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 $f = \sum_{n=0}^{\infinity} a_n x^n$.

    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.