Atiyah-Macdonald Chapter 1

Atiyah-Macdonald Commutative Algebra

Problem 1 (1.2). Let $A$ be a (commutative) ring and let $A[x]$ be the ring of polynomials in an indeterminate $x$, with coefficients in $A$. Let $f=a_0+a_{1}x+\cdots +a_n{x}^n\in A[x]$. Prove that i) $f$ is a unit in $A[x]$ $⟺$ $a_0$ is a unit in $A$ and $a_1,\dots ,a_n$ are nilpotent. ii) $f$ is nilpotent $⟺$ $a_0,a_1,\dots ,a_n$ are nilpotent. iii) $f$ is a zero-divisor $⟺$ there exists $a\ne 0$ in $A$ such that $af=0$. iv) $f$ is said to be primitive if $(a_0,a_1,\dots ,a_n)=(1)$. Prove that if $f,g\in A[x]$, then $fg$ is primitive $⟺$ $f$ and $g$ are primitive.

Proof) Note that for any ring $R$, if $x\in R$ is nilpotent then $1+x$ is a unit, and for any unit $u\in R$, $u+x$ is also a unit. (i) We use induction on $n$; the $n=0$ case is trivial, and we assume $<n$ holds. $\Leftarrow$ follows from the fact that if $a_n$ is nilpotent $a_n{x}^n$ is nilpotent, and by the $\Leftarrow$ of $n-1$, $f$ is the sum of a nilpotent element with a unit, which is a unit. We now show the $\Rightarrow$ direction. Let $b_0+b_{1}x+\dots +b_m{x}^m$ be the inverse of $f$. Clearly $a_n{b}_m=0$. Furthermore, $a_n{b}_{m-1}+b_m{a}_{n-1}=0$, implying $a_n^2{b}_{m-1}=0$. We can continue this; $a_n{b}_{m-2}+a_{n-1}{b}_{m-1}+a_{n-2}{b}_m=0\Rightarrow a_n^3{b}_{m-2}=0$. We can conclude that $a_n^m{b}_1=0$, but as $b_1$ must be a unit, this makes $a_n$ nilpotent. This implies $a_n{x}^n$ is nilpotent, which forces $a_0+a_{1}x+\dots +a_{n-1}{x}^{n-1}=f-a_n{x}^n=(unit)+(nilpotent)$ which is a unit. The theorem thus follows from the $\Rightarrow$ of the $n-1$ case. (ii) Trivial. (iii) Let $g=b_0+b_{1}x+\dots +b_m{x}^m$ be a polynomial of minimal degree such that $fg=0$. If $n=0$ we are done; otherwise, it is clear that $b_0{a}_0=0$, $b_0{a}_1+b_1{a}_0=0\Rightarrow a_1{b}_0^2=0$. Analogously, we get $a_2{b}_0^3=0,\dots ,a_n{b}_0^{n+1}=0$. This implies that $b_0^{n+1}f=0$, an immediate contradiction of minimality. (iv) $\Rightarrow$ is trivial as the coefficient ideal of $fg$ is a subset of those of $f$ and $g$. For $\Leftarrow$, assume that the coefficient ideal of $fg$ is some proper ideal $I$ in $A$. Take a maximal ideal $J$ containing $I$; in the polynomial ring $(A/J)[x]$, $fg=0$, which implies there is some $k\in A/J$ such that $k\ne 0$ and $kf=0$(note $f$ nor $g$ can vanish in $(A/J)[x]$ by hypothesis). As $A/J$ is a field, this is only possible if $f=0$ in $(A/J)[x]$, which is a contradiction as the coefficient ideal of $f$ must then be a subset of $J$.

Remark: In algebraic geometry (iv) is really the only useful one since we always work with fields.

Problem 2 (1.4). In the ring $A[x]$, the Jacobson radical $J$ is equivalent to the nilradical $N$.

Proof) Note that the nilradical is always a subset of the Jacobson radical as all maximal ideals are prime. Assume $f=a_0+a_{1}x+\dots +a_n{x}^n$ is such that $1-fg$ is a unit for all $g\in A[x]$. $1-f$ being a unit implies $a_1,\dots ,a_n$ are nilpotent and $1-a_0$ is a unit. We also have that $1-r{a}_0$ is a unit for all $r\in A$, which means $a_0$ is in the Jacobson radical of $A$. This forces $a_0$ to be in the nilradical of $A$ and thus $a_0$ is nilpotent. We are done by 1(ii).

