How does the nth Symmetric Power result from the quotient of the group of rank N tensors?

Question

In "Basic Algebra" by Knapp, the author gives this definition of the symmetric power over a vector space E over a field K and tensor algebra T(E) where I is the ideal generated by all $u \otimes v - v \otimes u$ in T(E) where $u,v \in E$:

Sⁿ(E) = Tⁿ(E) / (I $\cap$ Tⁿ(E)

My question is how does this quotient result in Sⁿ(E)?

For example in the case of n = 2, T²(E) would be the tensors of rank 2 and I $\cap$ T²(E) would be the antisymmetric tensors of rank 2 (could be represented as antisymmetric matrices). My understanding is that the quotient algebra generated should consist of the cosets of I $\cap$ T²(E) in T²(E) and I'm not sure how these cosets give us S²(E).

Answer 1

The way I see it is this. $T(E) = \bigoplus_{n \geq 0} T^n(E)$ is a non-commutative graded $K$-algebra. The two sided ideal $I$ generated by $(x \otimes y - y \otimes x)$ for $x,y \in E$ is generated by homogeneous elements so it is a homogeneous ideal of $T(E)$.

We know that the quotient of a graded ring by a homogeneous ideal has a natural grading so that we can write $T(E)/I = \bigoplus_{n \geq 0} S^n(E)$ where we define $S^n(E)$ to be the homogeneous elements of degree $n$ in $T(E)/I$. By the linked question this is indeed the quotient $T^n(E)/(I \cap T^n(E))$. Hence we can realize $S^n(E)$ as the vector space generated by cosets $v_1 \otimes v_2 \otimes \cdots \otimes v_k + I$. Then, $$(v_1 \otimes v_2 \otimes \cdots \otimes v_k + I) - (v_2 \otimes v_1 \otimes \cdots \otimes v_k + I) = (v_1 \otimes v_2 - v_2\otimes v_1) \otimes v_3 \otimes \cdots \otimes v_n + I = I$$ since $(v_1 \otimes v_2 - v_2\otimes v_1) \in I$. As such, $$v_1 \otimes v_2 \otimes \cdots \otimes v_k + I = v_2 \otimes v_1 \otimes \cdots \otimes v_k + I$$ so that $T^n(E)/(J \cap T^n(E))$ has the characteristic property you'd expect from $S^n(E)$, that you can switch around any two tensors.