integrality under base change (Vakil 7.3.N) Vakil 7.3.N goes as follows:

Suppose $B \to A$ is integral. Show that for any ring homomorphism $B \to C$, the induced map $C \to A \otimes_B C$ is integral (Hint: We wish to show that any $\sum_{i=1}^n a_i \otimes c_i \in A \otimes_B C$ is integral over $B$, and then Exercise 7.2.D)

Now Exercise 7.2.D just says the elements of A integral over B form a subalgebra of A, so I'm not sure if it's truly needed here. Additionally, I don't think we need to work over arbitary sums $\sum_{i=1}^n a_i \otimes c_i$ but instead pure tensors. That is, if $a \otimes c \in A \otimes_B C$ then as $a \in A$ we can find $n > 0$ and $b_i \in B$ such that $a^n + b_{n-1}a^{n-1} + \dots + b_0 = 0$. Then
$$
0 = 0 \otimes c = (a^n + b_{n-1}a^{n-1} + \dots + b_0) \otimes c = (a^n \otimes c) + b_{n-1}(a^{n-1} \otimes c) + \dots + b_0(1 \otimes 1)
$$
The problem is this doesnt agree with the ring structure on $A \otimes_B C$ in terms of $a \otimes c$ being a root of a polynomial in $A \otimes_B C$ — we would need
$$
0 =(a^n \otimes c^n) + b_{n-1}(a^{n-1} \otimes c^{n-1}) + \dots + b_0(1 \otimes 1)
$$
and I'm not sure why the latter is true
 A: Just to answer my own question here from @CraniumClamp 's  hint, I should note that I originally wasn't being careful to show that $A \otimes_BC$ is integral over $C$ instead of $B$ — keeping track of maps here proved to be helpful by going back to the definitions of our tensor product of algebras.
If we label the maps $\phi : B \to A$ and $\psi : B \to C$, then $A \otimes_B C$ has the structure of a $B$-algebra from the commutative diagram
$$
\require{AMScd}
\begin{CD}
                  B @>{\phi}>> A \\
                  @V{\psi}VV @VV{\textrm{id}_A \otimes 1_C}V \\
                  C @>>{1_A \otimes \textrm{id}_C}> A \otimes_B C
                  \end{CD}
$$
In particular, for every $ b \in B $ we have that $ \phi(b) \otimes 1 = 1 \otimes \psi(b) $. Now let $ a \otimes c \in A \otimes_B C$ be a pure tensor; as $ a $ is integral over $ B $, there exist $ n > 0 $ and $ b_i \in B $ such that $ a^n + \phi(b_{n-1})a^{n- 1} + \dots + \phi(b_0)=  0 $. Tensoring with $ c^n $, this tells us
$$
                  \begin{align}
                  0 = 0 \otimes c^n &= (a^n \otimes c^n) + (\phi(b_{n-1}) a^{n-1} \otimes c^n ) + \dots + (\phi(b_0) \otimes c^n)
                  \\&= (a^n \otimes c^n) + (\phi(b_{n-1}) \otimes c)(a^{n-1} \otimes c^{n-1}) + \dots + (\phi(b_0) \otimes c^n)(1 \otimes 1)
                  \\&= ( a^n \otimes c^n) + (1 \otimes \psi(b_{n-1})c )( a^{n-1} \otimes c^{n-1} ) + \dots + (1 \otimes \psi(b_0)c^n)(1 \otimes 1)
                  \end{align}
                  $$
Since each $1 \otimes \psi(b_i)c^{n-i} $ lies in the image of the natural map $ C \to A \otimes_B C $, we have that $ a \otimes c $ is integral over $C$