Problem 3 (1.5). Let $A$ be a ring and let $A[[x]]$ be the ring of formal power series $f={\sum }_{n=0}^{\infty }{a}_n{x}^n$ with coefficients in $A$. Show that (i) $f$ is a unit in $A[[x]]⟺a_0$ is a unit in $A$. (ii) If $f$ is nilpotent, then $a_n$ is nilpotent for all $n\ge 0$. (iii) $f$ belongs to the Jacobson radical of $A[[x]]⟺$ $a_0$ belongs to the Jacobson radical of $A$. (iv) The contraction of a maximal ideal $m$ of $A[[x]]$ is a maximal ideal of $A$, and $m$ is generated by $m^c$ and $x$. (v) Every prime ideal of $A$ is the contraction of a prime ideal of $A[[x]]$.

Proof) We first prove some elementary facts. If $R,S$ are commutative rings and $\varphi :R\to S$ is a surjective ring homomorphism, the following holds: (1) The image of an ideal is an ideal. (2) The preimage of a maximal ideal is maximal. (3) If $I$ is an ideal of $R$ and $\ker \varphi \subset I$, then $R/I\cong S/\varphi (I)$. Consider the projection $\pi :S\to S/\varphi (I)$ and the composition ${\varphi }^{\prime }=\pi \circ \varphi$. The kernel of ${\varphi }^{\prime }$ is clearly $I$ and ${\varphi }^{\prime }$ is surjective, proving the isomorphism. Note without the $\ker \varphi \subset I$ hypothesis, we cannot say that the kernel of ${\varphi }^{\prime }$ is $I$. (4) If $I$ is a maximal/prime ideal of $R$ and $\ker \varphi \subset I$, then the image $\varphi (I)$ is maximal/prime, respectively. Note that if $I$ is maximal and $\ker \varphi \notin I$, then $\varphi (I)=\varphi (I+\ker \varphi )=\varphi (R)=S$. We now proceed with the problem at hand. (i) is trivial. (ii) We assume $f$ is nilpotent and that $f^k=0$ for some $k$. Clearly, this implies $a_0^k=0$ so $a_0$ is nilpotent. Then, ${\sum }_{n=1}^{\infty }{a}_n{x}^n$ is nilpotent as it is the difference of nilpotent elements (note this argument is valid though $A$ is non-commutative as the two nilpotent elements commute). This forces ${\sum }_{n=0}^{\infty }{a}_{n+1}{x}^n$ to be nilpotent as well, as $x$ is not a zero divisor in $A[[x]]$. Therefore, $a_1$ is nilpotent. Repeating this argument gives us the desired result. (iii) Let $\varphi :A[[x]]\to A$ be such that $f\mapsto a_0$ (the $x^0$ coefficient map). This is a surjective ring homomorphism, which means the preimage of maximal ideals are maximal ideals. Thus, the Jacobson radical of $A[[x]]$ is a subset of the preimage of the Jacobson radical of $A$. $x$ is in the Jacobson radical of $A[[x]]$ as for any $g\in A[[x]]$, $1-xg$ is a unit by (1). As $\ker \varphi =(x)$ and every maximal ideal contains $(x)$, this implies the image of any maximal ideal is a maximal ideal of $A$. Therefore, the Jacobson radical of $A$ is contained in the image of the Jacobson radical of $A[[x]]$. This gives us the desired result. (iv) is a trivial corollary of what we derived in (iii). (v) Let $I$ be a prime ideal of $A$. $I$ is the contraction of $I+(x)$ in $A[[x]]$, which is obviously prime.

Problem 4 (1.14). In a ring $A$, let $\Sigma$ be the set of all ideals in which every element is a zero-divisor. Show that the set $\Sigma$ has maximal elements and that every maximal element of $\Sigma$ is a prime ideal. Hence the set of zero-divisors in $A$ is the union of prime ideals.

Proof) $\Sigma$ has maximal elements by Zorn’s lemma; let $I$ be one such ideal. Assume, FTSOC, that $ab\in I$ while $a,b\notin I$. Take $c,d\in I$ and $x,y\in A$ such that $xa+c$ and $yb+d$ are not zero divisors; then $(xa+c)(yb+d)\in I$ and must be a zero divisor. As the product of non-zero divisors must be a non-zero divisor, we are done.

