Rings with nontrivial idempotents decompose as a direct product

Let R be a ring with 1 \neq 0. An element e \in R is called idempotent if e^2 = e. Assume e is an idempotent in R and that er = re for all r \in R. (That is, e \in Z(R).) Prove that Re and R(1-e) are two-sided ideals in R and that R \cong Re \times R(1-e). Show that e and 1-e are identities in Re and R(1-e) respectively.

Note first that (1-e)^2 = 1 -2e + e = 1-e.

First we’ll show that Re is a two-sided ideal. First 0 \cdot e = 0 \in Re, so Re is nonempty. Let re, se \in Re. Then re-se = (r-s)e \in Re. Now if t \in R, then tre \in Re and ret = rte \in Re. Thus Re is a two-sided ideal of R.

Next we’ll show that R(1-e) is a two-sided ideal. First 0 \cdot (1-e) = 0 \in R(1-e), so R(1-e) is nonempty. Let r(1-e), s(1-e) \in R(1-e); then r(1-e) - s(1-e) = (r-s)(1-e) \in R(1-e). Now if t \in R, then tr(1-e) \in R and r(1-e)t = r(t-et) = r(t-te) = rt(1-e) \in R(1-e). Thus R(1-e) is a two-sided ideal in R.

Now suppose x \in Re \cap R(1-e). Then we have re = s(1-e) for some r,s \in R. Now re = s - se, and right multiplying by e, we see that re^2 = se - se^2, so that re = se - se = 0. Thus x = 0, and so Re \cap R(1-e) = 0 is trivial.

Let r \in R; since re + (1-r)e = r, R \subseteq Re + R(1-e). By the recognition theorem for direct products of groups, the map \varphi : Re \times R(1-e) \rightarrow R given by \varphi(a,b) = a+b is a group isomorphism. Now let (r_1e, s_1(1-e)), (r_2e,s_2(1-e)) \in Re \times R(1-e). Then \varphi((r_1e,s_1(1-e))(r_2e,s_2(1-e))) = \varphi(r_1r_2e,s_1s_2(1-e)) = r_1r_2e + s_1s_2(1-e) = r_1r_2e^2 + s_1s_2(1-e)^2 = r_1er_2e + r_1s_2(e-e^2) + r_2s_1(e-e^2) + s_1(1-e)s_2(1-e) = (r_1e + s_1(1-e))(r_2e + s_2(1-e)) \varphi(r_1e,s_1(1-e)) \cdot \varphi(r_2e,s_2(1-e)). Thus R \cong Re \times R(1-e) as rings.

Note that for all re \in Re, ere = re^2 = re and re^2 = re. Thus e \in Re is a multiplicative identity. Similarly, r(1-e)^2 = r(1-e) and (1-e)r(1-e) = (r - er)(1-e) = r(1-e)^2 = r(1-e), so that 1-e is a multiplicative identity in R(1-e).

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: