On proving that the symmetric algebra is isomorphic to the polynomial ring.
问题内容
Let $A$ be a commutative ring and $M$ an $A$-module. Suppose $M$ is free of rank $n$ with basis $\{ x_1, \ldots , x_n \}$. The $A$-module homomorphism $f \colon M \to A[X_1, \ldots, X_n]$ given by $x_i \mapsto X_i$ induces a unique $A$-algebra homomorphism $F \colon \operatorname{TS}_A(M) \to A[X_1, \ldots, X_n]$ by the universal property of $A$-algebras, with $\operatorname{Ker}(F) = I$ where $I$ is the ideal generated by all elements $x \otimes y - y \otimes x \in T^2_A(M)$. Hence, $\operatorname{TS}_A(M)/I \cong \operatorname{Im}(F)$, i.e $\operatorname{Sym}(M) \cong A[X_1, \ldots X_n]$ since $F$ is surjective.
Its obvious that $I \subset \operatorname{Ker}(F)$ but I cant find an elegant argument for $ \operatorname{Ker}(F) \subset I$. Any ideas?
EDIT: We define the tensor algebra of $M$ over $A$, $\operatorname{TS}_A(M)$, as the direct sum $$\operatorname{TS}_A(M)=\bigoplus_{n\geq 0}T^n_A(M)$$ where $T^{0}_A = A$ and $T^i_A(M)=\underbrace{M\otimes_A \ldots \otimes _AM}_{i \ \text{copies}}$ for $n > 0$.
回答 (2)
What you want to prove is equivalent to showing that the map defined on the quotient $TS_A(M)/I$ is injective. As this algebra is commutative, a general coset can be written in the form $\sum a_{i_1...i_n}x_1^{i_1}\otimes...\otimes x_n^{i_n}+I$. If it is mapped to the zero polynomial then it means the coefficients $a_{i_1...i_n}$ are all zeros, which clearly implies this is the zero element.
A few small comments.
- Mark's way to finish is the standard one in terms of coordinates.
- You can do a coordinate-free proof in a couple of ways. One is to show $\mathrm{TS}_A(M)/I$ satisfies the same universal property as $A[X_1, \ldots, X_n]$ in the category of commutative $A$-algebras and conclude they're naturally isomorphic by Yoneda's lemma.
- Alternatively, if $n! \neq 0$ in $A$, you can split the quotient map by applying the averaging operator $\frac{1}{n!} \sum_{\sigma \in S_n} \sigma$ where $S_n$ is the symmetric group permuting the $x_i$'s. This operator sends the full $n$-th tensor power onto the subspace of symmetric tensors, which is directly isomorphic to degree $n$ polynomials.
- If $n! = 0$ in $A$, then the above can be replaced by a slightly more involved construction. The $n$th upper symmetric power is the quotient you're working with. The $n$th lower symmetric power is the space of $S_n$-invariant tensors. The previous point shows they're isomorphic when $n! \neq 0$, indeed even as $\mathrm{GL}(M)$-modules, but this generally fails when $n! = 0$. The fix is instead that there's a natural non-degenerate pairing between the upper power of $M$ and the lower power of the dual $M^*$. This gives a natural isomorphism between the upper power of $M$ and the dual of the lower power of $M^*$. The latter is $A$-linearly isomorphic to the $n$th degree piece of the polynomial ring, so stringing these together gives the result afresh.