**Proof.**
Let us first prove $I_1 \cap \ldots \cap I_ r = I_1 \ldots I_ r$ as this will also imply the injectivity of the induced ring homomorphism $R/(I_1 \ldots I_ r) \rightarrow R/I_1 \times \ldots \times R/I_ r$. The inclusion $I_1 \cap \ldots \cap I_ r \supset I_1 \ldots I_ r$ is always fulfilled since ideals are closed under multiplication with arbitrary ring elements. To prove the other inclusion, we claim that the ideals

\[ I_1 \ldots \hat I_ i \ldots I_ r,\quad i = 1, \ldots , r \]

generate the ring $R$. We prove this by induction on $r$. It holds when $r = 2$. If $r > 2$, then we see that $R$ is the sum of the ideals $I_1 \ldots \hat I_ i \ldots I_{r - 1}$, $i = 1, \ldots , r - 1$. Hence $I_ r$ is the sum of the ideals $I_1 \ldots \hat I_ i \ldots I_ r$, $i = 1, \ldots , r - 1$. Applying the same argument with the reverse ordering on the ideals we see that $I_1$ is the sum of the ideals $I_1 \ldots \hat I_ i \ldots I_ r$, $i = 2, \ldots , r$. Since $R = I_1 + I_ r$ by assumption we see that $R$ is the sum of the ideals displayed above. Therefore we can find elements $a_ i \in I_1 \ldots \hat I_ i \ldots I_ r$ such that their sum is one. Multiplying this equation by an element of $I_1 \cap \ldots \cap I_ r$ gives the other inclusion. It remains to show that the canonical map $R/(I_1 \ldots I_ r) \rightarrow R/I_1 \times \ldots \times R/I_ r$ is surjective. For this, consider its action on the equation $1 = \sum _{i=1}^ r a_ i$ we derived above. On the one hand, a ring morphism sends 1 to 1 and on the other hand, the image of any $a_ i$ is zero in $R/I_ j$ for $j \neq i$. Therefore, the image of $a_ i$ in $R/I_ i$ is the identity. So given any element $(\bar{b_1}, \ldots , \bar{b_ r}) \in R/I_1 \times \ldots \times R/I_ r$, the element $\sum _{i=1}^ r a_ i \cdot b_ i$ is an inverse image in $R$.

To see (2), by the very definition of being distinct maximal ideals, we have $\mathfrak {m}_ a + \mathfrak {m}_ b = R$ for $a \neq b$ and so the above applies.
$\square$

## Comments (0)

There are also: