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The Tensor Product and Its Universal Property

We build $M \otimes N$ and prove it does the one job we asked of it: it linearizes bilinear maps. The universal property is the real definition; the construction is just an existence proof. We compute with pure tensors, learn the relations $m \otimes n$ must obey, and meet extension of scalars as the first big payoff.

The universal property is the definition

The tensor product of $R$-modules $M$ and $N$ is a module $M \otimes_R N$ together with a bilinear map $\otimes: M \times N \to M \otimes_R N$, $(m,n) \mapsto m \otimes n$, satisfying the universal property: for every $R$-module $P$ and every bilinear map $b: M \times N \to P$, there is a unique linear map $\bar b: M \otimes_R N \to P$ with $\bar b(m \otimes n) = b(m,n)$. In a slogan: bilinear maps out of $M \times N$ = linear maps out of $M \otimes N$. That bijection is the whole reason the object exists.

Building it, and the relations you must respect

Existence is a quotient construction. Start with the gigantic free module $F$ on the set of symbols $\{(m,n)\}$, then quotient by the submodule generated by all bilinearity relations. The classes of the symbols are written $m \otimes n$ and called pure (or simple) tensors. Crucially, not every element of $M \otimes N$ is a pure tensor — the general element is a finite sum $\sum_i m_i \otimes n_i$, and rewriting it can collapse a long sum to a short one.

  1. Additivity in slot 1: $(m + m') \otimes n = m \otimes n + m' \otimes n$.
  2. Additivity in slot 2: $m \otimes (n + n') = m \otimes n + m \otimes n'$.
  3. Scalars slide across: $(rm) \otimes n = r(m \otimes n) = m \otimes (rn)$, for $r \in R$.
  4. Consequently $m \otimes 0 = 0 = 0 \otimes n$, and you may freely move ring elements between the two factors.
Basis of a tensor product (over a field k).  If (e_i) is a basis of M
and (f_j) a basis of N, then (e_i (x) f_j) is a basis of M (x) N.
Hence   dim(M (x) N) = (dim M)(dim N).        <- the mn from guide 1!

Worked relation in R^2 (x) R^2.  Let v = e_1 + e_2, w = e_1 - e_2.
  v (x) w = (e_1 + e_2) (x) (e_1 - e_2)
          = e_1(x)e_1 - e_1(x)e_2 + e_2(x)e_1 - e_2(x)e_2.
So the single pure tensor v (x) w expands to 4 basis tensors.

Conversely, e_1(x)e_1 + e_2(x)e_2 is NOT a pure tensor: no a(x)b equals it
(its 'matrix' [1,0;0,1] has rank 2, while a pure tensor a(x)b has rank 1).
Pure tensors are the rank-$\le 1$ ‘matrices’; a sum of pure tensors of higher rank cannot be written as a single $a \otimes b$. The basis count recovers the $mn$ from guide 1.

First payoff: extension of scalars

If $R \to S$ is a ring map and $M$ is an $R$-module, then $S \otimes_R M$ is naturally an $S$-module — you have performed extension of scalars, promoting $M$ to live over the bigger ring $S$. Complexifying a real vector space is exactly $\mathbb{C} \otimes_{\mathbb{R}} V$. This single move underlies base change in geometry, representation theory over different fields, and the Kronecker product of matrices.