symmetric power

Question

Let F be a free R-module of rank r. Prove that $\mathbb{S}_R^l(F)$ is free. I know that this is talking about the $l^{th}$-symmetric power but how do I show it free.

Answer 1

It suffices to exhibit a basis of $\Bbb S_R^\ell(F)$. Given a basis $\{f_1,\dots,f_r\}$ of $F$, denote $$ S = \{f_{i_1} \vee \cdots \vee f_{i_{\ell}} : 1 \leq i_1 \leq \cdots \leq i_\ell \leq r\}. $$ It is easy to see that this set spans $\Bbb S_R^\ell(F)$. To see that it is linearly independent, it suffices to show that any choice of $\phi$

We want to show that for any function $\phi:S \to N$ for an $R$-module $N$, there exists a unique homomorphism $\bar \phi:\Bbb S_r^\ell(F)\to N$ with $\bar \phi(x) = \phi(x)$ for all $x \in S$. By the universal property, it is equivalent to show that there exists a unique symmetric mulilinear map $\Phi:F^\ell \to N$ for which $$ \Phi(f_{i_1},\dots,f_{i_\ell}) = \phi(f_{i_1} \otimes \cdots \otimes f_{i_\ell}) $$ for all choices of $1 \leq i_1 \leq \cdots \leq i_\ell \leq r$.

It is easy to see that if such a map exists, then this map is unique (or equivalently that the elements of $S$ span $\Bbb S_R^\ell(F)$). It remains to be shown that any choice of $\Phi(f_{i_1},\dots,f_{i_\ell}) \in N$ can be extended to a linear map (or equivalently that the elements of $S$ are linearly independent). Indeed: if we take $n_{i_1,\dots,i_\ell} := \Phi(f_{i_1},\dots,f_{i_\ell})$, then we can define a multilinear map by $$ \Phi(x_1,\dots,x_n) = \Phi \left(\sum_{j_1=1}^r a_{1j_{1}}f_{j_1}, \dots, \sum_{j_\ell=1}^r a_{\ell j_{\ell}}f_{j_\ell} \right) \\ := \sum_{j_1=1}^r\cdots \sum_{j_\ell=1}^r a_{1j_{1}}\cdots a_{\ell j_{\ell}} \Phi\left(f_{j_1}, \dots, f_{j_\ell} \right). $$ It is clear that this function is symmetric and well defined; it now suffices to argue that this function is linear in each argument. Indeed, we can see that $$ \Phi \left(\sum_{j_1=1}^r a_{1j_{1}}f_{j_1} + k\cdot\sum_{j_1=1}^r b_{1j_{1}}f_{j_1}, \dots, \sum_{j_\ell=1}^r a_{\ell j_{\ell}}f_{j_\ell} \right) = \\ \sum_{j_1=1}^r\cdots \sum_{j_\ell=1}^r (a_{1j_{1}} + kb_{1 j_1})\cdots a_{\ell j_{\ell}} \Phi\left(f_{j_1}, \dots, f_{j_\ell} \right) = \\ \sum_{j_1=1}^r\cdots \sum_{j_\ell=1}^r a_{1j_{1}}\cdots a_{\ell j_{\ell}} \Phi\left(f_{j_1}, \dots, f_{j_\ell} \right) \\ \qquad + k\Phi \sum_{j_1=1}^r\cdots \sum_{j_\ell=1}^r b_{1 j_1}\cdots a_{\ell j_{\ell}} \Phi\left(f_{j_1}, \dots, f_{j_\ell} \right)= \\ \Phi \left(\sum_{j_1=1}^r a_{1j_{1}}f_{j_1}, \dots, \sum_{j_\ell=1}^r a_{\ell j_{\ell}}f_{j_\ell} \right) \\ \quad + k \Phi \left(\sum_{j_1=1}^r b_{1j_{1}}f_{j_1}, \dots, \sum_{j_\ell=1}^r a_{\ell j_{\ell}}f_{j_\ell} \right), $$ as desired.