Commutative Algebra

• The important assumption is \(a(x)\not\in R[x]\), we’ll assume \(R\) is a UFD and try to contradict this.
• Write \(f(x) = a(x)b(x)\in F[x]\), then if \(R\) is a UFD we have \(r,s\in F\) such that \(f(x) = ra(x) sb(x) \in R[x]\).
• Since \(a(x), b(x)\) are monic and \(f=ab\), \(f\) is monic, and by the factorization in \(R[x]\) we have \(rs=1\). So \(r,s\in R^{\times}\).
• Then using that \(ra(x)\in R[x]\), we have \(r^{-1}ra(x) = a(x)\in R[x]\). \(\contradiction\)

• Set \(R = {\mathbf{Z}}[2\sqrt 2], F = {\mathbf{Q}}[2\sqrt 2]\).
• Let \(p(x) \coloneqq x^2-2 \in R[x]\) which splits as \(p(x) = (x+ \sqrt{2} )(x - \sqrt{2} ) \coloneqq a(x) b(x) \in F[x]\).
• Note neither \(a(x), b(x)\) are in \(R[x]\).
  • Explicitly, every monic linear \(p\in R[x]\) is of the form \(x + 2t\sqrt{2}\) with \(t\in {\mathbf{Z}}\), and \(\pm \sqrt{2} \neq 2t\sqrt{2}\) for any \(t\).
• So we have \(p(x) \in R[x]\) splitting as \(p=ab\) in \(F[x]\) with \(a\not \in R[x]\), so part (a) applies.

Define the set of proper ideals \begin{align*} S = \left\{{J {~\mathrel{\Big\vert}~}I \subseteq J < R}\right\} ,\end{align*}

which is a poset under set inclusion.

Given a chain \(J_1 \subseteq \cdots\), there is an upper bound \(J \coloneqq\cup J_i\), so Zorn’s lemma applies.

\(\implies\):

• We will show that \(x\in J(R) \implies 1+x \in R^{\times}\), from which the result follows by letting \(x=rx\).

• Let \(x\in J(R)\), so it is in every maximal ideal, and suppose toward a contradiction that \(1+x\) is not a unit.

• Then consider \(I = \left\langle{1+x}\right\rangle {~\trianglelefteq~}R\). Since \(1+x\) is not a unit, we can’t write \(s(1+x) = 1\) for any \(s\in R\), and so \(1 \not\in I\) and \(I\neq R\)

• So \(I < R\) is proper and thus contained in some maximal proper ideal \(\mathfrak{m} < R\) by part (1), and so we have \(1+x \in \mathfrak{m}\). Since \(x\in J(R)\), \(x\in \mathfrak{m}\) as well.

• But then \((1+x) - x = 1 \in \mathfrak{m}\) which forces \(\mathfrak{m} = R\).

\(\impliedby\)

• Fix \(x\in R\), and suppose \(1+rx\) is a unit for all \(r\in R\).

• Suppose towards a contradiction that there is a maximal ideal \(\mathfrak{m}\) such that \(x\not \in \mathfrak{m}\) and thus \(x\not\in J(R)\).

• Consider \begin{align*} M' \coloneqq\left\{{rx + m {~\mathrel{\Big\vert}~}r\in R,~ m\in M}\right\} .\end{align*}

• Since \(\mathfrak{m}\) was maximal, \(\mathfrak{m} \subsetneq M'\) and so \(M' = R\).

• So every element in \(R\) can be written as \(rx + m\) for some \(r\in R, m\in M\). But \(1\in R\), so we have \begin{align*} 1 = rx + m .\end{align*}

• So let \(s = -r\) and write \(1 = sx - m\), and so \(m = 1 + sx\).

• Since \(s\in R\) by assumption \(1+sx\) is a unit and thus \(m \in \mathfrak{m}\) is a unit, a contradiction.

• So \(x\in \mathfrak{m}\) for every \(\mathfrak{m}\) and thus \(x\in J(R)\).

