3
$\begingroup$

Given $A$ commutative ring with unity. If we have rings $A_1,\dots, A_n$such that $A\simeq A_1\times \dots \times A_n$ there are $e_1,\dots,e_n$ idempotent elements of $A$ such that $e_1+\dots+e_n = 1$ and $e_i e_j =0$ if $i\neq j$.

I'm not sure how general the proof should be written...

The isomorphic relation suggest that if $f$ defines the isomorphism then there is a $(a_1,\dots, a_n)\in \prod A_i$ such that $f(a_1,\dots,a_n)=1\in A$. Could it work thinking of the idemmpotent terms as $e_i=(0,\dots,0, a_i,0, \dots,0)?$ This should satisfy $e_i e_j =0$ if $i\neq j$. But I have two problems with it (1) I don't see why $e_i$ should idempotent as no assumptions in regard to the existence of idempotent elements in $A_i$ has been made. (2) The sum $\sum e_i=1$ has no reason to be true.

$\endgroup$
  • 2
    $\begingroup$ I'm not exactly clear on what you're trying to prove? The wording in your first paragraph is a little vague about what you're given and what you have to prove... but anyway, if $\phi:A \rightarrow A_1 \times \cdots \times A_n$ is an isomorphism, then you should set $e_i = \phi^{-1}(0,0,\ldots,0,1_i,0,\ldots,0)$, where $1_i \in A_i$ is the identity element. $\endgroup$ – Alex Zorn Sep 20 '17 at 17:56
  • $\begingroup$ @AlexZorn I apologize. Is it more clear now? $\endgroup$ – Cure Sep 20 '17 at 18:25
  • $\begingroup$ I edited your question, replace occurences of $e_ie_i = 0$ with $e_ie_j = 0$. $\endgroup$ – Robert Lewis Sep 23 '17 at 17:10
1
$\begingroup$

In this answer, I am going to assume that no $A_k$ is the zero ring $\{0\}$:

$\forall k, 1 \le k \le n, A_k \ne \{0\}; \tag 1$

this assumption is made without loss of generality, since it is easy to see that if $A_k = \{0\}$ for some $k$, there is a ring isomorphism

$\phi: A_1 \times A_2 \times \ldots A_{k - 1} \times A_k \times A_{k + 1} \times \ldots A_n \simeq A_1 \times A_2 \times A_{k - 1} \times A_{k + 1} \times \ldots \times A_n \tag 2$

given by

$\phi(a_1, a_2, \ldots, a_{k - 1}, 0, a_{k + 1}, \ldots, a_n) = (a_1, a_2, \ldots, a_{k - 1}, a_{k + 1}, \ldots, a_n); \tag 3$

by repeatedly employing such isomorphisims we may reduce any finite product $A_1 \times A_2 \times \ldots \times A_n$ to one which either has no zero factors or is the ring $\{0\}$ itself; in this latter case we have

$A \simeq \{0\}, \tag 4$

$1 = 0$ in $A$, and the result is trivial. Therefore (1) imposes no essential restriction on the result we seek to prove.

In the following I will deploy the notation

$P = A_1 \times A_2 \times \ldots \times A_n = \displaystyle \prod_1^n A_i \tag 5$

($P$ for product), which saves me a many a keystroke when typing $\LaTeX$.

If $f: A \to A_1 \times A_2 \times \ldots \times A_n = \displaystyle \prod_1^n A_i = P, \tag 6$

denotes an isomorphism 'twixt $A$ and $\prod_1^n A_i = P$, then $f(\mathbf 1_A)$, where $\mathbf 1_A$ is the unit of $A$, must be the unique multiplicative unit $\mathbf 1_P$ of $P$; thus $P$ is unital with unit $\mathbf 1_P = f(\mathbf 1_A)$.

