Proposition 2

Let $\mathbf{A}$ be a ring, $a$ a non-empty set of $\mathbf{A}$ which is closed under addition and multiplication and $(p_i)_{i \in I}$ a non-empty finite family of ideals of $\mathbf{A}$. Suppose that $a$ is contained in the union of the $p_i$ and that at most two of the $p_i$ are not prime. Then $a$ is contained in one of the $p_i$.

We argue by induction on $n = \mathrm{Card}(I)$; the proposition is trivial if $n = 1$. Suppose that $n \geq 2$; if there exists an index $j$ such that $a \cap p_j \subset \bigcup_{i \neq j} p_i$, the set $a$, which is the union of the $a \cap p_i$ where $i \in I$, is contained in $\bigcup_{i \neq j} p_i$ and hence in one of the $p_i$ by the induction hypothesis. Suppose then that such an index does not exist; for every $j \in I$ let $y_j$ be an element of $a \cap p_j$ not belonging to any $p_i$ for $i \neq j$. Let $k$ be an element of $I$ chosen in such a way that $p_k$ is prime if $n > 2$ and chosen arbitrarily if $n = 2$; let $z = y_k + \prod_{i \neq k} y_i$. Then $z \in a$, since $a$ is closed under addition and multiplication; if $j \neq k$, $\prod_{i \neq k} y_i$ belongs to $p_j$, but $y_k \notin p_j$, whence $z \notin p_j$. On the other hand, $\prod_{i \neq k} y_i$ does not belong to $p_k$, as none of the factors $y_i \ (i \neq k)$ belongs to it and $p_k$ is prime if $n - 1 > 1$; as $y_k \in p_k$, $z$ does not belong to $p_k$ and the proposition is established.