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

\(f\) is nilpotent \(\iff\) \(a_0, a_1,...,a_n\) are nilpotent are nilpotent.

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

\(f\) 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.
Exercises
Exercises from the end of chapter 1 of Atiyah 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 \(1 = 1 + x^n =(1+x)(1x +... + x^{n1})\) 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 \(f = a_0 + a_1 x +...+ a_n x^n \in R[x]\). Prove that
Solution:

(\(\Leftarrow\)) 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
\(f^{1} = 1 / (1  g) = 1 + g + g^2 + ... g^{r1}\)
(\(\Rightarrow\)) 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 \(a_n^{i+1} b_{mi} =0\) for all \(i \lt r\). By considering the coefficient of \(x^{m+nr}\) in the above product we have
\( a_n b_{mr} + a_{n1} b_{mr+1} + ... + a_{nk} b_{mr+k} = 0 \)
for some \(k\). Multiplying by \(a_n^r\) and using the inductive hypothesis gives \(a_n^{r+1} b_{mr} = 0\) 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, \(f  a_n x^n = a_0 +...+a_{n1} x^{n1}\) is also a unit so by induction \(a_1,...,a_{n1}\) are nilpotent.

(\(\Leftarrow\)) 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.

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 \(a_{ni} g = 0\) for all \(0 \le i \lt r\). The coefficient of \(x^{m+nr}\) in \(f g\) is
\[ a_{nr} b_m + ... + a_n b_{mr} = 0 \]By the inductive hypothesis this simplifies to \(a_{nr} b_m = 0\), thus \(a_{nr} g = 0\). 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 \(f = \sum_{n=0}^{\infty} a_n x^n\).

\(f\) is a unit in \(R[[x]]\) \(\iff\) \(a_0\) is a unit in \(R\).

If \(f\) is nilpotent then \(a_n\) is nilpotent for \(n\ge 0\).

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

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

Every prime ideal of \(R\) is the contraction of a prime ideal of \(R[[x]]\).
Solution:

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 \(f g = 1\).

See 2.ii.

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.

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.

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 \(f, g \in J\).


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 \((1e)^2 = 1e\) is a unit, so \((1e)x = 1\) for some \(x\in R\). But squaring gives \(1 = (1e)^2 x^2 = (1e)x^2 = x\) which implies \(1e = 1\), a contradiction since \(e\) is nonzero.

Let \(R\) be a ring such that every \(x \in R\) satisfies \(x^n = x\) for some \(n \gt 1\). 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 \(x^{n1} = 1\) 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 \(I =r(I)\) \(\iff\) \(I\) 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 \(\sqrt{I\cap J} = \sqrt{I} \cap \sqrt{J}\).

Let \(R\) be a ring and \(N\) its nilradical. Show that the following are equivalent.

\(R\) has exactly one prime ideal

Every element of \(R\) is a unit or nilpotent

\(R/N\) 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 \(R\).


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

\(2x = 0\) for all \(x\in R\).

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

Every finitely generated ideal in \(R\) is principal.
Solution:

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

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

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 \(e \ne 0,1\). Its unique maximal ideal is \(J(R)\) since \(J(R)\) is the intersection of all maximal ideals. If \(e \in J(R)\) then \(1e\) is a unit, that is \((1e)x = 1\) for some \(x \in R\). But this implies \(1 = (1e)^2 x^2 = (1e)x^2 = x\), thus \(1e=1\) 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.