Let $k$ be a field. Assume that you already know that the category $\mathrm{Alg}(k)$ of $k$-algebras (everything here is commutative and unital) has a coproduct $\sqcup$. But you don't know that this actually comes from the tensor product of vector spaces over $k$. You just know the universal property of $A \sqcup B$ (aka $A \otimes_k B$). From this you can deduce:
- $\sqcup$ is commutative and associative up to natural isomorphisms
- $(A/I) \sqcup B \cong (A \sqcup B) / \langle i_A(I) \rangle$
- $A \sqcup k[x_1,\dotsc,x_n] \cong A[x_1,\dotsc,x_n]$
- $A \sqcup -$ commutes with colimits
In particular, we can compute the tensor product of arbitrary algebras using presentations: $k[\{x_i\}]/I \sqcup k[\{y_j\}]/J \cong k[\{x_i\},\{y_j\}]/\langle I,J \rangle$
Question. How can we prove that for every injective homomorphism $\phi : A \to B$ of $k$-algebras the induced homomorphism $\phi \sqcup \mathrm{id} : A \sqcup C \to B \sqcup C$ is also injective for every $k$-algebra $C$?
For example, this is clear when $C$ is a polynomial algebra over $k$. In general, $C$ is free as a module over $k$, but we cannot use the isomorphism $C \cong k^{(I)}$ since this leaves the category of $k$-algebras.
Background: I assist a lecture where the students have just learned what the tensor product of algebras is, without knowing the tensor product of modules. Now they have to believe somehow some of the well-known properties, because they are usually proven with the help of the tensor product of modules. But perhaps we can do it with algebras alone. Since the lecture is about elementary algebraic geometry, you may assume that $k$ is algebraically closed and some basic results about affine varieties (but not about their fiber products ;)).
Appendix: Further properties which follow from the universal property:
1) Let us denote the coproduct inclusions by $i_A : A \to A \sqcup B$ and $i_B : B \to A \sqcup B$. It is easy to see that $\otimes : A \times B \to A \sqcup B, (a,b) \mapsto i_A(a) \cdot i_B(b)$ is $k$-bilinear and that the span of the image generates $A \sqcup B$ (since the image satisfies the same universal property).
2) When $A,B \neq 0$, then we also have $A \otimes_k B \neq 0$. Geometrically: The fiber product of two non-empty schemes is non-empty.
Proof: Since the zero algebra only maps to the zero algebra, we may replace $A,B$ by quotients. In particular, we may assume that $A,B$ are field extensions of $k$. Let $P$ be a transzendence basis of $A/k$ and $Q$ one of $B/k$. Let $M$ be a set containing disjoint copies of $P$ and $Q$ and let $C$ be the algebraic closure of $k(M)$. Then there are maps $A \to C$ and $B \to C$, which induce by the universal property a map $A \sqcup B \to C$. Since $C \neq 0$, we also have $A \sqcup B \neq 0$.