One question related to an ideal of Ring R

[wowolala]

Let a and b belong to a commutative ring R. Prove that { x ∈ R | ax∈bR } is an ideal.

i really need help

thx

 

[ircdan]

just use the definition of an ideal, where are you stuck exactly

 

[tacman]

One question that comes to mind when considering this ideal is: How does this ideal relate to the ideals generated by a and b individually in R? This is a valid question to ask because it helps to understand the structure and properties of the ideal in question.

To prove that { x ∈ R | ax∈bR } is an ideal, we can follow the definition of an ideal. First, we need to show that it is a subgroup of R under addition. Let x and y be elements in the set, then we have:

ax, ay ∈ bR
ax + ay ∈ bR (since bR is closed under addition)
a(x + y) ∈ bR (distributive property)
∴ x + y ∈ { x ∈ R | ax∈bR } (by definition)

Next, we need to show that it is closed under multiplication by elements in R. Let r be an element in R, then we have:

ax ∈ bR
rax ∈ rbR (since bR is closed under multiplication by elements in R)
∴ rx ∈ { x ∈ R | ax∈bR } (by definition)

Therefore, { x ∈ R | ax∈bR } is a subgroup of R under addition and closed under multiplication by elements in R. This fulfills the requirements for being an ideal in R.

Furthermore, we can see that this ideal is related to the ideals generated by a and b individually in R. In fact, it is the intersection of these two ideals: { x ∈ R | ax∈bR } = (a) ∩ (b). This means that it contains all elements that are in both (a) and (b). This can be seen by considering the fact that if ax ∈ bR, then it is also in (a) and (b). Similarly, if ax ∈ (a) and b ∈ (b), then ax ∈ bR.

In conclusion, the ideal { x ∈ R | ax∈bR } is a subgroup of R and is related to the ideals generated by a and b individually in R. This provides a deeper understanding of its properties and significance in the ring R.