Commutative Algebra

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

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.
Let R be a ring. Let f=a 0 +a 1 x+...+a nx n∈R[x]. Prove that
1. $f$ is a unit in $R [x]$ $\Leftrightarrow$ $a_{0}$ is a unit in $R$ and $a_{1}, . . ., a_{n}$ are nilpotent.
2. $f$ is nilpotent $\Leftrightarrow$ $a_{0}, a_{1}, . . ., a_{n}$ are nilpotent are nilpotent.
3. $f$ is a zero divisor $\Leftrightarrow$ there exists $a \neq 0 \in R$ such that $a f = 0$ .
4. $f$ is primitive if $⟨ a_{0}, . . ., a_{n} ⟩ = ⟨ 1 ⟩$ . Show that if $f, g \in R [x]$ then $f g$ is primitive $\Leftrightarrow$ $f, g$ 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 $i < r$ . 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 $i < r$ . 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 $i < r$ . 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 $0 \leq i < r$ . 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).
Generalize exercise 2 to a multinomial ring $R [x_{1}, . . ., x_{r}]$ .
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.
Let R be a ring and consider R[[x]], the ring of formal power series f=∑ n=0 ∞a nx n.
1. $f$ is a unit in $R [[x]]$ $\Leftrightarrow$ $a_{0}$ is a unit in $R$ .
2. If $f$ is nilpotent then $a_{n}$ is nilpotent for $n \geq 0$ .
3. $f \in J (R [[x]])$ $\Leftrightarrow$ $a_{0} \in J (R)$
4. The contraction of a maximal ideal $I ◃ R [[x]]$ 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 $c \in J ∖ I^{c}$ . 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 $I = ⟨ I^{c} \cup x ⟩$ , 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 $J = ⟨ I \cup x ⟩$ 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 $f, g \in J$ .
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.
Let $R$ be a ring such that every $x \in R$ satisfies $x^{n} = x$ for some $n > 1$ . 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.
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.
Let $I$ be a proper ideal of a ring $R$ . Show that $I = r (I)$ $\Leftrightarrow$ $I$ 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 $\sqrt{I \cap J} = \sqrt{I} \cap \sqrt{J}$ .
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 $x \in R ∖ N$ 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) $\Rightarrow$ (iii) $\Rightarrow$ (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 $R$ .
A ring R is Boolean if x 2 =x for all x∈R. Show that
1. $2 x = 0$ for all $x \in R$ .
2. Every prime ideal $P$ is maximal, and $R / P = ℤ_{2}$ .
3. Every finitely generated ideal in $R$ is principal.
Solution:
1. $1 + x = (1 + x)^{2} = 1 + x + 2 x$ implies $2 x = 0$ .
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.
A local ring contains no idempotent except $0$ and $1$ .

Solution: Suppose a local ring $R$ contained an idempotent $e \neq 0,1$ . 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.