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I'm studying tensor algebra and exterior algebra and having problems with the definition of $\bigwedge(\mathbb{V})$.

Let $\mathbb{V}$ be any vectorial space over a field $K$. Define $T^k\mathbb{V}:=\underbrace{\mathbb{V}\otimes \mathbb{V}\otimes\cdots\otimes\mathbb{V}}_{k-times}$, where $T^0\mathbb{V}=K$.

Define the tensorial algebra $T(\mathbb{V}):=\bigoplus_{k=0}^{\infty}T^k(\mathbb{V})=K\oplus\mathbb{V}\oplus(\mathbb{V}\otimes\mathbb{V})\oplus\cdots$.

Multiplication on $T(\mathbb{V})$ is determinated by the canonical isomorphism $T^k(\mathbb{V})\oplus T^j(\mathbb{V})\rightarrow T^{k+j}(\mathbb{V})$, so $T(\mathbb{V})$ is naturally a graded algebra.

Define exterior algebra (and here is my doubt) $\bigwedge(\mathbb{V})$ over a $K$-vector space $\mathbb{V}$ as the quotient algebra of the tensorial algebra $T(\mathbb{V})$ by the two sided ideal $I$ spanned by the elements of the form $x\otimes x$ with $x\in\mathbb{V}$,

$\bigwedge(\mathbb{V}):=\dfrac{T(\mathbb{V})}{I}$

My question is, first of all how are specifically the elements of $I$? It is a subspace of $T(\mathbb{V})$.

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Multiplication on $T(\mathbb{V})$ is determinated by the canonical isomorphism $T^k(\mathbb{V})\oplus T^j(\mathbb{V})\rightarrow T^{k+j}(\mathbb{V})$, so $T(\mathbb{V})$ is naturally a graded algebra.

I think that direct sum should be a tensor.

Define exterior algebra (and here is my doubt) $\bigwedge(\mathbb{V})$ over a $K$-vector space $\mathbb{V}$ as the quotient algebra of the tensorial algebra $T(\mathbb{V})$ by the two sided ideal $I$ spanned by the elements of the form $x\otimes x$ with $x\in\mathbb{V}$, $$\bigwedge(\mathbb{V}):=\dfrac{T(\mathbb{V})}{I}$$ My question is, first of all how are specifically the elements of $I$? It is a subspace of $T(\mathbb{V})$.

As it's mentioned in its name, the tensor algebra is an algebra, which means it's a vector space together with a product $\otimes: TV\times TV\to TV$. In particular it is a ring (under addition of vectors and tensor product) and that is where the notion of ideal comes from.

By definition, an ideal of a ring is a subset of a ring such that:

  • It's a subgroup under addition,

  • If you multiply any element of the ideal by ANY element of the ring, the result will be again an element of the ideal, for example, whenever you multiply an even integer by ANY integer, the result is again an even integer. Therefore, the even integers form an ideal of $\mathbb{Z}$.

Now, the question is how do I construct the ideal generated by a set? Let's do this for the example at hand. (I'll do the explicit construction, the less hands-on approach consists of intersecting all the ideals containing $S$.) Consider the set $S=\{x\otimes x|x\in V\}\subset V\otimes V$. The ideal generated by $S$ should satisfy the two properties we mentioned above:

  • It must be a subgroup under addition. (1) $0\in S$, (2) We must add inverses, that is, add all the elements of the form $-(x\otimes x)$ for all $x\in V$, (3) it must be closed under addition, so we must allow all possible finite sums of its elements.

With this in mind, what we have is that the ideal generated by $S$ must at least containg the set $$ S\subset S'=\{\sum_{i}(\pm 1)x_i\otimes x_i|\text{the sum is finite and }x_i\in V\}. $$

  • Tensoring any element of the ideal must be an element of the ideal again! The way to solve this is rather obvious, we just add all the possibilities to $S'$! The explicit formula of the ideal then is: $$ \boxed{ I=\{\sum_i r_i\otimes (x_i\otimes x_i)\otimes s_i|\text{the sum is finite and }x_i\in V, r_i,s_i\in TV\}. }$$

I hope I didn't end up making it look more complicated than it is, feel free to ask questions.

The essential idea behind this construction is that quotienting by this ideal is basically imposing the relation $x\otimes x=0$ (in the quotient) or equivalently (try to show it), $$ x\otimes y + y\otimes x\sim0\iff x\otimes y \sim -y\otimes x, $$ which turns out to be very useful in a lot of contexts, for example, notice that the determinant function somehow behaves like this: if you exchange two columns it changes signs.

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