# Let $\lbrace R_i \rbrace _{i=1}^{n}$ be a set of rings. Show that $e=(e_1,…,e_n)$ is the identity of $R$

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?

• 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. – Marc van Leeuwen Mar 24 '13 at 7:36

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 $i$.
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 $R_i$.
• 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? – Idonknow Mar 24 '13 at 15:10
• For the element $y$, how do we know for sure $y$ exists? – Idonknow Apr 16 '13 at 14:49