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wowolala
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Let a and b belong to a commutative ring R. Prove that { x ∈ R | ax∈bR } is an ideal.
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i really need help
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An ideal of a ring is a subset of the ring that satisfies certain properties. Specifically, it is a non-empty subset that is closed under addition and multiplication by elements of the ring. In other words, if you take any element from the ideal and multiply it by any element from the ring, the result will also be in the ideal. Additionally, the ideal must contain the additive identity element of the ring.
An ideal is a subset of a ring, while a subgroup is a subset of a group. While both are defined as non-empty subsets that are closed under certain operations, the operations are different. In a subgroup, the operation is usually addition or multiplication, while in an ideal, the operation is multiplication. Additionally, a subgroup must contain the inverse of each of its elements, while an ideal does not necessarily have this property.
Yes, an ideal can contain non-commutative elements. This is because the definition of an ideal only requires closure under multiplication, not commutativity. Therefore, an ideal can contain elements that do not commute with each other.
Yes, an ideal is always a subring. This is because an ideal is a subset of a ring that is closed under multiplication, and a subring is defined as a subset of a ring that is closed under both addition and multiplication. Since an ideal is already closed under multiplication, it automatically satisfies the definition of a subring.
No, a field cannot have any non-trivial ideals. This is because in a field, every element has a multiplicative inverse, which means that the only subsets that are closed under multiplication are the trivial subsets (containing just the identity element) and the entire field itself. Therefore, there are no non-trivial ideals in a field.