On proving that the symmetric algebra is isomorphic to the polynomial ring.

Question

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1. What is being asked?

We have

a commutative ring (A);
a free (A)-module (M) of rank (n) with basis ({x_1,\dots ,x_n});
the tensor algebra

[ T_A(M)=\bigoplus_{r\ge 0}T_A^{r}(M),\qquad T_A^{0}(M)=A,\; T_A^{r}(M)=\underbrace{M\otimes_A\cdots\otimes_AM}_{r\text{ copies}} . ]

Inside (T_A(M)) we consider the two‑sided ideal

[ I:=\big\langle\,x\otimes y-y\otimes x\mid x,y\in M\,\big\rangle . ]

The symmetric algebra of (M) is the quotient

[ \operatorname{Sym}_A(M)=T_A(M)/I . ]

The map

[ f\colon M\longrightarrow A[X_1,\dots ,X_n],\qquad x_i\longmapsto X_i ]

is (A)-linear, hence by the universal property of the tensor algebra it extends uniquely to an (A)-algebra homomorphism

[ F\colon T_A(M)\longrightarrow A[X_1,\dots ,X_n]. ]

Because the polynomial ring is commutative we already know (I\subseteq\ker(F)). The problem is to show the reverse inclusion

[ \ker(F)\subseteq I, ]

i.e. that (F) factors through the quotient and the induced map

[ \bar F\colon \operatorname{Sym}_A(M)=T_A(M)/I\;\longrightarrow\;A[X_1,\dots ,X_n] ]

is injective. Equivalently we must prove that (\operatorname{Sym}_A(M)) is (canonically) isomorphic to the polynomial ring.

2. Full solution, step by step

Step 1 – The homomorphism (F) and its surjectivity

The assignment (x_i\mapsto X_i) defines an (A)-linear map (f:M\to A[X_1,\dots ,X_n]). By the universal property of the tensor algebra there is a unique algebra homomorphism

[ F:T_A(M)\longrightarrow A[X_1,\dots ,X_n] ,\qquad F(x_{i_1}\otimes\cdots\otimes x_{i_r})=X_{i_1}\cdots X_{i_r}. ]

Every monomial (X_{i_1}\cdots X_{i_r}) occurs as the image of a pure tensor, therefore (F) is surjective.

Step 2 – The ideal (I) is contained in (\ker(F))

For any (x,y\in M),

[ F(x\otimes y-y\otimes x)=F(x)F(y)-F(y)F(x)=0 ]

because the polynomial ring is commutative. Hence the generators of (I) map to (0), and therefore

[ I\subseteq\ker(F). ]

Thus (F) factors through the quotient and we obtain a well‑defined (A)-algebra homomorphism

[ \bar F:\operatorname{Sym}_A(M)=T_A(M)/I\longrightarrow A[X_1,\dots ,X_n]. ]

(\bar F) is still surjective (it has the same image as (F)).

Step 3 – Constructing a map in the opposite direction

Define an (A)-algebra homomorphism

[ \psi:A[X_1,\dots ,X_n]\longrightarrow \operatorname{Sym}_A(M) ]

by sending each indeterminate to the class of the corresponding basis element:

[ \psi(X_i)=\overline{x_i}\in \operatorname{Sym}_A(M)\qquad (i=1,\dots ,n). ]

Because (\operatorname{Sym}_A(M)) is commutative (the ideal (I) forces all tensors to commute), this assignment respects the relations among the (X_i) (there are none besides commutativity), so (\psi) extends uniquely to an algebra homomorphism.

Step 4 – (\psi) is a two‑sided inverse of (\bar F)

On generators of (\operatorname{Sym}_A(M)).
Let (\overline{x_{i_1}\otimes\cdots\otimes x_{i_r}}) be the class in (\operatorname{Sym}_A(M)) of a pure tensor. Then

[ (\psi\circ\bar F)\bigl(\overline{x_{i_1}\otimes\cdots\otimes x_{i_r}}\bigr) =\psi\bigl(F(x_{i_1}\otimes\cdots\otimes x_{i_r})\bigr) =\psi(X_{i_1}\cdots X_{i_r}) =\overline{x_{i_1}}\cdots\overline{x_{i_r}} =\overline{x_{i_1}\otimes\cdots\otimes x_{i_r}} . ]

Thus (\psi\circ\bar F =\operatorname{id}_{\operatorname{Sym}_A(M)}).

On generators of the polynomial ring.
For each (i),

[ (\bar F\circ\psi)(X_i)=\bar F(\overline{x_i})=F(x_i)=X_i . ]

Since both maps are algebra homomorphisms, they agree on all polynomials, hence

[ \bar F\circ\psi=\operatorname{id}_{A[X_1,\dots ,X_n]} . ]

Consequently (\bar F) is bijective and (\psi) is its inverse.

Step 5 – The kernel of (F) is exactly (I)

Because (\bar F) is injective, (\ker(\bar F)={0}). By definition

[ \ker(\bar F)=\bigl(\ker(F)+I\bigr)/I . ]

Thus (\ker(F)+I=I), i.e. (\ker(F)\subseteq I). Combined with the inclusion proved in Step 2 we obtain

[ \ker(F)=I . ]

Equivalently, the induced map (\bar F) is an isomorphism

[ \boxed{\operatorname{Sym}_A(M)\;\cong\;A[X_1,\dots ,X_n]} . ]

3. Final answer

For a free (A)-module (M) of rank (n) with basis ({x_1,\dots ,x_n}) the symmetric algebra (\operatorname{Sym}_A(M)) is canonically isomorphic to the polynomial ring in (n) variables over (A):

[ \operatorname{Sym}_A(M)\;\xrightarrow{\;\;\cong\;\;}\;A[X_1,\dots ,X_n], \qquad \overline{x_i}\longmapsto X_i . ]

The kernel of the canonical map (F:T_A(M)\to A[X_1,\dots ,X_n]) is exactly the ideal (I=\langle x\otimes y-y\otimes x\mid x,y\in M\rangle).

4. Common mistakes to avoid

Mistake	Why it is wrong	How to correct it
Thinking that surjectivity of (F) already gives an isomorphism.	Surjectivity alone does not control the kernel; one must also prove injectivity.	Build the inverse map (\psi) or show the kernel is exactly (I).
Ignoring the commutativity of the target ring.	The ideal (I) is precisely what forces commutativity in the quotient. Without using it, you cannot define (\psi).	Explicitly use that (\operatorname{Sym}_A(M)=T_A(M)/I) is commutative when sending the indeterminates to (\overline{x_i}).
Working with arbitrary elements of (T_A(M)) without grading.	Elements are sums of tensors of different degrees; the map (F) respects the natural grading, which is crucial for the kernel argument.	Decompose any element into homogeneous components and treat each degree separately (or use the inverse map argument above).
Assuming the polynomial ring is a domain and concluding (\ker(F)=0).	Over a general commutative ring (A) the polynomial ring may have zero‑divisors, so “no non‑zero polynomial can map to zero” is false.	Use the explicit inverse (\psi) or the basis‑monomial argument; no hypothesis on (A) is needed.
Forgetting to mod out by the ideal before defining (\psi).	Directly sending (X_i) to (x_i\in T_A(M)) would not respect the relations (x\otimes y-y\otimes x).	Define (\psi) with target (\operatorname{Sym}_A(M)=T_A(M)/I); then the relations are already satisfied.

Keeping these points in mind ensures a clean and rigorous proof that the symmetric algebra of a free module is the usual polynomial ring.

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