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Let $P$ be a projective module and $P=P_1+N$, where $P_1$ is a direct summand of $P$ and $N$ is a submodule. Show that there is $P_2\subseteq N$ such that $P=P_1\oplus P_2$.

I know that there is a submodule $P'$ of $P$ such that $P=P_1\oplus P'$. I wanted to consider the projection from this to $P_1$ and use the definition of being projective. But I would also need a map from $P=P_1+N$ to $P_1$ and I don't know how to get a well defined map there because it is not a direct sum.

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  • $\begingroup$ Let $i_1\colon P_1 \to P_1 + N$ and $i_2 \colon P' \to P_1 + N$ be the canonical inclusions. This factors out to a unique mapping $(i_1, i_2)\colon P_1 \oplus P' \to P_1 + N$. But the identity mapping also makes everything commute, so $(i_1, i_2) = \text{Id}$. I'm still not sure how to go on from here.. Maybe given that $P_1 \cap P' = \{0\}$, this is also true of their images. This already proves that $\text{Im}(P') \subseteq N$, I think $\endgroup$ Nov 13, 2017 at 0:25

2 Answers 2

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Since $P_1$ is a direct summand, there exists $P_2$ such that $P=P_1\oplus P_2$. Consider $f:P\rightarrow P_2$ such that for every $x\in P$, write $x=x_1+x_2, x_1\in P_1,x_2\in P_2$, $f(x)=x_2$. Let $x\in P_2$, we can write $x=x_1'+n, x_1'\in P_1, n\in N$, we have $x=f(x)=f(x_1'+n)=f(n)$. This implies that the restriction $g$ of $f$ to $N$ is surjective, there exists $h:P\rightarrow N$ such that $f=g\circ h$; $P=P_1\oplus h(P_2)$.

Let $x\in P$, write $x=x_1+x_2, x_1\in P_1, x_2\in P_2$, $f(x-h(x_2))=f(x)-f(h(x_2))=x_2-x_2=0$ since $x_2=f(x_2)=f(h(x_2))$. This implies that $x-h(x_2)\in P_1$, we can write $x=x-h(x_2)+h(x_2)$ and we deduce that $P=P_1+h(P_2)$.

Let $x\in P_1\cap h(P_2)$, we can write $x=h(x_2), x_2\in P_2$, $f(x)=f(h(x_2))=f(x_2)=x_2=0$ because $x\in P_1$. This implies that $x_2=x=0$.

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The condition $P = P_1 + N$ implies that the natural map $N \to P/P_1$ is surjective. As $P$ is projective this means the quotient map $P \to P/P_1$ factors through $N$, so there is a homomorphism $P \to N$ such that the composition $P \to N \to P/P_1$ is the quotient map. Since the quotient map gives an isomorphism $P' \simeq P/P_1$ we can define $P_2 \subseteq N$ to be the image of $P'$ under the map $P \to N$. Now $P_1$ and $P_2$ don't intersect because the quotient map was injective on $P'$ and $P = P_1 + P_2$ because $P'$ surjects onto $P/P_1$. Thus $P = P_1 \oplus P_2$.

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  • $\begingroup$ Nice and clean. Great proof. $\endgroup$
    – user494247
    Nov 15, 2017 at 4:35

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