Let $R$ be a ring with identity, $R=R_1\oplus R_2 \oplus \dots \oplus R_n$, where $R_i \cap R_j =\emptyset$, $I$ is an ideal of $R$.
Prove that $$I=(I\cap R_1)\oplus (I\cap R_2)\oplus \dots \oplus (I\cap R_n)$$
I was learning ring theory. Please tell me please.
To prove that an ideal $I$ of a ring $R = R_1 \oplus R_2 \oplus \dots \oplus R_n$ is equal to the direct sum of the intersections of $I$ with each $R_i$, we'll proceed step by step.
Step 1: Define the Direct Sum
The ring $R$ is the direct sum $R_1 \oplus R_2 \oplus \dots \oplus R_n$, meaning each element of $R$ can be uniquely expressed as a tuple $(r_1, r_2, \dots, r_n)$ where $r_i \in R_i$.
Step 2: Components of $I$ in $R_i$
Since $I$ is an ideal, for any $(r_1, r_2, \dots, r_n) \in I$, the elements $r_i$ must belong to $I \cap R_i$. This follows from the ability to multiply $(r_1, r_2, \dots, r_n)$ by $(1, 0, \dots, 0)$, and similar elements, to isolate each component.
We examine how the elements of the ideal $I$ are expressed within the context of the ring $R$ being a direct sum of other rings $R_1, R_2, \ldots, R_n$.
Every element in $I$ can be represented as a tuple $(r_1, r_2, \ldots, r_n)$, with each $r_i$ belonging to the corresponding component ring $R_i$. An important property of an ideal is that if you multiply any element from the entire ring $R$ by an element of $I$, the product is also in $I$.
Why is this significant? Consider the element $(1, 0, \ldots, 0)$ in $R$. This element, when multiplied by any tuple in $I$, isolates the first component of that tuple. Multiplying any element $(r_1, r_2, \ldots, r_n)$ in $I$ by $(1, 0, \ldots, 0)$ yields $(r_1, 0, \ldots, 0)$. This resulting tuple must also belong to $I$, confirming that $r_1$ is in $I \cap R_1$.
Applying this logic to each component with elements like $(0, \ldots, 1, \ldots, 0)$—where 1 is in the $i$-th position—demonstrates that each $r_i$ must be in $I \cap R_i$.
This understanding is key because it reveals that each component of an element in $I$ lies within the intersection of $I$ with the respective subring $R_i$. Thus, the entire ideal $I$ can be "reconstructed" by taking the direct sum of these intersections of $I$ with each $R_i$.
This leads to the conclusion that $I$ is precisely the direct sum of the intersections $(I \cap R_1) \oplus (I \cap R_2) \oplus \ldots \oplus (I \cap R_n)$.
Step 3: Intersection with $I$ is Necessary
For any $r_i \in R_i$, the element $(0, \dots, r_i, \dots, 0)$ is in $R$. If $r_i$ is also in $I$, then $(0, \dots, r_i, \dots, 0)$ is in $I$, confirming that $r_i$ must be in $I \cap R_i$.
Step 4: Conclusion
Combining the results from Steps 2 and 3, we conclude that $I$ consists of tuples where each component is from $I \cap R_i$, and thus, $I = (I \cap R_1) \oplus (I \cap R_2) \oplus \dots \oplus (I \cap R_n)$.
Simple Explanation:
Imagine $R$ as a collection of different strands of yarn, each color representing one of the $R_i$'s. An ideal $I$ is like a pattern made from specific selections of these strands. What we've shown is that if you separate the pattern by color, and then put these colored sections back together, you re-create the original pattern. This means the original pattern ($I$) can be fully described by looking at each color (each $R_i$) separately.