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First I have defined the exterior algebra of a module $M$ as the quotient $T(M)/A(M)$ where $T(M)$ is the tensor algebra of $M$ and $A(M)$ is the ideal generated by all elements of the form $m\otimes m$ for $m\in M$.

The exterior algebra is graded, with k'th homogeneous component $\bigwedge ^k(M)=T^k(M)/A^k(M)$ where $A^k(M)=A(M)\cap T^k(M)$. Elements of $A^k(M)$ are finite sums of elements of the form $m_1\otimes ...\otimes m_{i-1}\otimes m\otimes m\otimes m_{i+2}\otimes ...\otimes m_k$ for some $1\leq i<k$ (right?)

Now I want to show that the k'th exterior power of M is equal to $T^k(M)/J^k(M)$, where $J^k(M)$ is the submodule of $T^k(M)$ generated by all elements of the form $m_1\otimes ...\otimes m_k$ where $m_i=m_j$ for some $i\neq j$.

The proof (Dummit & Foote) says this:

The k-tensors $A^k(M)$ inn the ideal $A(M)$ are finite sums of elements of the form $m_1\otimes ...\otimes m_{i-1}\otimes m\otimes m\otimes m_{i+2}\otimes ...\otimes m_k$ for some $1\leq i<k$, which is a k-tensor with two equal entries, so is an element in $J^k(M)$, i.e. $A^k(M)\subseteq J^k(M)$. I understand up to here. What I am struggling to see is how this implies that $T^k(M)/A^k(M)\subseteq T^k(M)/J^k(M)$. I feel like I've misinterpreted something or am missing something obvious.

Many thanks!

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  • $\begingroup$ It seems to me that if you have a natural inclusion there then it really ought to be an isomorphism/equality. Don't have the book on hand, unfortunately. $\endgroup$ – Hoot Dec 11 '14 at 17:00
  • $\begingroup$ As always, no calcualation is necessary when you work with the universal property instead. Then it's immediate from the statement that the symmetric group is generated by adjacent transpositions. $\endgroup$ – Martin Brandenburg Dec 11 '14 at 17:15
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$B \subset C$ doesn't imply $A/B \subset A/C$. It implies $A/B$ surjects to $A/C$. The argument in Dummit & Foote must not be finished.

To show what you want (which presumably comes later in Dummit & Foote):

The key is to use $(x+y) \otimes (x+y) = x \otimes y + y \otimes x + x\otimes x + y\otimes x$. This shows that $x\otimes y + y \otimes x$ is in $A(M)$, which will allow you to "move" (modulo $A(M)$) the two $m$'s next to each other.

For example, $x \otimes y \otimes x = - x \otimes x \otimes y + x \otimes (stuff)$

Where $stuff = (x+y)\otimes(x+y) - x\otimes x - y\otimes y \in A(M)$.

Thus $x \otimes y \otimes x \in A(M)$.

This reasoning in general shows $A(M) = J(M)$.

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