Solucionario

Solutions to Atiyah-Macdonald, Chapter 1
Dave Karpuk May 19, 2010
Exercise 1. Let x be a nilpotent element of a ring A. Show that 1 + x is a unit of A. Deduce that the sum of a nilpotent element and a unit is a unit. Proof. Note that x is nilpotent iff −x is nilpotent, so we may replace 1 + x with 1 − x. For all n ≥ 1, we have 1 + xn = (1 − x)(1 + x + · · · + xn−1 ). Thus if xn = 0, the element1 + x + · · · + xn−1 is the inverse of 1 − x. If u is any unit and u−1 its inverse, then u−1 x is nilpotent, so 1 + u−1 x is a unit. Since the units form a group under multiplication, we have that u + x = u(1 + u−1 x) is a unit. Exercise 2. Let A be a ring and let A[x] be the ring of polynomials in an indeterminate x, with coefficients in A. Let f = a0 + a1 x + · · · + an xn ∈ A[x]. Prove that (i) fis a unit in A[x] ⇐⇒ a0 is a unit in A and a1 , . . . , an are nilpotent. (ii) f is nilpotent ⇐⇒ a0 , a1 , . . . , an are nilpotent. (iii) f is a zero-divisor ⇐⇒ there exists a = 0 in A such that af = 0. (iv) f is said to be primitive if (a0 , a1 , . . . , an ) = (1). Prove that if f, g ∈ A[x], then f g is primitive ⇐⇒ f and g are primitive. Proof. (i) Suppose that a0 is a unit and that the otherai are all nilpotent. Then clearly ai xi is nilpotent, so a1 x + . . . + an xn is nilpotent. As a0 is a unit, it follows that f = a0 + a1 x + . . . + an xn is a unit by exercise 1. Conversely, suppose that f is a unit. Then we can find some g = b0 + . . . + bm xm such that f g = 1. This immediately implies that a0 is a unit, the hard part is proving the other ai are nilpotent. The equation f g = 1implies that 0 = an bm 0 = an bm−1 + an−1 bm . . . 0 = an b0 + an−1 b1 + . . . + a0 bn (set bj = 0 if j > m) Multiplying the second equation from the top by an gives 0 = a2 bm−1 . Multiplying the third n equation by a2 gives a3 bm−2 = 0. Continuing in this fashion, we see that there is some power aj n n n of an we can multiply the last equation by to obtain 0 = aj+1 b0 . n 1

Multiplying bothsides of this equation by b−1 (which exists) shows that an is nilpotent. 0 Now an xn is nilpotent, hence f − an xn , the sum of a unit and a nilpotent element, is a unit. But f − an xn has degree less than that of f , so we are done by induction on the degree of f . (ii) If all of the coefficients of f are nilpotent, then f is nilpotent as it is the sum of nilpotents. For the converse, note that f m= 0 for some m implies that am xnm = 0, hence an xn is nilpotent. n Thus f − an xn is the sum of nilpotent elements and hence nilpotent, the result now follows by induction on the degree of f . (iii) Let f = an xn +. . .+a0 be a zero-divisor of A[x], such that f g = 0, where g = bm xm +. . .+b0 is of minimal degree. Writing out f g immediately yields an bm = 0. We have an gf = 0, but since ankills the leading coefficient of g, we must have that an g is of degree smaller than g, unless an g = 0. Now suppose that an g = an−1 g = · · · = an−i g = 0. We have 0 = fg = an gxn + . . . + an−i gxn−i + an−i−1 gxn−i−1 + . . . + a0 g = an−i−1 gxn−i−1 + . . . + a0 g The leading coefficient of the above polynomial is an−i−1 bm = 0. Thus an−i−1 gf = 0, as before. By induction, we conclude that ai g = 0 forall i = 0, . . . , n. In particular, ai bm = 0 for all i, which implies that bm g = 0. (iv) Let f and g be as above, and let a = (a0 , . . . , an ), b = (b0 , . . . , bm ). Let c = (a0 b0 , a0 b1 + b0 a1 , . . . , an bm ) be the corresponding ideal for the product f g. Then the problem becomes to show that a = (1) = b if and only if c = (1). For one direction, note that c ⊆ a and c ⊆ b, so c = 1implies a = (1) = b trivially. For the other direction, suppose that a = (1) = b. Exercise 3. Generalize the results of Exercise 2 to a polynomial ring A[x1 , . . . , xn ] in several indeterminates. Proof. Exercise 4. In the ring A[x], the Jacobson radical is equal to the nilradical. Proof. We need only to show that the Jacobson radical is contained in the nilradical, as the other inclusion...