91Ó°ÊÓ

First, we need to prove that \(N+P\) is a submodule of \(M\). In order for this to be a submodule, it must be nonempty, closed under addition, and closed under scalar multiplication. (i) Nonempty: Since \(N, P\) are submodules of \(M\), they both contain the zero element of \(M\). This means that \(0 + 0 = 0 \in N+P\), so \(N+P\) is nonempty. (ii) Closed under addition: Let \(n_1 + p_1, n_2 + p_2 \in N+P\), where \(n_1, n_2 \in N\), and \(p_1, p_2 \in P\). The sum of these elements is \((n_1 + p_1) + (n_2 + p_2) = (n_1 + n_2) + (p_1 + p_2)\). Since \(N\) and \(P\) are submodules and closed under addition, \(n_1 + n_2 \in N\) and \(p_1 + p_2 \in P\). Therefore, \((n_1 + n_2) + (p_1 + p_2) \in N+P\), so \(N+P\) is closed under addition. (iii) Closed under scalar multiplication: Let \(r \in R\) and \(n + p \in N+P\), where \(n \in N\) and \(p \in P\). Since \(N\) and \(P\) are submodules, \(rn \in N\) and \(rp \in P\). Thus, their sum \(rn + rp = r(n + p) \in N+P\), so \(N+P\) is closed under scalar multiplication. Since \(N+P\) satisfies these three criteria, it is a submodule of \(M\).

We define the map \(\phi: N \rightarrow N+P/P\) by \(\phi(n) = n + P\) for each \(n \in N\). We will show that this map is an \(R\)-module isomorphism.

To show that \(\phi\) is well-defined, we need to verify that each element in \(N / N \cap P\) maps to exactly one element in \(N + P / P\). Let \(n + N\cap P \in N / N \cap P\). Since \(n \in N\), we know that \(\phi(n) = n + P \in N+P/P\). Therefore, \(\phi\) is well-defined.

We need to show that \(\phi\) preserves addition and scalar multiplication to be an \(R\)-module homomorphism. (i) Addition: Let \(n_1, n_2 \in N\). Then $$\phi(n_1 + n_2) = (n_1 + n_2) + P = (n_1 + P) + (n_2 + P) = \phi(n_1) + \phi(n_2).$$ (ii) Scalar multiplication: Let \(n \in N\) and \(r \in R\). Then $$\phi(rn) = rn + P = r(n + P) = r\phi(n).$$ Since \(\phi\) preserves both addition and scalar multiplication, it is an \(R\)-module homomorphism.

To show that \(\phi\) is surjective, we need to prove that for every \(x + P \in N + P /P\), there exists an \(n \in N\) such that \(\phi(n)=x+P\). For any \(x+P \in N+P/P\), we have \(x=n+p\) for some \(n \in N\) and \(p \in P\). Thus, \(\phi(n) = n+P = x + P\), which implies that \(\phi\) is surjective.

To show that \(\phi\) is injective, we have to prove that if \(\phi(n_1) = \phi(n_2)\) for some \(n_1, n_2 \in N\), then \(n_1 = n_2\). Suppose \(\phi(n_1) = \phi(n_2)\). Then \(n_1 + P = n_2 + P \Rightarrow n_1 - n_2 \in P\). However, since \(n_1, n_2 \in N\), we have \(n_1 - n_2 \in N\). Therefore, \(n_1 - n_2 \in N \cap P\), and since \(N\cap P\) is a submodule, it contains the zero element of \(M\). So, \(n_1 - n_2 = 0 \Rightarrow n_1 = n_2\). Thus, \(\phi\) is injective.

We have shown that \(\phi\) is a well-defined, homomorphic, surjective, and injective map between the \(R\)-modules \(N\) and \(N+P/P\), making \(\phi\) an \(R\)-module isomorphism. This proves the First Noether Isomorphism Theorem, which states that there is a natural \(R\)-module isomorphism between \(N/N\cap P\) and \(N+P/P\).