Idempotents and Product Decomposition

Previously, in §Chinese Remainder Theorem, ⁋Proposition 6, we saw that pairwise comaximal two-sided ideals whose intersection is $0$ decompose a given ring into a product of quotient rings. In that decomposition, the elements that served to pick out each factor were those whose $i$th component is $1$ and whose remaining components are $0$, and such elements are characterized as idempotents lying in the center of the ring.

In this post, we formally define these idempotents, prove that a complete orthogonal set of central idempotents is in bijection with a direct product decomposition of the ring, and then examine how the Chinese remainder theorem in general rings is recovered from this.

Idempotents

First, we make the following definition.

Definition 1 For a ring $A$, an element $e\in A$ satisfying $e^2=e$ is called an idempotent. Two idempotents $e,f$ are said to be orthogonal if $ef=fe=0$. Also, if an idempotent $e$ belongs to the center of $A$, $Z(A)$, it is called a central idempotent.

Any ring always has the two idempotents $0$ and $1$, and these two are always central. We call them trivial idempotents. Given an idempotent $e$, the element $1-e$ is also an idempotent, because

\[(1-e)^2=1-2e+e^2=1-2e+e=1-e\]

Furthermore, since

\[e(1-e)=e-e^2=0,\qquad (1-e)e=e-e^2=0\]

$e$ and $1-e$ are orthogonal to each other. We will treat $e$ paired with its complementary idempotent $1-e$, and $e$ being central is equivalent to $1-e$ being central.

Example 2 By a simple calculation, one can check that the idempotents of the ring $A=\mathbb{Z}/6\mathbb{Z}$ are $0,1,3,4$; here we can verify that $3,4$ are complementary to each other, and $0,1$ are complementary to each other. Furthermore, since $A$ is commutative, these four idempotents automatically become central idempotents. On the other hand, in $\mathbb{Z}/4\mathbb{Z}$, the only elements satisfying $x^2=x$ are $0,1$, so this ring has only trivial idempotents.

Complete set of orthogonal idempotents

We now introduce the following concept in order to handle multiple idempotents at once.

Definition 3 In a ring $A$, if idempotents $e_1,\ldots, e_n$ satisfy both of the following two conditions

\[e_1+\cdots+e_n=1,\qquad e_ie_j=0\quad\text{for $i\neq j$}\]

then $\{e_1,\ldots, e_n\}$ is called a complete set of orthogonal idempotents. If each $e_i$ is central, it is called a central complete set.

The simplest complete set for $n=1$ is $\{1\}$, and in Example 2, for $\mathbb{Z}/6\mathbb{Z}$, $\{0,1\}$ and $\{3,4\}$ are examples of central complete sets with $n=2$. More generally, given a single idempotent $e$, the set $\{e,1-e\}$ is always a complete set of orthogonal idempotents. Thus a complete set can be viewed as a generalization of partitioning an idempotent into multiple pieces. We now examine how this partition decomposes the ring.

Proposition 4 Let a ring $A$ and a complete set of orthogonal idempotents of $A$, $\{e_1,\ldots, e_n\}$, be given. Then the decomposition as a left $A$-module

\[A=Ae_1\oplus\cdots\oplus Ae_n\]

holds. Moreover, if each $e_i$ is central, then each $Ae_i$ is a two-sided ideal of $A$.

Proof

For any $x\in A$, from $e_1+\cdots+e_n=1$ we have

\[x=x\cdot 1=xe_1+\cdots+xe_n\]

and since each $xe_i\in Ae_i$, we have $A=Ae_1+\cdots+Ae_n$. Hence what remains to be shown is that this sum is a direct sum.

To this end, suppose some element is expressed in two ways. That is, let $a_1e_1+\cdots+a_ne_n=0$. Then multiplying this equation on the right by $e_j$, since for $i\neq j$ we have $e_ie_j=0$ and $e_j^2=e_j$, we obtain

\[0=(a_1e_1+\cdots+a_ne_n)e_j=\sum_{i=1}^n a_i(e_ie_j)=a_je_j\]

That is, the $j$th component $a_je_j$ is $0$, and since this holds for every $j$, the given sum is a direct sum.

