# Exercises

Exercises from the end of chapter 1 of Atiyah and MacDonald.

1. Let $$R$$ be a ring. Suppose $$x \in R$$ is nilpotent. Show $$1+x$$ is a unit of $$R$$. Deduce that the sum of a nilpotent element and a unit is a unit.

Solution: For a sufficiently large odd positive integer $$n$$ we have $$x^n = 0$$, hence $$1 = 1 + x^n =(1+x)(1-x +... + x^{n-1})$$ showing that $$1+x$$ has an inverse. Now suppose $$x$$ is nilpotent and $$u$$ is a unit. Note $$u^{-1} x$$ is also nilpotent, hence $$1 + u^{-1} x$$ is a unit. Multiplying by $$u$$ shows that $$u+x$$ is also a unit.

2. Let $$R$$ be a ring. Let $$f = a_0 + a_1 x +...+ a_n x^n \in R[x]$$. Prove that

1. $$f$$ is a unit in $$R[x]$$ $$\iff$$ $$a_0$$ is a unit in $$R$$ and $$a_1,...,a_n$$ are nilpotent.

2. $$f$$ is nilpotent $$\iff$$ $$a_0, a_1,...,a_n$$ are nilpotent are nilpotent.

3. $$f$$ is a zero divisor $$\iff$$ there exists $$a \ne 0 \in R$$ such that $$a f = 0$$.

4. $$f$$ is primitive if $$\langle a_0,...,a_n \rangle = \langle 1 \rangle$$. Show that if $$f,g \in R[x]$$ then $$f g$$ is primitive $$\iff$$ $$f,g$$ are primitive.

Solution:

1. ($$\Leftarrow$$) Suppose $$a_0$$ is a unit and $$a_1,...,a_n$$ are nilpotent. Without loss of generality $$a_0 = 1$$. Set $$g = - (a_1 x +...+a_n x^n)$$. Then $$g^r = 0$$ for some $$r$$ since sums and products of nilpotents are nilpotent. Then

$$f^{-1} = 1 / (1 - g) = 1 + g + g^2 + ... g^{r-1}$$

($$\Rightarrow$$) Write $$f^{-1} = b_0 + b_1 x +...+ b_m x^m$$, so that

$$(b_0 +...+b_m x^m)(a_0 +...+ a_n x^n) = 1$$

Thus $$a_0 b_0 = 1$$ so both $$a_0, b_0$$ are units. We also have $$a_n b_m = 0$$. Inductively assume $$a_n^{i+1} b_{m-i} =0$$ for all $$i \lt r$$. By considering the coefficient of $$x^{m+n-r}$$ in the above product we have

$$a_n b_{m-r} + a_{n-1} b_{m-r+1} + ... + a_{n-k} b_{m-r+k} = 0$$

for some $$k$$. Multiplying by $$a_n^r$$ and using the inductive hypothesis gives $$a_n^{r+1} b_{m-r} = 0$$ for all $$r$$. In particular $$a_n^{m+1} b_0 = 0$$, and since $$b_0$$ is a unit we have that $$a_n$$ is nilpotent. From exercise 1, $$f - a_n x^n = a_0 +...+a_{n-1} x^{n-1}$$ is also a unit so by induction $$a_1,...,a_{n-1}$$ are nilpotent.

2. ($$\Leftarrow$$) Since sums and products of nilpotents are nilpotent, if $$a_0,...,a_n$$ are nilpotent then so is $$f$$.

($$\Rightarrow$$) Let $$f^k = 0$$. Then $$a_0^k = 0$$. Inductively assume $$a_i$$ is nilpotent for $$i \lt r$$. By considering the coefficient of $$x^{r k}$$ we have $$a_r^k + T = 0$$ where every term in $$T$$ contains $$a_i$$ for some $$i \lt r$$. Thus $$a_r^k$$ is nilpotent, hence $$a_r$$ is.

3. One way is trivial. For the converse, choose $$g = b_0 + ... +b_m x^m$$ of least degree such that $$f g = 0$$. Then $$a_n b_m = 0$$. But since $$a_n g f = 0$$, $$a_n g$$ has degree less than $$m$$, and $$g$$ is of minimal degree, we must have $$a_n g = 0$$. Inductively assume $$a_{n-i} g = 0$$ for all $$0 \le i \lt r$$. The coefficient of $$x^{m+n-r}$$ in $$f g$$ is

$a_{n-r} b_m + ... + a_n b_{m-r} = 0$

By the inductive hypothesis this simplifies to $$a_{n-r} b_m = 0$$, thus $$a_{n-r} g = 0$$. Hence $$a_i b_j = 0$$ for all $$i,j$$, thus any of the $$b_j$$ will annihilate $$f$$ (so in fact $$g$$ has degree zero).