Problem 5 (1.15). Let $A$ be a ring and let $X$ be the set of all prime ideals of $A$. For each subset $E$ of $A$, let $V(E)$ denote the set of all prime ideals of $A$ which contain $E$. Prove that i) if $\mathfrak{a}$ is the ideal generated by $E$, then $V(E)=V(\mathfrak{a})=V(r(\mathfrak{a}))$. ii) $V(0)=X$, $V(1)=\varnothing$. iii) if $(E_i{)}_{i\in I}$ is any family of subsets of $A$, then

$$V\left(\bigcup _{i\in I}{E}_i\right)=\bigcap _{i\in I}V(E_i).$$

iv) $V(\mathfrak{a}\cap \mathfrak{b})=V(\mathfrak{ab})=V(\mathfrak{a})\cup V(\mathfrak{b})$ for any ideals $\mathfrak{a}$, $\mathfrak{b}$ of $A$. These results show that the sets $V(E)$ satisfy the axioms for closed sets in a topological space. The resulting topology is called the Zariski topology. The topological space $X$ is called the prime spectrum of $A$, and is written $\mathrm{Spec}(A)$.

Proof) (i), (ii), and (iii) are all trivial. For (iv), it follows from definition that $V(\mathfrak{a})\cup V(\mathfrak{b})\subset V(\mathfrak{a}\cap \mathfrak{b})$ and $V(\mathfrak{ab})\supset V(\mathfrak{a}\cap \mathfrak{b})$; therefore, it suffices to show $V(\mathfrak{ab})\subset V(\mathfrak{a})\cup V(\mathfrak{b})$. Let $I$ be a prime ideal that contains $\mathfrak{ab}$. For any $a\in \mathfrak{a}$ and $b\in \mathfrak{b}$, $ab\in I$ implies $a\in I$ or $b\in I$. Therefore, $I$ must contain either $\mathfrak{a}$ or $\mathfrak{b}$, as desired.

Problem 6 (1.16). What are prime spectra of $\mathbb{Z}$, $\mathbb{R}$, $\mathbb{C}[x]$, and $\mathbb{R}[x]$?

Proof) $\mathbb{Z}$, $\mathbb{R}$, $\mathbb{C}[x]$, and $\mathbb{R}[x]$ are all PIDs, so the prime ideals are the principal ideals generated by irreducible elements along with the zero ideal, and all nonzero prime ideals are maximal. This means that all nonzero prime ideals are closed sets, as if $I$ is a nonzero prime ideal, $V(I)=\{I\}$. Finite collections of nonzero prime ideals are automatically closed, as well. Furthermore, the only closed set that can contain the zero ideal is the entire set $X$, by the way we defined the Zariski topology. For any subset $E$, if $E$ is empty, zero, or all units then $V(E)$ becomes the entire set $X$ or $\varnothing$. Otherwise, $E$ contains some non-unit element $f$, which has finitely many irreducible divisors as a PID is a UFD. Thus, $V(E)$ is a finite collection of nonzero prime ideals. This exhausts the closed sets of the spectra, being $\varnothing$, $X$, and finite collections of nonzero prime ideals.

Problem 7. (1.17) Let $A$ be a ring and $X_f$ denote the complement of $V(f)$ in $\mathrm{Spec}(A)$. The sets $X_f$ are open. Show that they form a basis of open sets for the Zariski topology, and that i) $X_f\cap X_g=X_{fg}$ ii) $X_f=\varnothing ⟺f$ is nilpotent iii) $X_f=X⟺$ $f$ is a unit iv) $X_f=X_g⟺$ $r((f))=r((g))$

Proof) (i) follows from the fact that a prime ideal that doesn’t contain $f$ or $g$ cannot contain $fg$, and a prime ideal that doesn’t contain $fg$ cannot contain $f$ or $g$. (ii) is equivalent to every prime ideal containing $f$, which is true iff $f$ is in the nilradical. (iii) is equivalent to no prime ideal containing $f$; if $f$ is a unit, clearly $X_f=X$. If not, by Zorn’s lemma there is a maximal ideal containing $(f)$. (iv) is equivalent to a prime ideal containing $f$ iff it contains $g$. Note that a prime ideal contains $f$ iff it contains $r((f))$, and the same applies to $g$. The theorem follows from the fact that, for any ideal $I\subset A$, the radical $\sqrt{I}$ is precisely the intersection of all prime ideals that contain $I$, the correspondence between prime ideals containing $I$ and prime ideals in $A/I$.