Algebraic tensor product as a quotient space I am trying to get a better understanding of algebraic tensor products. The following definition comes from page 47 of Weidmann's Linear Operators in Hilbert Spaces.

Let $H_1$ and $H_2$ be vector spaces over $\mathbb K$. We denote by $F(H_1,H_2)$ the vector space of formal linear combinations of the pairs $(f,g)$ with $f\in H_1$, $g\in H_2$, i.e.,
  $$
F(H_1,H_2)=\Biggl\{\sum_{j=1}^nc_j(f_j,g_j):c_j\in\mathbb K,f_j\in H_1, g_j\in H_2,j=1,2,\ldots,n;n\in\mathbb N\Biggr\}.
$$
  Let $N$ be the subspace of $F(H_1,H_2)$ spanned by the elements of the form
  $$
\sum_{j=1}^n\sum_{k=1}^ma_jb_k(f_j,g_k)-1\times\biggl(\sum_{j=1}^na_jf_j,\sum_{k=1}^mb_kg_k\biggr).
$$
  The quotient space 
  $$
H_1\otimes H_2=F(H_1,H_2)/N
$$
  is called the algebraic tensor product of $H_1$ and $H_2$.

What happens here is that the elements of $F(H_1,H_2)$ are put into equivalence classes. Roughly speaking, $N$ is the new zero vector. What is the intuition behind $N$? Why do we need to take this particular $N$? Why do we need to take the quotient space $F(H_1,H_2)/N$ instead of just $F(H_1,H_2)$?
How would $N$ and the algebraic tensor product look like if $H_1=H_2=\mathbb R^d$?
Any help is much appreciated!
 A: As you pointed out correctly, taking the quotient with respect to $N$ effectively achieves $[X]=0$ whenever $X\in N$, where $[X]$ denotes the equivalence class of some $X\in F(H_1,H_2)$ in $F(H_1, H_2)/N$. In particular
$$
\left[\sum_{j=1}^n\sum_{k=1}^ma_jb_k(f_j,g_k)-1\cdot\biggl(\sum_{j=1}^na_jf_j,\sum_{k=1}^mb_kg_k\biggr)\right] = 0.
$$
Let us denote by $f\otimes g := [(f,g)]$ the equivalence class of $(f,g)$ in $F(H_1, H_2)/N$, then we can rewrite this equation as
$$
\sum_{j=1}^n\sum_{k=1}^ma_jb_k (f_j\otimes g_k)-1\cdot\left( \left(\sum_{j=1}^na_jf_j\right)\otimes \left(\sum_{k=1}^mb_kg_k\right)\right) = 0
$$
which after some rearranging becomes
$$
\left(\sum_{j=1}^na_jf_j\right)\otimes \left(\sum_{k=1}^mb_kg_k\right) = 
\sum_{j=1}^n\sum_{k=1}^ma_jb_k (f_j\otimes g_k).
$$
Now this equation amounts to saying that the map $H_1\times H_2\to F(H_1, H_2)/N$ given by $(f,g)\mapsto f\otimes g$ is bilinear.
This is the essence of the tensor product $H_1\otimes H_2$: it is spanned by the simple tensors $f\otimes g$ and the only relations are that the product $-\otimes-$ behaves as a bilinear map $H_1\times H_2 \to H_1\otimes H_2$.
