Question

Math

Posted 3 months ago

irstly, note that 0 ∈ S and 0 ∈ I, so 0 ∈ I ∩ S. Let x, y ∈ S ∩ I. Then
x, y ∈ S and x, y ∈ I. So −x ∈ S and −x ∈ I, since S is a subring of R and I is
an ideal of R. So −x ∈ S ∩ I. We also have x + y ∈ S and x + y ∈ I for the same
reason, so x + y ∈ S ∩

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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
```

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Answer from Sia

Posted 3 months ago

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 $I \cap S$ 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

Since $I \cap S$ is a subring of $S$ and is closed under multiplication by elements of $S$, $I \cap S$ is an ideal of $S$

Answer

$I \cap S$ is an ideal of $S$.

Key Concept

Intersection of Ideal and Subring

Explanation

The intersection of an ideal of a ring and a subring of the same ring is an ideal of the subring.

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