Logic problem: Atiyah-Macdonald 1.11

Question

Proposition 1.11 in Atiyah-Macdonald's "Introduction to commutative algebra" states the following:

"Given an ideal $I$ in a ring $A$ and $p_1, \dots p_n$ prime ideals, then $I \subset \cup_i p_i$ implies that $I\subset p_j$ for some $j\in \{1, \dots, n\}.$ "

This is clearly equivalent to saying that "if $I$ is not a subset of $p_i$ for any $i,$ then $I$ is not a subset of $\cup_i p_i.$"

But they then claim that this last statement is logically equivalent to the following:

"For each $i,$ there exists $x_i \in I$ such that $x_i \notin p_k$ for all $k\neq i.$"

Can anyone help me to see why this statement is logically a consequence of the first? I know this is elementary, but I'm truly stuck and I'm hoping that someone can phrase things in a way that makes them more transparent.

Answer 1

The sentence you are citing, "For each $i$, there exists $x_i\in I$ such that $x_i\notin p_k$ for all $k\neq i$", is not the logical negation of the statement, but the application of induction. That is, they assume that the statement is true for $n-1$ prime ideals and are applying to all subsets of size $n-1$ of $\{p_1,\ldots, p_n\}$, i.e., the set of the form $\{p_j\,|\,j\neq i\}$ with $i$between $1$ and $n$

Answer 2

They do not really claim that "$I$ not in $p_i$ for every $i$ implies $I$ not in $\cup_i p_i$" (1) is logically equivalent to "For each $i,$ there exists $x_i \in I$ such that $x_i \notin p_k$ for all $k\neq i.$"(2). First of all, you haven't specified what $k$, so how could these two be equivalent?

What is trully happening is that in trying to prove (1), they use (2) in an induction argument. In particular, they assume that the statement is true whenever we have $n-1$ prime ideals. So suppose that $I \not\subset \cup_{i=1}^n p_i$. Then certainly for every $k= 1, ..., n$ we have that $I \not\subset \cup_{j \neq k} p_j$. Then apply the induction hypothesis to get (2).