Why does the tensor product of rings correspond to the product of their spectra? I've been grinding away at understanding algebraic geometry on my own for a long while and things are starting to fall into place (a month before I start my first formal class in it). I answered some of my own questions in typing this up, but one thing is still nagging at me.
Start with the standard setup: let $k$ be algebraically closed and $A$ and $B$ f.g. $k$-algebras. Here is the knowledge we have coming in:


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*$\text{Spec}(A\times B)=\text{Spec}(A)\sqcup\text{Spec}(B)$

*$A\otimes_k B$ is the quotient of $A\times B$ by the ideal generated by... (not gonna type out the whole definition of tensor)

*For a ring $R$ with ideal $I$, the primes of $R/I$ are in correspondence with the ideals of $R$ containing $I$ (Correspondence theorem)


By the correspondence theorem it ought to be the case that $\text{Spec}(A\otimes_k B)$ be smaller than $\text{Spec}(A\times B)$, when in fact we have $\text{Spec}(A\otimes_k B)=\text{Spec}(A)\times\text{Spec}(B)$. I understand the categorical explanation in that $\text{Spec}$ is a contravariant functor and thus must send limits to colimits, but I don't understand why this disagrees with the intuition I have presented above. My best guess is that it has something to do with the tensor being in the category of $k$-algebras rather than commutative rings, but I'm failing at putting the pieces together specifically. Any insights would be greatly appreciated!
 A: Let $C$ be a commutative ring and $A$ and $B$ be $C$-algebras. You have $C = k$, but I’ll be a bit more general.
As Tobias Kildetoft already said, $A \otimes_C B$ is not a quotient of $A × B$, but instead a quotient of the very much larger structure 
$$F(A×B) = \bigoplus_{(a,b) ∈ A × B} C·(a,b)$$
as a $C$-module – not as a ring. That module is free $C$-module on the set $A×B$ (hence the notation ‘$F(…)$’).
It then just turns out that the tensor product $A \otimes_C B$ of this structure can also be endowed with a ring structure, turning it to a $C$-algebra. The structure inherently comes from the universal property of tensor products as $C$-module, since multiplication in $C$-algebras are $C$-bilinear maps.
So you can’t easily compare the ideals of $A \otimes_C B$ with the ideals of $A × B$, that’s why your intuition fails.
Another thing: The high-level reason that $\operatorname{Spec} \colon \mathrm{C\,Rings}^\mathrm{op} → \mathrm{Schemes}$ turns the tensor product into the fibre product is not that it “is a contravariant functor”, but that it is, as such, a right-adjoint functor to the global section $Γ\colon \mathrm{Schemes} → \mathrm{C\,Rings}^\mathrm{op}$ and category theory tells us that right adjoints always preserve limits. Since the tensor product of $C$-algebras is the coproduct in $\mathrm{C\, Rings}$, it’s the product in the opposite category, so it is preserved by $\mathrm{Spec}$.
To help your intuition, maybe think of the very intuitive statements
$$\mathbb A^m_C × \mathbb A^n_C = \mathbb A^{m+n}_C \quad\text{and}\quad C[X_1,…,X_m] \otimes_C C[Y_1,…,Y_n] = C[X_1,…,X_m,Y_1,…Y_n].$$ 
