[proofplan] We define the map $\varphi: M \otimes_R V \to V$ by $m \otimes v \mapsto mv$, where the product $mv$ uses the $K$-module structure of $V$ and the embedding $M \subseteq K$. Surjectivity is immediate: fixing any nonzero $m_0 \in M$, every $v \in V$ equals $\varphi(m_0 \otimes m_0^{-1}v)$. The key to injectivity is that every element of $M \otimes_R V$ is a pure tensor: since all $m_i \in M \subseteq K$ share a common denominator relative to a fixed nonzero $m_0 \in M$, any finite sum $\sum m_i \otimes v_i$ reduces to $m_0 \otimes w$ for some $w \in V$. Injectivity on pure tensors then follows from the fact that $m_0 \neq 0$ acts invertibly on $V$. [/proofplan] [step:Define the map $\varphi: M \otimes_R V \to V$ via the universal property] Define the map \begin{align*} \mu: M \times V &\to V \\ (m, v) &\mapsto mv, \end{align*} where $mv$ denotes the product of $m \in M \subseteq K$ with $v \in V$ using the $K$-module structure of $V$. We verify that $\mu$ is $R$-bilinear: - **Additivity in $M$:** For $m_1, m_2 \in M$ and $v \in V$, $\mu(m_1 + m_2, v) = (m_1 + m_2)v = m_1 v + m_2 v = \mu(m_1, v) + \mu(m_2, v)$. - **Additivity in $V$:** For $m \in M$ and $v_1, v_2 \in V$, $\mu(m, v_1 + v_2) = m(v_1 + v_2) = mv_1 + mv_2 = \mu(m, v_1) + \mu(m, v_2)$. - **$R$-balance:** For $r \in R$, $m \in M$, and $v \in V$, $\mu(rm, v) = (rm)v = r(mv) = m(rv) = \mu(m, rv)$, where the third equality uses $r \in R \subseteq K$ and the commutativity of $K$. By the universal property of the tensor product, there exists a unique $R$-module homomorphism \begin{align*} \varphi: M \otimes_R V \to V, \quad m \otimes v \mapsto mv. \end{align*} [/step] [step:Prove surjectivity of $\varphi$ using a fixed nonzero element of $M$] Since $M \neq 0$, fix $m_0 \in M \setminus \{0\}$. Since $M \subseteq K$ and $m_0 \neq 0$, the element $m_0$ is a nonzero element of the field $K$, hence $m_0^{-1} \in K$ exists. For any $v \in V$, the element $m_0^{-1}v$ is well-defined in $V$ (using the $K$-module structure), and \begin{align*} \varphi(m_0 \otimes m_0^{-1}v) = m_0 \cdot (m_0^{-1}v) = (m_0 m_0^{-1})v = 1_K \cdot v = v. \end{align*} Therefore $\varphi$ is surjective. [guided] Why does $m_0^{-1}v$ make sense? The module $V$ is a $K$-module and $m_0^{-1} \in K$, so $m_0^{-1}v$ is the scalar multiplication of $m_0^{-1}$ with $v$ in the $K$-module $V$. Also, $m_0 \otimes m_0^{-1}v$ is a legitimate element of $M \otimes_R V$ because $m_0 \in M$ and $m_0^{-1}v \in V$. The computation uses the associativity of scalar multiplication in $V$: $m_0 \cdot (m_0^{-1} v) = (m_0 m_0^{-1}) v = 1_K v = v$. This shows every element of $V$ is in the image of $\varphi$, so $\varphi$ is surjective. [/guided] [/step] [step:Show every element of $M \otimes_R V$ is a pure tensor by extracting a common denominator] We claim that every element of $M \otimes_R V$ can be written as a single pure tensor $m_0 \otimes w$ for some $w \in V$, where $m_0 \in M$ is the fixed nonzero element from the previous step. Let $\sum_{i=1}^\ell m_i \otimes v_i$ be an arbitrary element of $M \otimes_R V$, with $m_i \in M$ and $v_i \in V$. Each $m_i \in M \subseteq K$, so $m_i / m_0 \in K$ (this quotient is well-defined since $m_0 \neq 0$ in $K$). We compute in $M \otimes_R V$: \begin{align*} \sum_{i=1}^\ell m_i \otimes v_i &= \sum_{i=1}^\ell \left(\frac{m_i}{m_0} \cdot m_0\right) \otimes v_i. \end{align*} Here $\frac{m_i}{m_0} \in K$, but we need the scalar in the $R$-balance relation to lie in $R$. Since $V$ is a $K$-module, we can instead move the $K$-scalar to the other side: \begin{align*} m_i \otimes v_i = \left(\frac{m_i}{m_0} \cdot m_0\right) \otimes v_i = m_0 \otimes \left(\frac{m_i}{m_0} \cdot v_i\right). \end{align*} The last equality requires justification. We use the $K$-module structure of $V$: the $R$-balance relation in the tensor product gives $rm_0 \otimes v_i = m_0 \otimes rv_i$ for $r \in R$. More generally, since $V$ is a $K$-module, we establish the identity $m_i \otimes v_i = m_0 \otimes (m_i / m_0) v_i$ as follows. Write $m_i / m_0 = a / b$ where $a, b \in R$ and $b \neq 0$ (this is possible since $m_i / m_0 \in K = \operatorname{Frac}(R)$), so $m_i = (a/b) m_0$, i.e., $b m_i = a m_0$. Then: \begin{align*} b(m_i \otimes v_i) &= bm_i \otimes v_i = am_0 \otimes v_i = m_0 \otimes av_i, \\ b\left(m_0 \otimes \frac{m_i}{m_0} v_i\right) &= m_0 \otimes b \cdot \frac{m_i}{m_0} v_i = m_0 \otimes \frac{bm_i}{m_0} v_i = m_0 \otimes \frac{am_0}{m_0} v_i = m_0 \otimes av_i. \end{align*} So $b(m_i \otimes v_i) = b(m_0 \otimes \frac{m_i}{m_0}v_i)$, i.e., $b(m_i \otimes v_i - m_0 \otimes \frac{m_i}{m_0}v_i) = 0$. Since $V$ is a $K$-module, the tensor product $M \otimes_R V$ is also a $K$-module (via the action on $V$), and $b \neq 0$ is a unit of $K$, so $m_i \otimes v_i = m_0 \otimes \frac{m_i}{m_0}v_i$. Summing over $i$: \begin{align*} \sum_{i=1}^\ell m_i \otimes v_i = \sum_{i=1}^\ell m_0 \otimes \frac{m_i}{m_0} v_i = m_0 \otimes \underbrace{\sum_{i=1}^\ell \frac{m_i}{m_0} v_i}_{=:\, w \,\in\, V}. \end{align*} [guided] The goal is to show that every element of $M \otimes_R V$ is a pure tensor, which is a special property coming from the embedding $M \subseteq K$ and the $K$-module structure of $V$. In a typical tensor product, elements are sums of pure tensors and cannot be simplified. Here, the fractions $m_i / m_0 \in K$ allow us to "transfer" all the $m_i$ to $m_0$ at the cost of modifying $v_i$. The key subtlety is justifying the identity $m_i \otimes v_i = m_0 \otimes (m_i / m_0) v_i$ in $M \otimes_R V$. The $R$-balance relation gives $rm \otimes v = m \otimes rv$ for $r \in R$, but here $m_i / m_0 \in K$ may not lie in $R$. We cannot directly use the $R$-balance relation with a $K$-scalar. The way around this is to note that $M \otimes_R V$ carries a natural $K$-module structure (inherited from the $K$-module structure of $V$). Concretely, $K$ acts on $M \otimes_R V$ via $k \cdot (m \otimes v) = m \otimes (kv)$, and this is well-defined because the $K$-action on $V$ is compatible with the $R$-action. In this $K$-module, every nonzero $b \in R$ acts invertibly (since $b^{-1} \in K$ provides the inverse). So from $b(m_i \otimes v_i) = b(m_0 \otimes \frac{m_i}{m_0}v_i)$ and the fact that $b$ acts as a unit, we deduce $m_i \otimes v_i = m_0 \otimes \frac{m_i}{m_0}v_i$. Alternatively, one can express this directly: write $m_i / m_0 = a/b$ with $a, b \in R$, $b \neq 0$, so $bm_i = am_0$. Then $m_i \otimes v_i = m_i \otimes (b \cdot b^{-1}v_i) = bm_i \otimes b^{-1}v_i = am_0 \otimes b^{-1}v_i = m_0 \otimes a b^{-1} v_i = m_0 \otimes \frac{m_i}{m_0}v_i$. (The step $m_i \otimes bw = bm_i \otimes w$ uses the $R$-balance with $r = b \in R$, and $b^{-1}v_i \in V$ since $V$ is a $K$-module.) Either way, every finite sum collapses to a single pure tensor $m_0 \otimes w$. [/guided] [/step] [step:Conclude injectivity of $\varphi$ from purity of all tensors] Let $x \in \ker(\varphi)$. By the previous step, $x = m_0 \otimes w$ for some $w \in V$. Then \begin{align*} 0 = \varphi(x) = \varphi(m_0 \otimes w) = m_0 w. \end{align*} Since $m_0 \neq 0$ in $K$ and $V$ is a $K$-module, multiplication by $m_0$ is an injective $R$-linear map $V \to V$ (its inverse is multiplication by $m_0^{-1} \in K$). Therefore $m_0 w = 0$ implies $w = 0$, and hence $x = m_0 \otimes 0 = 0$. Since $\ker(\varphi) = 0$, the map $\varphi$ is injective. Combined with surjectivity from the second step, $\varphi: M \otimes_R V \xrightarrow{\sim} V$ is an $R$-module isomorphism. [guided] This is where the purity result pays off. If we only knew that $\ker(\varphi)$ contained some finite sum $\sum m_i \otimes v_i$, we could not directly conclude it is zero. But since every element is a pure tensor $m_0 \otimes w$, the kernel computation reduces to a single equation $m_0 w = 0$ in the $K$-module $V$. The decisive property is that $m_0$ is a nonzero element of the field $K$, hence acts invertibly on any $K$-module. Multiplication by $m_0$ on $V$ has the two-sided inverse "multiplication by $m_0^{-1}$", so $m_0 w = 0$ forces $w = m_0^{-1}(m_0 w) = m_0^{-1} \cdot 0 = 0$. Therefore $\varphi$ is both surjective and injective, hence an isomorphism of $R$-modules $M \otimes_R V \cong V$. [/guided] [/step]