If we donote the projection onto the $j$-th component of $\prod_1^n A_i$ by

$\pi_j: \displaystyle \prod_1^n A_i \to A_j, \tag 7$

that is, for

$a = (a_1, a_2, \ldots, a_j, \ldots, a_n) \in P \tag 8$

we have

$\pi_j(a) = \pi_j(a_1, a_2, \ldots, a_j, \ldots, a_n) = a_j \in A_j, \tag 9$

then we may also define the injection

$\iota_j: A_j \to P = \displaystyle \prod_1^n A_i \tag{10}$

by

$\iota_j(a_j) = (0, 0, \ldots, a_j, \ldots, 0) = (\delta_{1j}a_j, \delta_{2j}a_j, \ldots, \delta_{ij}a_j, \ldots, \delta_{nj}a_j), \tag{11}$

where $\delta_{ij}$ is the usual Kronecker symbol; that is

$\delta_{ij} = 1, \; i = j, \tag{12}$

$\delta_{ij} = 0, \; i \ne j; \tag{13}$

thus (7) is equivalent to

$(\iota_j(a_j))_i = \delta_{ij} a_j, \tag{14}$

that is, the $i$-th component of $\iota_j(a_j)$ is $0$ when $i \ne j$, and $a_j$ for $i = j$. It is easy to see that both the $\pi_j$ and the $i_j$ are homomorphisms, and that

$a_j \in A_j \Longrightarrow \pi_j\iota_j(a_j) = a_j, \tag {15}$

that is,

$\pi_j \iota_j = \mathbf I_j, \tag{16}$

where $\mathbf I_j$ is the identity map on $A_j$; we also have, with $a$ as in (8),

$\iota_j \pi_j(a) = \iota_j(a_j) = (0, 0, \ldots, a_j, \ldots, 0) = (\delta_{1j}a_j, \delta_{2j}a_j, \ldots, \delta_{ij}a_j, \ldots, \delta_{nj}a_j). \tag{17}$

We note that it is easy to see that

$\text{Range}(\pi_j) = A_j, \tag{18}$

and

$\ker \iota_j = \{0\}. \tag{19}$

Bearing these preliminary remarks in mind, we set

$\pi_j(\mathbf 1_P) = \mathbf 1_j \in A_j, \tag{20}$

and thus we see that $A_j$ is unital with unit $\mathbf 1_j$; this follows from the fact that $\pi_j$ is surjective, hence for any $c_j \in A_j$ there exists $b_j \in P$ such that

$\pi_j(b_j) = c_j; \tag{21}$

we have

$c_j = \pi_j(b_j) = \pi_j(\mathbf 1_P b_j) = \pi_j(\mathbf 1_P) \pi_j(b_j) = \pi_j(\mathbf 1_P)c_j \tag{22}$

and

$c_j = \pi_j(b_j) = \pi_j(b_j \mathbf 1_P) = \pi_j(b_j) \pi_j(\mathbf 1_P) = c_j \pi_j(\mathbf 1_P); \tag{23}$

(22) and (23) together show that $\pi_j(\mathbf 1_P)$ is either $0_j$ or the multiplicative unit of $A_j$; we can rule out $\pi_j(\mathbf 1_P) = 0_j$ since this implies (via (22) and (23)) that $c_j = 0_j$ for all $c_j \in A_j$, impossible by (1); thus we see that $\mathbf 1_j$ as defined by (20) is the (non-zero) multiplicative unit of $A_j$.

Now set

$e_j = \iota_j(\mathbf 1_j) \in P; \tag{24}$

we have

$e_j^2 = (\iota_j(\mathbf 1_j))^2 =\iota_j(\mathbf 1_j^2) = \iota_j(\mathbf 1_j) = e_j; \tag{25}$

thus $e_j$ is an idempotent in $P$;

$e_j \ne 0 \tag{26}$

since $\iota_j$ is injective and $\mathbf 1_j \ne 0_j$ in $A_j$. We have now established the existence of $n$ non-zero idempotents $e_i$, $1 \le i \le n$, in $P$.

Further progress towards our goal will be facilitated via the aid of the following

Fact:

$\forall a, b \in P ([a = b] \Longleftrightarrow [\forall j, 1 \le j \le n, \pi_j(a) = \pi_j(b)]), \tag{27}$

for which we offer the following

Proof of Fact: It is clear that

$[a = b] \Longrightarrow [\forall j, 1 \le j \le n, \pi_j(a) = \pi_j(b)], \tag{28}$

so suppose

$\forall j, 1 \le j \le n, \pi_j(a) = \pi_j(b), \tag{29}$

then

$\forall j, 1 \le j \le n, \pi_j(a - b) = \pi_j(a) - \pi_j(b) = 0, \tag{30}$

whence

$\forall j, 1 \le j \le n, \iota_j \pi_j(a - b) = 0; \tag{31}$

via (17), this leads to

$\forall j, 1 \le j \le n, a_j -b_j = (a - b)_j = 0, \tag{32}$

that is,

$\forall j, 1 \le j \le n, a_j = b_j, \tag{33}$

which is manifestly equivalent to $a = b$. This proves

$[\forall j, 1 \le j \le n, \pi_j(a) = \pi_j(b)]\Longrightarrow [a = b], \tag{34}$

and so we see that (27) binds. End of Proof of Fact.