\(\mathfrak N(R) \subseteq J(R)\):

• Use the fact \(x\in \mathfrak N(R) \implies x^n = 0 \implies 1 + rx\) is a unit \(\iff x\in J(R)\) by (b): \begin{align*} \sum_{k=1}^{n-1} (-x)^k = \frac{1 - (-x)^n}{1- (-x)} = (1+x)^{-1} .\end{align*}

\(J(R) \subseteq \mathfrak N(R)\):

• Let \(x \in J(R) \setminus \mathfrak N(R)\).

• Since \(R\) is finite, \(x^m = x\) for some \(m > 0\).

• Without loss of generality, we can suppose \(x^2 = x\) by replacing \(x^m\) with \(x^{2m}\).

• If \(1-x\) is not a unit, then \(\left\langle{1-x}\right\rangle\) is a nontrivial proper ideal, which by (a) is contained in some maximal ideal \({\mathfrak{m}}\). But then \(x\in {\mathfrak{m}}\) and \(1-x \in {\mathfrak{m}}\implies x + (1-x) = 1 \in {\mathfrak{m}}\), a contradiction.

• So \(1-x\) is a unit, so let \(u = (1-x)^{-1}\).

• Then \begin{align*} (1-x)x &= x - x^2 = x - x = 0 \\ &\implies u (1-x)x = x = 0 \\ &\implies x=0 .\end{align*}

• Suppose \(c\coloneqq f(a)\neq 0\), noting that \(c\) may not be positive.
• By continuity, pick \({\varepsilon}\) small enough so that \({\left\lvert {x-a} \right\rvert}<{\varepsilon}\implies {\left\lvert {f(x) - f(a)} \right\rvert} < c/2\). Since we’re on the interval, we have \(f(x) \in (f(a) - c/2, f(a) + c/2) = (c/2, 3c/2)\) which is a ball of radius \(c/2\) about \(c\), which thus doesn’t intersect \(0\).
• So \(f(x) \neq 0\) on this ball, and \(g \coloneqq f^2 > 0\) on it. Note that ideals are closed under products, so \(g\in I\)
• Moreover \(f^2(x) \geq 0\) since squares are non-negative, so \(g\geq 0\) on \([0, 1]\).

• By contrapositive, suppose \(V(I)= \emptyset\), we’ll show \(I\) contains a unit and thus \(I=R\).
• For each fixed \(x\in [0, 1]\), since \(V(I)\) is empty there is some \(f_x\) such that \(f_x(x) \neq 0\).
• By (a), there is some \(g_x\) with \(g_x(x) > 0\) on a neighborhood \(U_x\ni x\) and \(g_x \geq 0\) everywhere.
• Ranging over all \(x\) yields a collection \(\left\{{(g_x, U_x) {~\mathrel{\Big\vert}~}x\in [0, 1]}\right\}\) where \(\left\{{U_x}\right\}\rightrightarrows[0, 1]\).
• By compactness there is a finite subcover, yielding a finite collection \(\left\{{(g_k, U_k)}\right\}_{k=1}^n\) for some \(n\).
• Define the candidate unit as \begin{align*} G(x) \coloneqq{1\over \sum_{k=1}^n g_k(x)} .\end{align*}
• This is well-defined: fix an \(x\), then the denominator is zero at \(x\) iff \(g_k(x) = 0\) for all \(k\). But since the \(U_k\) form an open cover, \(x\in U_\ell\) for some \(\ell\), and \(g_\ell > 0\) on \(U_\ell\).
• Since ideals are closed under sums, \(H\coloneqq{1\over G} \coloneqq\sum g_k \in I\). But \(H\) is clearly a unit since \(HG = \operatorname{id}\).

• If \(I{~\trianglelefteq~}R\) is maximal, \(I\neq R\), and so by (b) we have \(V(I) \neq \emptyset\).
• So there is some \(x_0\in[0, 1]\) with \(f(x_0) = 0\) for all \(f\in I\).
• Define \({\mathfrak{m}}_{x_0} \coloneqq\left\{{f\in R {~\mathrel{\Big\vert}~}f(x_0) = 0}\right\}\), which is clearly an ideal.
  • This is a proper ideal, since constant nonzero functions are continuous and thus in \(R\), not not \({\mathfrak{m}}_{x_0}\).
• We thus have \(I \subseteq {\mathfrak{m}}_{x_0}\), and by maximality they are equal.