Now to show the last part, assume each $e_i$ is central. Then that $Ae_i$ is a left ideal is trivial by definition, so it suffices to show it is a right ideal. For this we need to check that for any $ae_i\in Ae_i$ and $x\in A$, we have $(ae_i)x\in Ae_i$; but since $e_i$ is central,

\[(ae_i)x=a(e_ix)=a(xe_i)=(ax)e_i\in Ae_i\]

so $Ae_i$ is a two-sided ideal.

When each $e_i$ is central, $Ae_i$ itself carries the structure of a ring. In particular, $Ae_i$ has an identity element, and that identity element is $e_i$. Indeed, for any $ae_i\in Ae_i$, since $e_i$ is idempotent and central,

\[e_i\cdot(ae_i)=(e_ia)e_i=(ae_i)e_i=ae_i^2=ae_i\]

and likewise $(ae_i)\cdot e_i=ae_i$.

Central idempotents and product decomposition

From the argument above we know that each $Ae_i$ is a ring with identity $e_i$ and that $A$ is their direct sum. The following theorem shows that this direct sum is in fact a direct product of rings, and moreover that such a decomposition is in bijection with a central complete set.

Theorem 5 For a ring $A$, there is a one-to-one correspondence between the following two data:

A complete set of central orthogonal idempotents in $A$, $\{e_1,\ldots, e_n\}$.
A decomposition of $A$ as a direct product of $n$ two-sided ideals $A=\mathfrak{a}_1\times\cdots\times\mathfrak{a}_n$.

This correspondence is given by $\mathfrak{a}_i=Ae_i$.

Proof

First, suppose the first datum is given. Setting $\mathfrak{a}_i:=Ae_i$ as given by the correspondence above, by Proposition 4, each $\mathfrak{a}_i$ is a two-sided ideal and $A=\mathfrak{a}_1\oplus\cdots\oplus\mathfrak{a}_n$ is a direct sum as a left module. Therefore, what remains is to show that this direct sum is also a direct product of rings. Consider the morphism

\[\varphi:A\rightarrow \prod_{i=1}^n Ae_i;\quad x\mapsto (xe_1,\ldots, xe_n)\]

This preserves addition, and since the $e_i$ are central and orthogonal,

\[\varphi(x)\varphi(y)=(xe_1\cdot ye_1,\ldots, xe_n\cdot ye_n)=(xye_1,\ldots, xye_n)=\varphi(xy)\]

shows that it also preserves multiplication. Moreover, $\varphi(1)=(e_1,\ldots, e_n)$ and each $e_i$ is the identity element of $Ae_i$, so $\varphi$ is a ring homomorphism. That this is an isomorphism follows because the direct sum decomposition of Proposition 4 is an isomorphism.

Conversely, suppose the second datum is given. A direct product decomposition yields in particular a direct sum $A=\mathfrak{a}_1\oplus\cdots\oplus\mathfrak{a}_n$ with respect to addition, so the identity element can be written uniquely according to this decomposition as

\[1=e_1+\cdots+e_n,\qquad e_i\in\mathfrak{a}_i\]

For any $x\in\mathfrak{a}_j$, since $\mathfrak{a}_i$ is an ideal we have $xe_i\in\mathfrak{a}_i$, and from this we obtain the decomposition of $x$

\[x=x\cdot 1=xe_1+\cdots+xe_n\]

which is unique. However, the decomposition of $x\in\mathfrak{a}_j$ is the one where only the $j$th component is $x$ and the rest are $0$, so by this uniqueness, $xe_j=x$ and for $i\neq j$, $xe_i=0$. In particular, substituting $x=e_j$ yields $e_j^2=e_j$, and for all $i\neq j$, we obtain $e_je_i=0$. That is, $\{e_1,\ldots, e_n\}$ is a complete set of orthogonal idempotents. Now, what remains to be shown is only centrality. Writing any $y\in A$ as

\[y=y_1+\cdots+y_n,\qquad y_i\in\mathfrak{a}_i\]

from above, $y_ie_j$ is, when $i=j$, $y_i$, and $0$ otherwise, and the same holds for $e_jy_i$, so $ye_j=y_j=e_jy$. Therefore, each $e_j$ commutes with every $y$ and is central.

Finally, let us verify that these two constructions are inverses of each other. Starting from the first datum and setting $\mathfrak{a}_i=Ae_i$, since $1=e_1+\cdots+e_n$ is the unique expression of $1$ according to this decomposition, the idempotents returned by the second construction are again $e_i$. Conversely, starting from the second datum, from $e_i\in\mathfrak{a}_i$ we have $Ae_i\subseteq\mathfrak{a}_i$, and the equation $x=xe_i$ obtained above gives the reverse inclusion for any $x\in\mathfrak{a}_i$, so $\mathfrak{a}_i=Ae_i$. Therefore, this correspondence is one-to-one.

By Theorem 5, a ring having no nontrivial central idempotents is equivalent to it not decomposing as a product of two nonzero rings. Such a ring is called connected or indecomposable. For instance, a division ring $A$ is always indecomposable. If $A$ decomposed as a product of two nonzero rings, then $(1,0)$ and $(0,1)$ corresponding to the identity elements of the factors would both be nonzero while their product is $0$, so $A$ would have zero divisors.

The next example shows in particular what happens when the centrality condition is dropped. In this case, Theorem 5 cannot be applied as is, but the decomposition as a module is still guaranteed by Proposition 4.

Example 6 For a ring $A$, consider the $n\times n$ matrix ring $\Mat_n(A)$. If we let $E_{ij}$ be the matrix unit whose $(i,j)$ entry is $1$ and whose other entries are $0$, the diagonal entries $E_{11},\ldots, E_{nn}$ satisfy

\[E_{ii}^2=E_{ii},\qquad E_{ii}E_{jj}=0\ (i\neq j),\qquad E_{11}+\cdots+E_{nn}=I\]

and thus form a complete set of orthogonal idempotents. Hence by Proposition 4 we obtain the left module decomposition

\[\Mat_n(A)=\Mat_n(A)E_{11}\oplus\cdots\oplus \ Mat_n(A)E_{nn}\]

where $\Mat_n(A)E_{ii}$ is the set of matrices whose entries are only in the $i$th column.

Note that if $n\geq 2$, then the $E_{ii}$ are not central. For example, when $n=2$,

\[E_{11}E_{12}=E_{12}\neq 0=E_{12}E_{11}\]

so $E_{11}$ does not commute with $E_{12}$. Therefore this decomposition does not give a direct product decomposition of the ring.

In fact, for any ring $A$ one can compute the center of $\Mat_n(A)$ directly. If a matrix $M$ commutes with all matrix units, we know that

\[(E_{kl}M)_{i,j}=\delta_{ik}M_{lj},\qquad (ME_{kl})_{i,j}=M_{ik}\delta_{lj}\]

must be equal. Now consider the case $i=k$. Then

\[M_{lj}=M_{kk}\delta_{lj}\]

so if $j\neq l$ then $M_{lj}=0$, and where $j=l$ we have $M_{ll}=M_{kk}$, so $M$ must be of the form $cI$. On the other hand, such an $M$ must commute not only with the matrix units but also with matrices of the form $aI$, so this coefficient $c$ must belong to the center of $A$, $Z(A)$. Conversely, if $c\in Z(A)$, then for any $M$ we have $(cIM)_{i,j}=cM_{i,j}=M_{i,j}c=(McI)_{i,j}$, so $cI$ commutes with all matrices, and therefore

\[Z(\Mat_n(A))=\{cI\mid c\in Z(A)\}\]

Now since $(cI)^2=c^2I$, the central idempotents of $\Mat_n(A)$ are, for $c^2=c$ with $c\in Z(A)$, the elements $cI$, which simply carries the central idempotents of $A$ directly over to $\Mat_n(A)$. In particular, if $A$ is a division ring, then the only central idempotents of $A$ are $0,1$, so the only central idempotents of $\Mat_n(A)$ are $0$ and $I$.

