0
$\begingroup$

Assuming $\lbrace R_i \rbrace _{i=1}^{n}$ be a set of rings, show that $$e=(e_1,...,e_n)$$ is the identity of $$R=R_1 \times R_2 \times ... \times R_n$$ if and only if $e_i$ is the identity of $R_i$ for all $1 \leq i \leq n$. I come across this statement when $R$ is the internal direct product of $R_i$ but here, we don have this assumption. How do we tackle this problem?

$\endgroup$
  • $\begingroup$ Apparently your rings are assumed to come with an identity element (not everybody agrees on this definition). But then giving the identity is part of the definition of a direct product of rings. $\endgroup$ – Marc van Leeuwen Mar 24 '13 at 7:36
1
$\begingroup$

Backwards: If $e=(e_1,...,e_n)$ is the identity of $R \implies \forall x=(x_1,...,x_n)\in R, $ we have $ex=x$ which, since we're dealing with a Cartesian Product, means that $ \forall i$ we have $x_ie_i=x_i \implies e_i $ is the identity in $R_i$.

Forwards: If $e_i$ is the identity for $R_i \implies \forall x_i\in R_i$ we have $ e_ix_i=x_i \implies$ for arbitrary $x=(x_1,...,x_n)\in R$ we have $(x_1,...,x_n)(e_1,...,e_n)=(x_1e_1,...,x_ne_n)=(x_1,...,x_n)$ by definition of Cartesian Product and the fact that $e_i$ is the identity of $R_i$ for each $i$.

$\endgroup$
1
$\begingroup$

Hints: Suppose $e=(e_1,\cdots ,e_n)$ is the identity of $R$. Take now an arbitrary element $x\in R_i$ (so fix some $i$ as well). Consider the element $y=(0,0,\cdots , e_i,0,0,\cdots)\in R$ with $e_i$ at position $i$. Now, spell out what the identity property of $e$ means for $y$? Conclude from that that $e_i$ is the identity element in $R_i$.

Try doing the other direction now (should be easier).

$\endgroup$
  • $\begingroup$ for the forward direction , since $ye=(0,...,e_i e_i,...,0)=(0,...,e_i e_i,...,0)$, we have $e_i e_i=e_i \implies e_i=1$, is this correct? $\endgroup$ – Idonknow Mar 24 '13 at 15:10
  • $\begingroup$ For the element $y$, how do we know for sure $y$ exists? $\endgroup$ – Idonknow Apr 16 '13 at 14:49

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.