• Let \(\phi\) denote the map in question, it suffices to show that \(\phi\) is \(R{\hbox{-}}\)linear, i.e. \(\phi(s\mathbf{x} + \mathbf{y}) = s\phi(\mathbf{x}) + \phi(\mathbf{y})\): \begin{align*} \phi(s\mathbf{x} + \mathbf{y}) &= r(s\mathbf{x} + \mathbf{y}) \\ &= rs\mathbf{x} + r\mathbf{y} \\ &= s(r\mathbf{x}) + (r\mathbf{y}) \\ &= s\phi(\mathbf{x}) + \phi(\mathbf{y}) .\end{align*}

Let \(\phi_r(x) \coloneqq rx\) be the multiplication map.

• Let \(x\in \ker \phi_r \coloneqq\left\{{x\in R {~\mathrel{\Big\vert}~}rx = 0}\right\}\).

• Since \(R\) is commutative \(0 = rx = xr\), and so \(r\in \ker \phi_x\), so \(\ker \phi_x \neq 0\) and \(x\) is a zero divisor by definition.

• Let \(y\in \operatorname{im}\phi_r \coloneqq\left\{{y \coloneqq rx {~\mathrel{\Big\vert}~}x\in R}\right\}\), we want to show \(\ker \phi_y\) is nontrivial by producing some \(z\) such that \(yz=0\). Write \(y\coloneqq rx\) for some \(x\in R\).

• Since \(r\) is a zero divisor, we can produce some \(z\neq 0 \in \ker \phi_r\), so \(rz = 0\).

• Now using that \(R\) is commutative, we can compute \begin{align*} yz = (rx)z = (xr)z = x (rz) = x(0) = 0 ,\end{align*} so \(z\in \ker \phi_y\).

• Let \(Z \coloneqq\left\{{z_i}\right\}_{i=1}^n\) be the set of \(n\) zero divisors in \(R\).

• Let \(\phi_i\) be the \(n\) maps \(x \mapsto z_i x\), and let \(K_i = \ker \phi_i\) be the corresponding kernels.

• Fix an \(i\).

• By (b), \(K_i\) consists of zero divisors, so \begin{align*} {\left\lvert {K_i} \right\rvert} \leq n < \infty \quad \text{for each } i .\end{align*}

• Now consider \(R/K_i \coloneqq\left\{{r + K_i}\right\}\).

• By the first isomorphism theorem, \(R/K_i \cong \operatorname{im}\phi\), and by (b) every element in the image is a zero divisor, so \begin{align*} [R: K_i] = {\left\lvert {R/K_i} \right\rvert} = {\left\lvert {\operatorname{im}\phi_i} \right\rvert} \leq n .\end{align*}

• But then \begin{align*} {\left\lvert {R} \right\rvert} = [R:K_i]\cdot {\left\lvert {K_i} \right\rvert} \leq n\cdot n = n^2 .\end{align*}

• By (c), if there are exactly 2 zero divisors then \({\left\lvert {R} \right\rvert} \leq 4\). Since every element in a finite ring is either a unit or a zero divisor, and \({\left\lvert {R^{\times}} \right\rvert} \geq 2\) since \(\pm 1\) are always units, we must have \({\left\lvert {R} \right\rvert} = 4\).

• Since the characteristic of a ring must divide its size, we have \(\operatorname{ch}R = 2\) or \(4\).

• Using the hint, we see that only \({\mathbf{Z}}/(4)\) has characteristic 4, which has exactly 2 zero divisors given by \([0]_4\) and \([2]_4\).

• If \(R\) has characteristic 2, we can check the other 3 possibilities.

