Let $R$ be a direct sum of ideals $R=R_1\oplus R_2\oplus\dots\oplus R_k$. Each ideal $R_i$ is commutative of order $p_{i}^{n_{i}}$ ($p$ is prime), and has a unity. How to show that the direct sum of these ideals is commutative?
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4$\begingroup$ Multiplication is defined term-wise, so commutativity follows trivially from the commutativity of each component. $\endgroup$– Alex BeckerJul 26, 2013 at 9:27
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$\begingroup$ @YACP Let $R=\mathbb{Z}\oplus\mathbb{Z}$; the summands are ideals which have their unity (different from the unity in $R$, of course). $\endgroup$– egregJul 26, 2013 at 11:20
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1 Answer
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It's sufficient to prove the statement for $R=I\oplus J$, because the general one follows by induction.
The conditions on the ideals $I$ and $J$ are that $I+J=R$ and $I\cap J=\{0\}$; since $IJ\subseteq I\cap J$ for any ideals $I$ and $J$, we can say that $$ xy=0,\text{ for $x\in I$, $y\in J$}. $$
Therefore, if $a,b\in R$, we can write $$ a=x_1+y_1,\quad b=x_2+y_2,\quad(x_1,x_2\in I, y_1,y_2\in J) $$ and so \begin{align} ab&=(x_1+y_1)(x_2+y_2)=x_1x_2+y_1y_2.\\ ba&=(x_2+y_2)(x_1+y_1)=x_2x_1+y_2y_1. \end{align} Now, commutativity of $I$ and $J$ ends the proof.
