In a ring with multiplicative identity, If x^n is in an ideal then is x also in the ideal? with n a natural number.
The Attempt at a Solution
I can't find a proof. Which is likely to mean the statement is false?
pivoxa15 said:I can't find a proof. Which is likely to mean the statement is false?
matt grime said:It is perfectly possible for the ideal generated by x^2 to have x in it. It is just not *certain* to have x in it.
An ideal is a subset of a ring that satisfies certain properties. In simple terms, it is a collection of elements that can be multiplied by any element in the ring and still remain within the ideal.
If x^n is an element of an ideal, it means that the ideal contains all possible powers of x up to the nth power. This is because an ideal must contain all products of elements within the ideal, and x^n can be written as a product of x multiplied by itself n times.
X^n being in an ideal is significant because it allows for the ideal to generate all other powers of x through multiplication. This is useful in many mathematical applications, such as in polynomial rings and algebraic geometry.
Yes, x^n can still be in an ideal even if x is not in the ideal. This is because the ideal only needs to contain all possible powers of x, not necessarily x itself.
An ideal generated by x^n is the smallest ideal that contains x^n. It is denoted as