3. Generalize exercise 2 to a multinomial ring $$R[x_1,...,x_r]$$.

4. In $$R[x]$$, the Jacobson radical is equal to the nilradical.

Solution: Suppose $$f \in R[x]$$ lies in the Jacobson radical. Then $$1 - f x$$ is a unit, hence by exercise 2.i we have that $$a_0,...,a_n$$ are nilpotent, implying that $$f$$ is nilpotent by exercise 2.ii, thus $$f$$ also lies in the nilradical.

5. Let $$R$$ be a ring and consider $$R[[x]]$$, the ring of formal power series $$f = \sum_{n=0}^{\infty} a_n x^n$$.

1. $$f$$ is a unit in $$R[[x]]$$ $$\iff$$ $$a_0$$ is a unit in $$R$$.

2. If $$f$$ is nilpotent then $$a_n$$ is nilpotent for $$n\ge 0$$.

3. $$f \in J(R[[x]])$$ $$\iff$$ $$a_0 \in J(R)$$

4. The contraction of a maximal ideal $$I \triangleleft R[[x]]$$ is a maximal ideal of $$R$$ and $$I$$ is generated by $$I^c$$ and $$x$$.

5. Every prime ideal of $$R$$ is the contraction of a prime ideal of $$R[[x]]$$.

Solution:

1. If $$f$$ is a unit in $$R$$ then let $$g = \sum b_i$$ be its inverse. Then $$a_0 b_0 = 1$$ hence $$a_0$$ is a unit. Conversely, suppose $$a_0$$ is a unit. Then define the $$b_i$$ inductively by $$b_0 = a_0^{-1}$$, $$b_{i+1} = -b_0( a_1 b_i + ... + a_{i+1} b_0)$$ and we have $$f g = 1$$.

2. See 2.ii.

3. Suppose $$f \in J(R[[x]])$$ then for all $$b \in R$$ we have $$1 - f b$$ is a unit, thus by exercise 5.i we have $$1 - a_0 b$$ is a unit in $$R$$ for all $$b \in R$$, hence $$a_0 \in J(R)$$. Conversely, if $$a_0 \in J(R)$$ then $$1 - f g$$ for any $$g \in J(R)$$ is a unit again by 5.i.

4. Let $$I$$ be a maximal ideal of $$R[[x]]$$. We first show that $$x \in I$$. Suppose not. Then since $$I$$ is maximal, we must have $$f + x g = 1$$ for some $$f \in I, g \in R[[x]]$$. This implies the constant term of $$f$$ is 1, which means $$f$$ is a unit by exercise 5.i, a contradiction.

Now suppose $$I^c$$ is not maximal. Then let $$J$$ be some proper ideal of $$R$$ strictly containing $$I^c$$, and take $$c \in J \setminus I^c$$. Since $$I$$ is maximal, $$I$$ and $$c$$ generate all of $$R[[x]]$$. Thus $$f + c g = 1$$ for some $$f \in I, g \in R[[x]]$$. The constant terms therefore satisfy $$a_0 + c b_0 = 1$$ (where $$a_i, b_i$$ are the coefficients of $$f, g$$). But $$a_0 = f - x f'$$ where $$f' = a_1 + a_2 x +...$$, thus $$a_0 \in I^c$$. Hence $$1 = a_0 + c b_0 \in J$$, a contradiction.

Then $$I = \langle I^c \cup x \rangle$$, because the constant term of $$f \in I$$ must be contained in $$I^c$$, and there are no restrictions on the other coefficients.

5. Let $$I$$ be a prime ideal of $$R$$. Consider the ideal $$J = \langle I \cup x \rangle$$ in $$R[[x]]$$. Suppose $$f g \in J$$. Then if $$a_0, b_0$$ are the constant terms of $$f,g$$ then $$a_0 b_0 = c$$ where $$c \in I$$. Since $$I$$ is prime we have $$a_0, b_0 \in I$$, hence $$f, g \in J$$.

6. Let $$R$$ be a ring such that every ideal not contained in the nilradical contains a nonzero idempotent. Then $$N_R = J(R)$$.

