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(In the setting of number fields and algebraic integers) If $(a),(b)$ are two principle ideals then $(a)+(b)$ corresponds to $(\gcd(a,b))$, so while the natural definition of addition for ideals has a pleasing meaning, it does not correspond to addition of numbers.

My question is whether there is a "natural" operation $\oplus$ on ideals (one that can be generalized to non-principal ideals) such that $(a)\oplus(b)=(a+b)$? From what I've read I've come under the impression that there is not such operation but I'd like to make sure I'm not missing anything - and even better, I'd like to hear an explanation as to why such an operator is not likely to exist.

Gadi A
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2 Answers2

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It's not well-defined, even for $\mathbb{Z}$: because $(a)=(-a)$ for any $a\in\mathbb{Z}$, we have $$(a)\oplus(a)=(a+a)=(2a)$$ and $$(a)\oplus(a)=(a)\oplus(-a)=(a+(-a))=(0)$$ which is impossible if $a\neq0$.

The same problem occurs for any ring not of characteristic 2.

Here's an analogy to consider, maybe it will clarify why we wouldn't expect this to work:

Suppose we want to try to make an operation $\star$ on subspaces of a vector space $V$ such that for any $v,w\in V$, we have $$\text{span}(v)\star\text{span}(w)=\text{span}(v+w).$$ There's a problem because $\text{span}(w)=\text{span}(tw)$ for any $t\neq 0$, while $\text{span}(v+w)\neq\text{span}(v+tw)$ in general (specifically, when $v$ and $w$ are linearly independent).

Note that the ideals of a ring $R$ are precisely the $R$-submodules of $R$ (hence the analogy with subspaces of a vector space).

Community
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Zev Chonoles
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HINT $\rm\ (a)\oplus(b)\ $ is invariant under $\rm\ a\to u\:a,\ b\to v\:b\:$ for units $\rm\:u,v\:$ but $\rm\:(a+b)\:$ need not be.

Bill Dubuque
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