Show that the ring of upper triangular 2x2 matrices is a direct sum of two of its modules

Question

The definition I have for the direct sum of R-modules is the following:

$\bigoplus_{i \in I} M_i = \big\{f \in \Pi_{i\in I}M_i : \ \#\{i \in I:\ f(i) \neq 0 \} < \infty \big\}$.

Say k is some field, and R is the ring of upper triangular 2x2 matrices. I am asked to show that R is the direct sum of two R-modules, ie $R = P \oplus Q$. I'm a little bit confused as to why it is ok to put an equation sign between these two quantities, as it seems to me that an isomorphism sign would be more precise. My questions are:

1. Why can you equate R with a direct sum of (two) R-modules?
2. My intuition is telling me that P could be the strictly upper triangular (2x2) matrices, but I'm uncertain what Q should be. Any hints as to what P and Q could be?

Answer 1

As for why equality is used, here's the idea. The direct sum (coproduct) has a "Universal Property". Let $R$ be a ring, $M,N$ two $R$-modules. We denote by $\imath_M:M\to M\oplus N$ and $\imath_N:N\to M\oplus N$ the natural inclusion maps. Now here's the universal property

Let $P$ be another $R$-module together with $R$-linear maps $\jmath_M: M\to P$ and $\jmath_N: N\to P$. Then there exists a unique $R$-linear map $\psi: M\oplus N\to P$ such that $\jmath_M=\psi\circ \imath_M$ and $\jmath_N=\psi\circ \imath_N$.

Proof. Define $\psi: M\oplus N\to P$ via the mapping $(m,n)\mapsto \jmath_M(m)+\jmath_N(n)$. It's easy to check this is an $R$-linear map.

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Let's now get closer to why replacement of isomorphism and equality makes sense in your case...

Let $R$ be a ring, $M$ an $R$-module and $P,Q$ two sub-modules of $M$. Then, we have inclusions $\imath_P:P\to P\oplus Q$, $\imath_Q:Q\to P\oplus Q$ naturally and also inclusions $\jmath_P: P\to M$ and $\jmath_Q: Q\to M$ because $P,Q$ are sub-modules. Using the universal property we have a unique $R$-linear map $\psi: P\oplus Q\to M$ which satisfies the above.

If this $\psi$ an isomorphism, we say $M$ is naturally isomorphic to $P\oplus Q$.

Proposition: Let $R$ be a ring, $M$ an $R$-module and $P,Q$ two sub-modules of $M$. Then $M$ is naturally isomorphic to $P\oplus Q$ (i.e. $\psi(p,q)=p+q$ is an isomorphism) if and only if $M=P+Q$ and $P\cap Q=(0)$.

I leave the proof of this to you.

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Now as you can see, we can think of $P\oplus Q$ as the sub-module $P+Q\subset M$ provided that $P\cap Q = (0)$. Since $P+Q=M$ in sub-module sense of the equality, it is justified to replace the statement: "M is naturally isomorphic to $P\oplus Q$" by simply $M=P\oplus Q$.