On the other hand, whether an idempotent is central depends on which ring it is viewed in. If we partition $n=n_1+\cdots+n_r$, then in the subring of block-diagonal matrices arranged along the diagonal with the first block $n_1\times n_1$, the next block $n_2\times n_2$, and so on,

\[B=\left\{\diag(M_1,\ldots, M_r)\mid M_k\in \Mat_{n_k}(A)\right\}\cong\prod_{k=1}^r \Mat_{n_k}(A)\]

the element having the identity matrix only in the $k$th block, $P_k$, is central, so by the correspondence of Theorem 5 it gives this direct product decomposition. However, we have already verified that this $P_k$ is not central in the example ring $\Mat_n(A)$ above.

Connection with the Chinese Remainder Theorem

Meanwhile, even before Theorem 5, we had already seen a method of decomposing a given ring as a direct product. (§Chinese Remainder Theorem, ⁋Proposition 6) These two results are not independent, and in the remainder of this post we examine the relationship between them. The key point is that the pairwise comaximal condition pulls the natural idempotents of the product ring into $A$, and in the following theorem $\pi:A\rightarrow\prod_{i=1}^n A/\mathfrak{a}_i$ denotes the morphism

\[x\mapsto (x+\mathfrak{a}_1,\ldots, x+\mathfrak{a}_n)\]

induced by the projections onto each quotient.

Theorem 7 Let a ring $A$ and its two-sided ideals $\mathfrak{a}_1,\ldots, \mathfrak{a}_n$ be given. The following are all equivalent.

$\pi:A\rightarrow\prod_{i=1}^n A/\mathfrak{a}_i$ is an isomorphism.
For all $i\neq j$, $\mathfrak{a}_i+\mathfrak{a}_j=A$ and $\bigcap_i\mathfrak{a}_i=0$.
There exists in $A$ a central complete set of orthogonal idempotents $\{e_1,\ldots, e_n\}$ such that for all $i$, $\mathfrak{a}_i=A(1-e_i)$.

Proof

First assume the first condition and show the second. If $\pi$ is an isomorphism, then $\bigcap_i\mathfrak{a}_i=\ker\pi=0$. Also, for a fixed pair $i\neq j$, let the preimage of the element whose $i$th component is $1+\mathfrak{a}_i$ and whose remaining components are all $0$ be $e_i\in A$. Then from the $j$th component we have $e_i\in\mathfrak{a}_j$ and from the $i$th component $1-e_i\in\mathfrak{a}_i$, so

\[1=(1-e_i)+e_i\in\mathfrak{a}_i+\mathfrak{a}_j\]

and therefore $\mathfrak{a}_i+\mathfrak{a}_j=A$.

Now assume the second condition and show the third. From the pairwise comaximal condition and $\bigcap_i\mathfrak{a}_i=\ker\pi=0$, §Chinese Remainder Theorem, ⁋Proposition 6 implies that $\pi$ is an isomorphism. In the product ring $\prod_i A/\mathfrak{a}_i$, if we consider the elements whose $i$th component is $1+\mathfrak{a}_i$ and whose remaining components are $0$,

\[\bar{e}_i=(0,\ldots, 0,1,0,\ldots, 0)\]

we can see by a componentwise calculation that they form a central complete set of orthogonal idempotents, and since $\pi$ is a ring isomorphism, $e_i:=\pi^{-1}(\bar{e}_i)$ is also a central complete set of orthogonal idempotents in $A$. What remains is the equality $\mathfrak{a}_i=A(1-e_i)$; since for $\bar{e}_i$ the $i$th component is $1+\mathfrak{a}_i$, we have $1-e_i\in\mathfrak{a}_i$, and since $\mathfrak{a}_i$ is an ideal, $A(1-e_i)\subseteq\mathfrak{a}_i$ is trivial. Conversely, let $a\in\mathfrak{a}_i$; then in $\pi(a)$ the $i$th component is $0$ and in $\bar{e}_i$ all components except the $i$th are $0$, so $\pi(ae_i)=\pi(a)\bar{e}_i=0$, and since $\pi$ is injective, $ae_i=0$. Therefore

\[a=ae_i+a(1-e_i)=a(1-e_i)\in A(1-e_i)\]