We use this Fact to show

$\mathbf 1_P = \displaystyle \sum_{k = 1}^n e_k; \tag{35}$

in the light of (20) we need to evaluate $\pi_j(\sum_1^k e_k)$; we find

$\displaystyle \pi_j(\sum_{k = 1}^n e_k) = \sum_{k = 1}^n \pi_j(e_k) = \sum_{k = 1}^n \pi_j \iota_k(\mathbf 1_k) = \sum_{k = 1, k \ne j}^n \pi_j \iota_k(\mathbf 1_k) + \pi_j \iota_j(\mathbf 1_j), \tag{36}$

and by (16),

$\pi_j \iota_j(\mathbf 1_j) = \mathbf I_j(\mathbf 1_j) = \mathbf 1_j, \tag{37}$

whereas, for $j \ne k$,

$\pi_j \iota_k(\mathbf 1_k) = 0 \tag{38}$

in accord with (17); thus we have

$\displaystyle \pi_j(\sum_{k = 1}^n e_k) = \mathbf 1_j = \pi_j(\mathbf 1_P) \tag{39}$

for every $j$, $1 \le j \le n$; so in the light of our proven Fact, we have

$\mathbf 1_P = \displaystyle \sum_1^n e_k, \tag{40}$

and it remains to show that

$e_i e_j = 0, \; i \ne j; \tag{41}$

now,

$\pi_k(e_i e_j) = \pi_k(e_i)\pi_k(e_j) = \pi_k \iota_i(\mathbf 1_i) \pi_k \iota_j(\mathbf 1_j) = 0,\; 1 \le k \le n, \tag{42}$

since we have established in (38) that $\pi_k \iota_i = 0$ for $k \ne i$, and since $i \ne j$ at most one of $k = i$ or $k = j$ can hold. Thus

$\pi_k(e_i e_j) = \pi_k(0), \; 1 \le k \le n, \tag{43}$

and we may now once again invoke our Fact to affirm that

$e_ie_j = 0, \; i \ne j. \tag{44}$

We have thus shown the existence of $n$ orthogonal, non-zero idempotents which satisfy (40) in the ring $P = \prod_1^n A_i$; and hence in our original ring $A$ itself, since $f^{-1}: P \to A$ is an isomorphism.

$\endgroup$
2
$\begingroup$

Robert Lewis's answer gives a good bit of detail. If you are looking for a more terse answer:

Your idea is right, for each $j$ let $e_j = (0,\dotsc,0,a_j,0,\dotsc,0)$, where the $a_j$ are defined by $f(a_1,\dotsc,a_n)=1 \in A$.

The first key point is to identify the $a_i$'s. As Robert Lewis explains, each $a_i = \pi_i(1)$ is a multiplicative identity in $A_i$. In fact, since $f$ is an isomorphism, $(a_1,\dotsc,a_n)$ is a multiplicative identity in $P = \prod A_i$; now for any $x \in A_i$, consider the multiplication $$ (0,\dotsc,0,x,0,\dotsc,0) \cdot (a_1,\dotsc,a_i,\dotsc,a_n) = (0,\dotsc,a_i x, \dotsc 0). $$ Since $(a_1,\dotsc,a_n)$ is a multiplicative identity, then $a_i x = x \in A_i$. So each $a_i$ is a multiplicative identity in $A_i$. (If you're worried, you can check they are also right identities, or better yet show each $A_i$ is commutative.) We can henceforth denote each $a_i$ by $1_{A_i}$.

Now we follow your idea. For each $j$, let $e_j = (0,\dotsc,0,1_{A_j},0,\dotsc,0) \in \prod A_i$, the element with $1$ in the $j$th entry and all other entries $0$. You can verify that each $e_j^2 = e_j$; $e_i e_j = 0$ for $i \neq j$; and $\sum e_j = (1,1,\dotsc,1) = 1_P$ (where $P = \prod A_j$). You can verify these things by directly computing with the "vectors" for the $e_j$.

The $f(e_j)$ are the desired idempotent elements of $A$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.