Solution: Suppose not. Let $$e$$ be a nonzero idempotent in $$J(R)$$. Now $$(1-e)^2 = 1-e$$ is a unit, so $$(1-e)x = 1$$ for some $$x\in R$$. But squaring gives $$1 = (1-e)^2 x^2 = (1-e)x^2 = x$$ which implies $$1-e = 1$$, a contradiction since $$e$$ is nonzero.

7. Let $$R$$ be a ring such that every $$x \in R$$ satisfies $$x^n = x$$ for some $$n \gt 1$$. Show every prime ideal is maximal.

Solution: Let $$I$$ be a prime ideal. Consider the integral domain $$R/I$$. Then for all $$x \in R$$, $$x + R = x^n + R$$ for some $$n$$. Since $$R/I$$ is an integral domain, either $$x = 0$$ or $$x^{n-1} = 1$$ in which case $$x$$ has an inverse. Thus $$R/I$$ is a field showing that $$I$$ is maximal.

8. Let $$R$$ be a nonzero ring. Show that the set of prime ideals has minimal elements with respect to inclusion.

Solution: Apply Zorn’s Lemma.

9. Let $$I$$ be a proper ideal of a ring $$R$$. Show that $$I =r(I)$$ $$\iff$$ $$I$$ is an intersection of prime ideals.

Solution: Recall that the radical of an ideal is the intersection of the prime ideals containing it. For the other direction, we use the fact that $$\sqrt{I} = I$$ for prime $$I$$ and that $$\sqrt{I\cap J} = \sqrt{I} \cap \sqrt{J}$$.

10. Let $$R$$ be a ring and $$N$$ its nilradical. Show that the following are equivalent.

1. $$R$$ has exactly one prime ideal

2. Every element of $$R$$ is a unit or nilpotent

3. $$R/N$$ is a field.

Solution: If $$R$$ has exactly one prime ideal then $$N$$ must be that ideal because it is the intersection of all the prime ideals. Since maximal ideals are prime, $$N$$ is maximal as there is only one prime ideal. Thus $$R/N$$ is a field, so for any $$x \in R \setminus N$$ we have $$x y \in 1 + N$$ for some $$y$$. Since the sum of a nilpotent and a unit is itself a unit, we have that $$x y$$ and hence $$x$$ is a unit. Thus (i) $$\implies$$ (iii) $$\implies$$ (ii).

Lastly suppose every element is a unit or nilpotent. $$N$$ is maximal because every other unit is a unit, and since $$N$$ is prime (as it is maximal) and the intersection of all the prime ideals, it is in fact the only prime ideal of $$R$$.

11. A ring $$R$$ is Boolean if $$x^2 = x$$ for all $$x\in R$$. Show that

1. $$2x = 0$$ for all $$x\in R$$.

2. Every prime ideal $$P$$ is maximal, and $$R/P = \mathbb{Z}_2$$.

3. Every finitely generated ideal in $$R$$ is principal.

Solution:

1. $$1+x = (1+x)^2 = 1 + x +2x$$ implies $$2x = 0$$.

2. Every prime ideal is maximal from exercise 7. In $$R/P$$, we have $$x + P = x^2 + P$$, thus $$x = 0$$ or $$x = 1$$.

3. Consider the expression $$x + y - x y$$. Since $$x(x+y- x y) = x, y(x + y - x y) = y$$ we have that $$x+y- x y$$ generates everything that $$x, y$$ does. Thus by repeating this argument any finite set of generators for an ideal may be replaced by a single element.

12. A local ring contains no idempotent except $$0$$ and $$1$$.

Solution: Suppose a local ring $$R$$ contained an idempotent $$e \ne 0,1$$. Its unique maximal ideal is $$J(R)$$ since $$J(R)$$ is the intersection of all maximal ideals. If $$e \in J(R)$$ then $$1-e$$ is a unit, that is $$(1-e)x = 1$$ for some $$x \in R$$. But this implies $$1 = (1-e)^2 x^2 = (1-e)x^2 = x$$, thus $$1-e=1$$ which cannot be if $$e$$ is nonzero. On the other hand, if $$e \notin J(R)$$ then we have $$e^2 + R = e + R$$ in the field $$R/J(R)$$ implying that $$e$$ is $$0$$ or $$1$$, also a contradiction.

Ben Lynn blynn@cs.stanford.edu 💡