• We can write \({\mathbf{Z}}/(2)[t]/(t^2) = \left\{{a + bt {~\mathrel{\Big\vert}~}a,b\in {\mathbf{Z}}/(2)}\right\}\), and checking the multiplication table we have \begin{align*} \begin{array}{c|cccc} & 0 & 1 & t & 1+t \\ \hline 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & t & 1+t \\ t & 0 & t & \mathbf{0} & t \\ 1 + t & 0 & 1+t & t & 1 \\ \end{array} ,\end{align*}

  and so we find that \(t, 0\) are the zero divisors.

• In \({\mathbf{Z}}/(2)[t]/(t^2 - t)\), we can check that \(t^2 = t \implies t t^2 = t^2 \implies t(t^2 + 1) = 0 \implies t(t+1) = 0\), so both \(t\) and \(t+1\) are zero divisors, along with zero, so this is not a possibility.

• Similarly, in \({\mathbf{Z}}/(2)[t]/(t^2 + t + 1)\), we can check the bottom-right corner of the multiplication table to find \begin{align*} \left[\begin{array}{c|cc} & t & 1 +t \\ \hline t & 1+t & 1 \\ t & 1 & t \\ \end{array}\right] ,\end{align*}

  and so this ring only has one zero divisor.

• Thus the only possibilities are: \begin{align*} R &\cong {\mathbf{Z}}/(4) \\ R &\cong {\mathbf{Z}}/(2)[t] / (t^2) .\end{align*}

• Maximal: a proper ideal \(I{~\trianglelefteq~}R\), so \(I\neq R\), such that if \(J\supseteq I\) is any other ideal, \(J = R\).
• Existence of a maximal ideal: use that \(0\in \operatorname{Id}(R)\) always, so \(S\coloneqq\left\{{I\in \operatorname{Id}(R) {~\mathrel{\Big\vert}~}I\neq R}\right\}\) is a nonempty poset under subset inclusion. Applying the usual Zorn’s lemma argument produces a maximal element.

\(\impliedby\): By contrapositive: if \(r\in R\) is a unit and \({\mathfrak{m}}\) is maximal, then \(r\in {\mathfrak{m}}\implies {\mathfrak{m}}= R\), contradicting that \({\mathfrak{m}}\) is proper.

\(\implies\):

• Suppose \(a\) is not a unit, we’ll produce a maximal ideal containing it.
• Let \(I\coloneqq Ra\) be the principal ideal generated by \(a\), then \(Ra \neq R\) since \(a\) is not a unit.
• Take a poset \(S \coloneqq\left\{{J\in \operatorname{Id}(R) {~\mathrel{\Big\vert}~}J\supseteq Ra, J\neq R}\right\}\) ordered by set inclusion.

  • Let \(C_*\) be a chain in \(S\), set \(\widehat{C} \coloneqq\cup C_i\). Then \(\widehat{C} \in S\):

    • \(\widehat{C} \neq R\), since if so it contains a unit, forcing some \(C_i\) to contain a unit and thus equal \(R\).
    • \(\widehat{C} \supseteq Ra\), since e.g. \(\widehat{C} \supseteq C_1 \supseteq Ra\).
    • \(\widehat{C}\) is an ideal since \(xy\in \widehat{C} \implies x\in C_i, y\in C_j\) and \(C_i \subseteq C_j\) without loss of generality, so \(xy\in C_j \subseteq \widehat{C}\).
• Then \(Ra \subseteq \widehat{C}\), some maximal ideal.

