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In Bourbaki's Algebra there is the following proposition:


Let $A$ be a ring (with $1$), $(x_\lambda)_{\lambda\in L}$ a family of elements of $A$ and $\mathfrak{a}$ the set of sums $\sum_{\lambda\in L}a_\lambda x_\lambda b_\lambda$ where $(a_\lambda)_{\lambda\in L}$, $(b_\lambda)_{\lambda\in L}$ are families with finite support of elements of $A$. Then $\mathfrak{a}$ is the two-sided ideal of $A$ generated by the elements $x_\lambda$.


The proof, they say, is analogous to the corresponding statement for left ideals.

However, I wonder if that is true. Let's consider the case $|L|=1$. I don't see at all, how one can, for $x,a,a',b,b'\in A$, find $\alpha, \beta\in A$ such that $$axb+a'xb'=\alpha x\beta.$$

The form of the definition of two-sided principal ideal on Wikipedia strengthens my doubts.

Can somebody clear this up?

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3 Answers 3

up vote 3 down vote accepted

Your suspicion is correct. The set of elements of the form $axb, a,b\in A$ (fixed $x$) is not necessarily closed under addition. An example that comes to mind is to choose $x$ to be a rank one matrix in the ring $A$ of 2x2 matrices over a field. In that case all the matrices of the form $axb$ are of rank $\le 1$. Yet, a standard exercise shows that $A$ has no non-trivial 2-sided ideals, so the 2-sided ideal generated by $x$ must contain full rank matrices as well.

I suspect that Bourbaki allows the same element $x_\lambda$ to appear several times in the sum. The notation for such a double sum may quickly become a bit awkward, so it is understandble that they seek to compress it. That notation may have been introduced earlier in the book. It is unthinkable that they would make a mistake here.

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@Jyrki Lahtonen already posted an answer, but that answer does not appear in a clear form. So it behooves me to try to complement it here.
To say that $AxA$ is a two-sided ideal, not only does it need to check the multiplicative part, but it also needs to verify that it forms an abelian group. In fact, this concern is contained in the definition: the set contains all finite sums of expressions of the form: $\Sigma axb$ where a,b belong to A, so that, by definition, the left side of your equation is itself an element in that set.
Hope this does clarify the matter.

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The set $\{axb\,|\,a,b\in A\}$ is the semigroup-theoretic principal ideal generated by $x.$ $A$ with multiplication has a structure of a semigroup (a monoid even, because $A$ is unital). For a semigroup $S$ and its subset $I,$ we say that $I$ is an ideal in $S$ iff $$(\forall i\in I,s\in S)\;\;is\in I\text{ and }si\in I,$$ which is clearly satisfied by $A=:S\supset I:=\{axb\,|\,a,b\in A\}.$ It is also easy to see that $I$ is the smallest semigroup-theoretic ideal in $A$ containing $x$. (It contains $x$ because $A$ has an identity.) We call this a principal ideal generated by $x$.

This is certainly not what Bourbaki had in mind, but I thought it would be a good idea to point this out.

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