答案introduction mutative algebra solutions.pdf

mutative_algebra__solutionsSolutions to Atiyah and MacDonald’s Introduction to
Commutative Algebra
Athanasios Papaioannou
August 5, 2004
2
Chapter 1
Rings and Ideals

We see that x ∈ R implies x ∈ J (the Jacobson radical), hence 1 + xA ⊂ A∈ . In particular, 1 + x is a unit.
We can now easily deduce that the sum of a nilpotent element and a unit is a unit itself.

We have the following:
n m
(i) If f(x) = a0 + a1x + . . . + anx is a unit in A[x], let g(x) = b0 + b1x + . . . + bmx be its inverse.
We deduce that a0 is a unit. We use induction on n to prove that the coefficients are nilpotent. The case
r+1
n = 0 is a tautology. If the proposition holds for n
1, then we see that an bm0 r = 0 (we just write down
explicitly the relations that ensue from fg = 1 and then multiply each of them by increasing powers of an).
m+1 m+1
In particular, this implies that an b0 = 0 and, since b0 is a unit, we deduce that an = 0. Hence an is
nilpotent and we may apply the inductive hypothesis.
The converse follows from exercise 1 and exercise 2, (ii).
(ii) If f(x) is nilpotent, then we can apply induction to n to show that all its coefficients are nilpotent.
m
The case n = 0 is a tautology. In the general case, it’s apparent that the leading coefficient will be an for
suitable m ∈ N hence an is nilpotent. Now the inductive hypothesis applies.
N d
Conversely, if all the coefficients of f(x) are nilpotent and d ∈ is such that ai = 0, 0 ≤ i ≤ n (. let
d be the sum of the orders of all the orders of the coefficients), then we see tha