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Let $V$ be an $R$-module. Traditionally the symmetric algebra $S(V)$ is defined as a quotient of the tensor algebra $T(V)$, by the ideal generated by all $a\otimes b-b\otimes a$.

Can $S^n(V)$ also be viewed as a submodule of $T(V)$?

This is clearly true if $n!$ is invertible in $R$, by mapping $a_1\cdots a_n$ to the symmetrized tensor $1/n!\sum a_{\pi(1)}\otimes\cdots\otimes a_{\pi(n)}$. However, I'm interested in a universal construction, valid over all rings.

In particular, without the $1/n!$, the above map would not necessarily be injective: it maps $a^n$ to $n! a^n$ which could be $0$ if the characteristic of $R$ divides $n!$.

Note the related question MO69940 asking about splitting the natural map $T(V)\to S(V)$. I ask for something weaker, namely just an injective module map that is natural in $V$.

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  • $\begingroup$ If you want an injective natural homomorphism $\phi_V : S^n(V) \to T^n(V)$ (not $\to T(V)$), then I'm pretty sure that the answer is "no" whenever $n > 1$. Indeed, let $R$ be an infinite field whose characteristic divides $n$. Let $a \in R$ be the unique constant such that the $1$-dimensional $R$-module $R$ satisfies $\phi_R (1^n) = a \cdot 1^{\otimes n}$ in the $1$-dimensional $R$-module $R^{\otimes n}$. Then, the functoriality of $\phi_V$ shows that every $R$-module $V$ and every $v \in V$ satisfy $\phi_V (v^n) = a \cdot v^{\otimes n}$. Now, let $V = R^2$, and let $x$ and $y$ be ... $\endgroup$ Aug 7, 2018 at 21:29
  • $\begingroup$ ... the two elements of the standard basis of $V$. Then, the formula we just proved can be applied to $v = x+ty$ for every $t \in R$. We thus obtain $\phi_V ((x+ty)^n) = a \cdot (x+ty)^{\otimes n}$. This holds as a polynomial identity in $t$ (here, we are using the infiniteness of the field $R$); thus, we can extract the coefficients of $t^1$. We thus obtain $\phi_V (nx^{n-1}y) = a \cdot s$, where $s$ is the sum of all pure $n$-tensors which have one tensor factor equal to $y$ and the remaining $n-1$ tensor factors equal to $x$. The left hand side ... $\endgroup$ Aug 7, 2018 at 21:31
  • $\begingroup$ ... of this equality is $0$ (since $n$ divides $\operatorname{char} R$); thus, the right hand side is $0$, too. Hence, $a = 0$ (since $s \neq 0$). But this yields that the map $\phi_R$ is not injective, a contradiction. $\endgroup$ Aug 7, 2018 at 21:31
  • $\begingroup$ Note that this leaves the case when $\operatorname{char} R$ is divisible by $n!$ but not by $n$ unanswered. Parts of this case can be solved by extracting other coefficients (see math.stackexchange.com/questions/1461189/… for a useful fact). $\endgroup$ Aug 7, 2018 at 21:33
  • $\begingroup$ If you really want to send $S^n(V)$ into $T(V)$ instead of $T^n(V)$, the answer is still "no": To see this, it suffices to show that every natural homomorphism $\phi_V : S^n(V) \to T^m(V)$ with $m \neq n$ must be $0$. To prove this, let $R$ be any infinite field, and let $a\in R$ be the unique constant such that the $1$-dimensional $R$-module $R$ satisfies $\phi_R (1^n) = a \cdot 1^m$. Then, the functoriality of $\phi_V$ shows that every $R$-module $V$ and every $v \in V$ satisfy $\phi_V(v^n) = a \cdot v^{\otimes m}$. Now, let $V = R$, and apply this to $v = t$ for a varying $t \in R$. $\endgroup$ Aug 7, 2018 at 21:37

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