• Write \(I \coloneqq\operatorname{Ann}(u)\) for some \(u\), and toward a contradiction suppose \(ab\in I\) but \(a,b\not\in I\).
• Then \(abu=0\) but \(au\neq 0, bu\neq 0\).
• Since \(abu=0\), we have \(a\in \operatorname{Ann}(bu)\). Note that \(\operatorname{Ann}(bu) \supseteq\operatorname{Ann}(u)\), since \(su = 0\implies bsu=sbu=0\).
• We can’t have \(\operatorname{Ann}(bu) = R\), since if \(sbu=0\) for all \(s\in R\), so we could take \(s=1\) to get \(bu=0\) and \(b\in \operatorname{Ann}(u)\).
• By maximality, this forces \(\operatorname{Ann}(u) = \operatorname{Ann}(bu)\), so \(sbu = 0 \implies su=0\) for any \(s\in R\).
• Now take \(s=a\), and since \(abu=0\) we get \(au=0\) and \(a\in \operatorname{Ann}(u)\). \(\contradiction\)

• Let \(\mathfrak{p}\) be prime and \(x\in N\).
• Then \(x^k = 0 \in \mathfrak{p}\) for some \(k\), and thus \(x^k = x x^{k-1} \in \mathfrak p\).
• Since \(\mathfrak p\) is prime, inductively we obtain \(x\in\mathfrak p\).

• Let \(S = \left\{{r^k \mathrel{\Big|}k\in {\mathbb{N}}}\right\}\) be the set of positive powers of \(r\).

• Then \(S^2 \subseteq S\), since \(r^{k_1}r^{k_2} = r^{k_1+k_2}\) is also a positive power of \(r\), and \(0\not\in S\) since \(r\neq 0\) and \(r\not\in N\).

• By the corollary, \(\left\{{I {~\trianglelefteq~}R {~\mathrel{\Big\vert}~}I\cap S = \emptyset}\right\}\) has a maximal element \(\mathfrak{p}\).

• Since \(R\) is commutative, \(\mathfrak{p}\) is prime.

• Suppose \(R\) has a unique prime ideal \(\mathfrak{p}\).

• Suppose \(r\in R\) is not a unit, and toward a contradiction, suppose that \(r\) is also not nilpotent.

• Since \(r\) is not a unit, \(r\) is contained in some maximal (and thus prime) ideal, and thus \(r \in \mathfrak{p}\).

• Since \(r\not\in N\), by (b) there is a maximal ideal \(\mathfrak{m}\) that avoids all positive powers of \(r\). Since \(\mathfrak{m}\) is prime, we must have \(\mathfrak{m} = \mathfrak{p}\). But then \(r\not\in \mathfrak{p}\), a contradiction.

\(\implies\):

• Suppose \(\operatorname{Id}(R) = \left\{{\left\langle{0}\right\rangle, \left\langle{1}\right\rangle}\right\}\). Then for any nonzero \(r\in R\), the ideal \(\left\langle{r}\right\rangle = \left\langle{1}\right\rangle\) is the entire ring.
• In particular, \(1\in \left\langle{r}\right\rangle\), so we can write \(a = tr\) for some \(t\in R\).
• But then \(r\in R^{\times}\) with \(t\coloneqq r^{-1}\).

\(\impliedby\):

• Suppose \(R\) is a field and \(I\in \operatorname{Id}(R)\) is an ideal.
• If \(I \neq \left\langle{0}\right\rangle\) is proper and nontrivial, then \(I\) contains at least one nonzero element \(a\in I\).
• Since \(R\) is a field, \(a^{-1}\in R\), and \(aa^{-1}= 1\in I\) forces \(I = \left\langle{1}\right\rangle\).

• Let \(J{~\trianglelefteq~}R\) be a nonzero two-sided ideal, noting that \(R\) is noncommutative – the claim is that \(J\) contains \(I_n\), the \(n\times n\) identity matrix, and thus \(J = R\).
• Pick a nonzero element \(M\in I\), then \(M\) has a nonzero entry \(m{ij}\).
• Let \(A_{ij}\) be the matrix with a 1 in the \(i,j\) position and zeros elsewhere.
  • Left-multiplying \(A_{ij}M\) moves row \(j\) to row \(i\) and zeros out the rest of \(M\).
  • Right-multiplying \(MA_{ij}\) moves column \(i\) to column \(j\) and zeros out the rest.
  • So \(A_{ij} M A_{kl}\) moves entry \(j, k\) to \(i, l\) and zeros out the rest.
• So define \(B \coloneqq A_{i, i}MA_{j, i}\), which movies \(m_{ij}\) to the \(i,i\) position on the diagonal and has zeros elsewhere.
• Then \(m_{ij}^{-1}{\varepsilon}_{ii} \coloneqq A_{ii}\) is a matrix with \(1\) in the \(i, i\) spot for any \(i\). Since \(I\) is an ideal, we have \({\varepsilon}_{ii}\in I\) for every \(i\).
• We can write the identity \(I_n\) as \(\sum_{i=1}^n {\varepsilon}_{ii}\), so \(I_n \in I\) and \(I=R\).

