4. Let R be a ring, S a subring of R and I an ideal of R. Prove that I ∩ S is an ideal
of S

Question

4. Let R be a ring, S a subring of R and I an ideal of R. Prove that I ∩ S is an ideal
of S

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ Let $R$ be a ring, $S$ a subring of $R$, and $I$ an ideal of $R$. We need to show that $I \cap S$ is an ideal of $S$. ‖ ‖ step 2 ⋮ To prove that $I \cap S$ is an ideal of $S$, we need to show two things: (1) $I \cap S$ is a subring of $S$, and (2) $I \cap S$ is closed under multiplication by elements of $S$. ‖ ‖ step 3 ⋮ First, we show that $I \cap S$ is a subring of $S$. Since $I$ is an ideal of $R$, it is a subring of $R$. Therefore, $I \cap S$ is a subring of $R$. Since $S$ is a subring of $R$, $I \cap S$ is also a subring of $S$. ‖ ‖ step 4 ⋮ Next, we show that $I \cap S$ is closed under multiplication by elements of $S$. Let $a \in I \cap S$ and $s \in S$. Since $a \in I$ and $I$ is an ideal of $R$, $sa \in I$. Since $a \in S$ and $s \in S$, $sa \in S$ because $S$ is a subring of $R$. Therefore, $sa \in I \cap S$. ‖ ‖ step 5 ⋮ We have shown that $I \cap S$ is a subring of $S$ and is closed under multiplication by elements of $S$. Therefore, $I \cap S$ is an ideal of $S$. ‖ ∻Answer∻ ⚹ $I \cap S$ is an ideal of $S$. ⚹ ∻Key Concept∻ ⚹Intersection of an ideal and a subring⚹ ∻Explanation∻ ⚹The intersection of an ideal of a ring and a subring of that ring is itself an ideal of the subring. This is because the intersection inherits the properties of being a subring and being closed under multiplication by elements of the subring.⚹◊What are the properties of an ideal of a subring in ring theory⍭ Generate me a similar question◊

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