Exercise 3.6.3 (Projection sets up generating submodules)

Question

Let R be a ring, M an R-module and $p \colon M 	o M$ an R-module homomorphism such that $p^2 = p$ (such a map is called a projection). Prove that $M ~\cong ~	ext{ker}~p ~\oplus~ 	ext{im}~ p$.

Shinkenjoe · Accepted Answer

\subsection*{Step 1: Decomposition of arbitrary element of $ M $}
Let $ z $ be an arbitrary element of $ M $. Then $ p(z) = m_1 $, where $ m_1 \in \operatorname{im} p $. Consider the element $ m_2 = z - m_1 $. Then:
$$
z = m_1 + m_2.
$$
Also,
$$
p(m_2) = p(z - m_1).
$$
Using the properties of a module homomorphism, we have:
$$
p(m_2) = p(z) - p(m_1).
$$
Substituting $ p(z) = m_1 $, we get:
$$
p(m_2) = m_1 - m_1 = 0.
$$
Thus, $ m_2 \in \ker p $.

Since $ z $ was arbitrary, every element of $ M $ can be written as a sum of elements of $ \operatorname{im} p $ and $ \ker p $.

\subsection*{Step 2: Properties of $ p $ on $ \operatorname{im} p $ and $ \ker p $}
Consider $ m_1 \in \operatorname{im} p $. Then $ p(m_1) = m_1 $. Assume to the contrary that $ m_1 \neq 0 $ and $ m_1 \in \ker p $. Then, on the one hand, $ p(m_1) = 0 $ (since $ m_1 \in \ker p $), and on the other hand, $ p(m_1) = m_1 $ (since $ m_1 \in \operatorname{im} p $). This is a contradiction. Therefore:
$$
\ker p \cap \operatorname{im} p = \{0\}.
$$

\subsection*{Step 3: Define the map $ f $}
Define $ f : M \to \ker p \oplus \operatorname{im} p $ by:
$$
f(z) = (z - p(z), p(z)).
$$
By the above, $ z - p(z) \in \ker p $, so $ f(z) \in \ker p \oplus \operatorname{im} p $.

\subsection*{Step 4: $ f $ is a module homomorphism}
\textbf{(i) Abelian group homomorphism:}  
$$
f(-z) = (-z - p(-z), p(-z)).
$$
Using $ p $ being a module homomorphism:
$$
f(-z) = (-z - (-p(z)), -p(z)) = (-(z - p(z)), -p(z)) = -f(z).
$$
Similarly, for $ y, z \in M $:
$$
f(y + z) = (y + z - p(y + z), p(y + z)).
$$
Using $ p $ being a module homomorphism:
$$
f(y + z) = ((y - p(y)) + (z - p(z)), p(y) + p(z)).
$$
By addition in $ \ker p \oplus \operatorname{im} p $:
$$
f(y + z) = (y - p(y), p(y)) + (z - p(z), p(z)) = f(y) + f(z).
$$

\textbf{(ii) Module homomorphism:}  
For $ r \in R $ and $ z \in M $:
$$
f(rz) = (rz - p(rz), p(rz)).
$$
Using $ p $ being a module homomorphism:
$$
f(rz) = (rz - rp(z), rp(z)).
$$
By $ M $ being an $ R $-module:
$$
f(rz) = r(z - p(z), p(z)) = rf(z).
$$

Thus, $ f $ is a module homomorphism.

\subsection*{Step 5: Injectivity of $ f $}
Let $ f(z) = f(y) $, i.e.,
$$
(z - p(z), p(z)) = (y - p(y), p(y)).
$$
Then:
$$
z - p(z) = y - p(y) \quad \text{and} \quad p(z) = p(y).
$$
Adding these, we get $ z = y $. Thus, $ f $ is injective.

\subsection*{Step 6: Surjectivity of $ f $}
Let $ m_1 \in \ker p $ and $ m_2 \in \operatorname{im} p $. Consider $ z = m_1 + m_2 $. Then:
$$
f(z) = (z - p(z), p(z)).
$$
Since $ z = m_1 + m_2 $ and $ p(z) = m_2 $, we have:
$$
f(z) = (m_1 + m_2 - m_2, m_2) = (m_1, m_2).
$$
Thus, $ f $ is surjective.

\subsection*{Conclusion}
Since $ f $ is a bijective module homomorphism, we have:
$$
M \cong \ker p \oplus \operatorname{im} p.
$$

Exercise 3.6.3 (Projection sets up generating submodules)

Answers

Step 1: Decomposition of arbitrary element of $M$

Step 2: Properties of $p$ on $im p$ and $\ker p$

Step 3: Define the map $f$

Step 4: $f$ is a module homomorphism

Step 5: Injectivity of $f$

Step 6: Surjectivity of $f$

Conclusion

Comments

Exercise 3.6.3 (Projection sets up generating submodules)

Answers

Step 1: Decomposition of arbitrary element of M

Step 2: Properties of p on im ⁡ p and ker ⁡ p

Step 3: Define the map f

Step 4: f is a module homomorphism

Step 5: Injectivity of f

Step 6: Surjectivity of f

Conclusion

Comments

Add answer

Step 1: Decomposition of arbitrary element of $M$

Step 2: Properties of $p$ on $im p$ and $\ker p$

Step 3: Define the map $f$

Step 4: $f$ is a module homomorphism

Step 5: Injectivity of $f$

Step 6: Surjectivity of $f$