• Existence: the claim is that \(2R \coloneqq\left\{{2y {~\mathrel{\Big\vert}~}y\in R}\right\}\) is a nontrivial two-sided ideal of \(R\), forcing \(2R = R\) by simpleness.
  • That \(2R \neq 0\) follows from condition (1): Provided \(y\neq 0\), we have \(2y\neq 0\), and so if \(R\neq 0\) then there exists some nonzero \(a\in R\), in which case \(2a\neq 0\) and \(2a\in 2R\).
  • That \(2R\) is a right ideal: clear, since \((2y)\cdot r = 2(yr)\in 2R\).
  • That \(2R\) is a left ideal: use that multiplication is distributive: \begin{align*} r\cdot 2y \coloneqq r(y+y) = ry + ry \coloneqq 2(ry) \in 2R .\end{align*}
• So \(2R = R\) by simpleness.
• Uniqueness:
  • Use the contrapositive of condition (1), so that \(2x = 0 \implies x=0\).
  • Suppose toward a contradiction that \(x=2y_1 = 2y_2\), then \begin{align*} 0 = x-x = 2y_1 - 2y_2 = 2(y_1 - y_2) \implies y_1 - y_2 = 0 \implies y_1 = y_2 .\end{align*}

• First we’ll show \(z=2(yz)\): \begin{align*} xy + xy &= x \\ \implies xy + xy - x &= 0 \\ \implies xyz + xyz - xz &= 0 \\ \implies x(yz + yz - z) &= 0 \\ \implies yz + yz - z &= 0 && \text{since } x\neq 0 \text{ and no zero divisors }\\ \implies 2(yz) &= z .\end{align*}

• Now we’ll show \(z=2(zy)\): \begin{align*} yz + yz &= z \\ \implies zyz + zyz &= zz \\ \implies zyz + zyz - zz &= 0 \\ \implies (zy + zy - z)z &= 0\\ \implies z=0 \text{ or } zy+zy-z &= 0 && \text{ no zero divisors } .\end{align*}

• Then if \(z=0\), we have \(yz = 0 = zy\) and we’re done.

• Otherwise, \(2(zy) = z\), and thus \begin{align*} 2(zy) = z = 2(yz) \implies 2(zy - yz) = 0 \implies zy-yz = 0 ,\end{align*} so \(zy=yz\).

• If \(1\in R\), \begin{align*} 2xy &= x \\ \implies 2xy-x &= 0 \\ \implies x(2y-1) &= 0 \\ \implies 2y-1 &= 0 && x\neq 0 \text{ and no zero divisors}\\ \implies 2y &= 1 .\end{align*}
• Now use \begin{align*} 1\cdot z &= z\cdot 1 \\ \implies (2y)z &= z(2y) \\ \implies (y+y)z &= z(y+y) \\ \implies yz+yz &= zy+zy \\ \implies 2(yz) &= 2(zy) \\ \implies 2(yz-zy) &= 0 \\ \implies yz-zy &= 0 \\ ,\end{align*} using condition (2).

\begin{align*} 2a = (2a)^2 = 4a^2 = 4a \implies 2a = 0 .\end{align*} Note that this implies \(x = -x\) for all \(x\in R\).

\begin{align*} a+b = (a+b)^2 &= a^2 + ab + ba + b^2 = a + ab + ba + b \\ &\implies ab + ba = 0 \\ &\implies ab = -ba \\ &\implies ab = ba \quad\text{by (a)} .\end{align*}