Check if the solution is correct. Correct the mistakes. Problems:Show that every nonzero ideal in the ring of formal power series F[[x]] in an indeterminate x over a field F is of the form (x^m) for some nonnegative integer m. Solution: Let I be a nonzero ideal in the ring of formal power series F[[x]]. Take an arbitrary nonzero element f(x) from I. Then I must contain all the polynomials that are divisible by f(x) without remainder. Let h(x) be a polynomial that belongs to I and is not a multiple of f(x), i.e. h(x) is not divisible by f(x) without remainder. Then, by the remainder division theorem for polynomials, there are polynomials q(x) and r(x) such that h(x) = f(x)q(x) + r(x), where either r(x) = 0, or the degree of r(x) is less than the degree of f(x). Since h(x) belongs to the ideal I, then f(x)q(x) must also belong to I, which means that the difference r(x) = h(x) - f(x)q(x) must also belong to I. If r(x) ≠ 0, then we can continue the division process with the remainder and obtain a sequence of polynomials h(x) > r(x) > r1(x) > r2(x) > ... , where the degrees of the polynomials decrease and r_i(x) is not divisible by f(x) without remainder. Since we are dealing with an infinite sequence of polynomials, then at some step the degree of the polynomial ri(x) must become negative. But this means that ri(x) = 0, since there are no zero divisors in the field F. Then we get that h(x) is divisible by f(x) without remainder. Thus, we have shown that for any nonzero element f(x) from the ideal I, all polynomials that are divisible by f(x) without remainder also belong to the ideal I. In particular, for any element g(x) from the ideal I, there exists such a nonnegative integer m that xm divides g(x) without remainder. That is, every nonzero ideal in the ring of formal power series F[[x]] has the form (xm) for some nonnegative integer m.
Check if the solution is correct. Correct the mistakes.
Problems:Show that every nonzero ideal in the ring of formal power series F[[x]] in an indeterminate x over a field F is of the form (x^m) for some nonnegative integer m.
Solution: Let I be a nonzero ideal in the ring of formal power series F[[x]]. Take an arbitrary nonzero element f(x) from I. Then I must contain all the polynomials that are divisible by f(x) without remainder.
Let h(x) be a polynomial that belongs to I and is not a multiple of f(x), i.e. h(x) is not divisible by f(x) without remainder. Then, by the remainder division theorem for polynomials, there are polynomials q(x) and r(x) such that h(x) = f(x)q(x) + r(x), where either r(x) = 0, or the degree of r(x) is less than the degree of f(x).
Since h(x) belongs to the ideal I, then f(x)q(x) must also belong to I, which means that the difference r(x) = h(x) - f(x)q(x) must also belong to I.
If r(x) ≠ 0, then we can continue the division process with the remainder and obtain a sequence of polynomials h(x) > r(x) > r1(x) > r2(x) > ... , where the degrees of the polynomials decrease and r_i(x) is not divisible by f(x) without remainder.
Since we are dealing with an infinite sequence of polynomials, then at some step the degree of the polynomial ri(x) must become negative. But this means that ri(x) = 0, since there are no zero divisors in the field F. Then we get that h(x) is divisible by f(x) without remainder.
Thus, we have shown that for any nonzero element f(x) from the ideal I, all polynomials that are divisible by f(x) without remainder also belong to the ideal I. In particular, for any element g(x) from the ideal I, there exists such a nonnegative integer m that xm divides g(x) without remainder. That is, every nonzero ideal in the ring of formal power series F[[x]] has the form (xm) for some nonnegative integer m.
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