Two Theorems:
Theorem 1 (Atiyah-Macdonald, Proposition 1.11(ii)): Let $A$ be a commutative ring, and let $\mathfrak p_1,\dots,\mathfrak p_n$ be prime ideals in $A$, and let $\mathfrak a$ be an ideal of $A$ such that $\mathfrak a\subset\mathfrak p_1\cup\cdots\cup\mathfrak p_n$. Then $\mathfrak a\subset\mathfrak p_i$ for some $i$.
Proof. Induction on $n$. The case $n=1$ is clear. Suppose $n>1$ and suppose that $\mathfrak a\not\subset\mathfrak p_i$ for $i=1,\dots,n$. By the inductive hypothesis, for each $j=1,\dots,n$ we have $\mathfrak a\not\subset \bigcup_{j\ne i}\mathfrak p_j$, so there is some $x_j\in\mathfrak a\setminus \bigcup_{j\ne i}\mathfrak p_j$ for each $j$. Now let
$$ y=\sum_{i=1}^n\prod_{j\ne i} x_j $$
Certainly $y\in\mathfrak a$. We claim that $y\not\in\mathfrak p_k$ for each $k$. Indeed, suppose $y\in\mathfrak p_k$ for some $k$. Then every term of the sum except the $i=k$ term is a multiple of $x_k$, so is contained in $\mathfrak p_k$. So, after subtracting these terms, we see that
$$ \prod_{j\ne k}x_j\in\mathfrak p_k $$
Therefore, since $\mathfrak p_k$ is a prime ideal, we must have $x_j\in\mathfrak p_k$ for some $j\ne k$, contradicting the definition of $x_j$.
Therefore, $y\in\mathfrak a\setminus\mathfrak p_1\cup\cdots\cup\mathfrak p_n$. $\Box$
Theorem 2 (Lagrange Interpolation): Let $k$ be a field and let $x_1,\dots,x_n,y_1,\dots,y_n\in k$. Then there exists a polynomial $f\in k[x]$ of degree at most $n-1$ such that $f(x_i)=y_i$ for each $i$.
Proof. The polynomial
$$ f=\sum_{i=1}^jy_j\prod_{j\ne i}\frac{(x-x_j)}{(x_i-x_j)} $$ will do. $\Box$
There's a striking similarity between these proofs - they both employ the same sum-product trick to get to their result. I therefore thought it would be easy to come up with a proof of Theorem 2 from Theorem 1, but I didn't manage to get there. Am I missing something obvious?
Note: I am aware that the Lagrange interpolation theorem can be proved using the Chinese remainder theorem - and indeed there are similar similarities between the proofs of those two theorems.