Finally, assume the third condition and show the first. By Theorem 5, the central complete set $\{e_1,\ldots, e_n\}$ gives, via the ring isomorphism $x\mapsto (xe_1,\ldots, xe_n)$, the direct product decomposition $A\cong\prod_i Ae_i$. On the other hand, for each $i$ the morphism

\[A\rightarrow Ae_i;\quad a\mapsto ae_i\]

is clearly surjective, and if $ae_i=0$ then $a=a(1-e_i)$ while conversely $a(1-e_i)e_i=0$, so its kernel is $A(1-e_i)=\mathfrak{a}_i$. Hence $A/\mathfrak{a}_i\cong Ae_i$, and since this isomorphism is given by $x+\mathfrak{a}_i\mapsto xe_i$, composing with the above direct product decomposition yields exactly $\pi$. Therefore $\pi$ is an isomorphism.

Let us apply this to the most familiar case, the ring of integers.

Example 8 Let the prime factorization of $n\geq 2$ be $n=p_1^{a_1}\cdots p_r^{a_r}$ (distinct primes $p_k$), and for $A=\mathbb{Z}/n\mathbb{Z}$, consider the ideals $\mathfrak{a}_k=p_k^{a_k}\mathbb{Z}/n\mathbb{Z}$. In $\mathbb{Z}$, the ideals $p_k^{a_k}\mathbb{Z}$ are pairwise comaximal and their intersection is $n\mathbb{Z}$, so in $A$ the $\mathfrak{a}_k$ are pairwise comaximal and $\bigcap_k\mathfrak{a}_k=0$. Hence Theorem 7 applies, yielding the classical Chinese remainder theorem

\[\mathbb{Z}/n\mathbb{Z}\cong\prod_{k=1}^r\mathbb{Z}/p_k^{a_k}\mathbb{Z}\]

together with the complete set of central orthogonal idempotents of $\mathbb{Z}/n\mathbb{Z}$ corresponding to this decomposition.

Concretely, for $n=6=2\cdot 3$ we have $\mathbb{Z}/6\mathbb{Z}\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$. If we pull back from the product ring $\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/3\mathbb{Z}$ the two idempotents $\bar{e}_1=(1,0)$, $\bar{e}_2=(0,1)$ along the isomorphism, the element of $\mathbb{Z}/6\mathbb{Z}$ corresponding to $(1,0)$ is $3$, and the one corresponding to $(0,1)$ is $4$. Indeed, $3\equiv 1\ (\mathrm{mod}\ 2)$, $3\equiv 0\ (\mathrm{mod}\ 3)$ and $4\equiv 0\ (\mathrm{mod}\ 2)$, $4\equiv 1\ (\mathrm{mod}\ 3)$. This matches exactly the idempotents $3,4$ found by hand in Example 2.

On the other hand, for a prime $p$ and $a\geq 1$, the ring $\mathbb{Z}/p^a\mathbb{Z}$ has no idempotents other than $0,1$. This is because $x^2\equiv x\ (\mathrm{mod}\ p^a)$ is equivalent to $x(x-1)\equiv 0\ (\mathrm{mod}\ p^a)$, and since $x$ and $x-1$ are coprime, $p^a$ must divide exactly one of them. Hence each factor in the above decomposition is indecomposable in the sense of Theorem 5, and the direct product decomposition according to prime factorization is the finest decomposition that cannot be split any further. That $\mathbb{Z}/4\mathbb{Z}$ had only trivial idempotents in Example 2 is a special case of this.

References

[AM] M. F. Atiyah and I. G. Macdonald, Introduction to commutative algebra, Addison–Wesley, 1969.
[DF] D. S. Dummit and R. M. Foote, Abstract algebra, 3rd ed., Wiley, 2004.
[Lam] T. Y. Lam, A first course in noncommutative rings, 2nd ed., Graduate Texts in Mathematics 131, Springer, 2001.

K

Idempotents and Product Decomposition

Idempotents

Complete set of orthogonal idempotents

Central idempotents and product decomposition

Connection with the Chinese Remainder Theorem

Share on

Comments

Leave a comment

찾아보기