdot 2 + n\cdot(1+\sqrt{-5}):m,n\in\mathbb{Z}\}$$ – Zev Chonoles Sep 30 '17 at 7:37
• oh, so the notation $(a,b,c,...)$ is essentially a linear basis for the ideal, with both basis values and coefficients in the ring $R$? – Jason S Sep 30 '17 at 20:17
• @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 $(a_1,\ldots,a_n)=(\gcd(a_1,\ldots,a_n))$. – Zev Chonoles Oct 1 '17 at 6:18
• How do prove that $(2,1+\sqrt{-5})$ is non principal in $\mathbb{Z}[\sqrt{-5}]$ ? – dafnahaktana Sep 28 '18 at 18:29

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".