Rings and modules

Let \[ J_1\supseteq J_2\supseteq J_3\supseteq\cdots \] be a descending chain of left ideals of \(R\). Since \(f:A\to R\) is surjective, each preimage \(f^{-1}(J_i)\) is a left ideal of \(A\), and \[ f^{-1}(J_1)\supseteq f^{-1}(J_2)\supseteq f^{-1}(J_3)\supseteq\cdots. \] Because \(A\) is left Artinian, there is \(N\) such that \[ f^{-1}(J_i)=f^{-1}(J_N) \qquad(i\ge N). \] Applying \(f\) and using surjectivity gives \[ J_i=f(f^{-1}(J_i))=f(f^{-1}(J_N))=J_N \qquad(i\ge N). \] Thus every descending chain of left ideals of \(R\) stabilizes.

Let \(A\) be an Artinian integral domain and let \(0\ne a\in A\). The descending chain \[ (a)\supseteq(a^2)\supseteq(a^3)\supseteq\cdots \] stabilizes, so for some \(n\ge1\), \[ (a^n)=(a^{n+1}). \] Hence \(a^n\in(a^{n+1})\), so there exists \(b\in A\) with \[ a^n=a^{n+1}b. \] Because \(A\) is a domain and \(a^n\ne0\), cancellation gives \[ 1=ab. \] Thus every nonzero element of \(A\) is a unit, so \(A\) is a field.

Let \(\mathfrak p\) be a prime ideal of the commutative Artinian ring \(A\). The quotient \(A/\mathfrak p\) is an integral domain. By <1>1, applied to the quotient map \[ A\twoheadrightarrow A/\mathfrak p, \] the quotient is Artinian. By <1>2 it is therefore a field. Hence \(\mathfrak p\) is maximal.

Coefficient reduction \(\mathbb Z[x]\to\mathbb F_2[x]\) is surjective and has kernel \((2)\). For any \(f\in\mathbb Z[x]\), the inverse image of \((\overline f)\) is \((2,f)\): if \(\overline h=\overline f\,\overline a\), lift \(\overline a\) to an integer polynomial \(a\); then \(h-fa\in(2)\). The reverse inclusion follows immediately by reduction. Thus the induced quotient maps are isomorphisms. The polynomials in the problem reduce to \(q\) and \(x^3+x^2+x=xq\), respectively.

The polynomial \(q\) has neither \(0\) nor \(1\) as a root. A reducible quadratic over a field has a linear factor and hence a root, so \(q\) is irreducible. Polynomial division gives unique representatives \(a+bx\), with \(a,b\in\mathbb F_2\), in \(\mathbb F_2[x]/(q)\). Let \(\alpha\) be the class of \(x\). Its defining relation is \(\alpha^2=\alpha+1\), and \(\alpha^3=\alpha(\alpha+1)=1\). The three nonzero elements are \(1,\alpha,\alpha^2\) and are all invertible, proving directly that \(E\) is a field of four elements.

It has characteristic \(2\), additive group \(C_2\times C_2\), and multiplicative group \(E^\times=\langle\alpha\rangle\cong C_3\). Its only ideals are \((0)\) and \(E\), since a nonzero ideal contains a unit. In particular \(I\) is maximal and prime. The only idempotents of \(E\) are \(0,1\), as \(e^2=e\) implies \(e(e-1)=0\) in a field; it has no nonzero zero divisors or nilpotents.

Define \[ \Theta:B\longrightarrow\mathbb F_2\times E, \qquad [h(x)]\longmapsto(h(0),h(\alpha)). \] Both evaluations are ring homomorphisms, and \(xq\) vanishes at \(0\) and \(\alpha\), so the map is well defined. If \(\Theta([h])=0\), then \(x\mid h\) and \(q\mid h\). The identity \(q-x(x+1)=1\) shows that \(x\) and \(q\) are coprime. More explicitly, write \(h=xk\). Modulo \(q\), the class of \(x\) is the unit \(\alpha\), so \(q\mid k\). Thus \(xq\mid h\), proving injectivity.

Let \[ e_0=[q],\qquad e_1=[x^2+x]. \] Then \(\Theta(e_0)=(1,0)\) and \(\Theta(e_1)=(0,1)\). For \(a,b,c\in\mathbb F_2\), the element \[ a e_0+(b+c[x])e_1 \] maps to \((a,b+c\alpha)\). This proves surjectivity and gives the inverse on every element of the product. Consequently \(B\cong\mathbb F_2\times E\) as rings.

The ring \(B\) has eight elements, characteristic \(2\), and additive group \(C_2^3\). Equivalently, polynomial division gives the basis \(1,[x],[x^2]\) over \(\mathbb F_2\). A pair \((a,b)\in\mathbb F_2\times E\) is a unit exactly when \(a=1\) and \(b\ne0\), so \(B^\times\cong E^\times\cong C_3\). A nonzero pair is a zero divisor exactly when at least one component is zero: multiply by a nonzero pair supported in that component to obtain zero; when both components are nonzero the pair is a unit. Thus there are four nonzero zero divisors, namely \((1,0)\) and the three \((0,b)\) with \(b\ne0\).

Every ideal \(L\) in the product decomposes into its two coordinate ideals: for \((a,b)\in L\), multiplication by \((1,0)\) and \((0,1)\) puts \((a,0)\) and \((0,b)\) in \(L\). Each coordinate ideal in a field is zero or the whole field. Therefore the complete ideal list in \(B\) is \[ (0),\quad (e_0),\quad(e_1),\quad B. \] The middle two are maximal, with quotients \(E\) and \(\mathbb F_2\), respectively, and are the only prime ideals: the zero ideal is not prime because \(e_0e_1=0\) with both factors nonzero. Hence \(B\) is neither a field nor a domain nor a local ring, and \(J\) is neither prime nor maximal.

The only idempotents are \(0,e_0,e_1,1\), because each coordinate of an idempotent in a field is \(0\) or \(1\). Finally, a nilpotent pair has nilpotent coordinates, which must both be zero. Thus \(B\) is reduced despite having nonzero zero divisors.

The coefficients \(b\) and \(-a\) are nonzero, so a basis of the ideal has at most one element; an ideal with basis \(\{f\}\) equals \((f)\).

\(R/I\cong\ZZ\) is an integral domain but not a field, so \(I\) is prime but not maximal; indeed \((x)\subsetneq(2,x)\subsetneq R\).

\(M\) is generated by \(x,y\), and it is torsion-free as a nonzero ideal of an integral domain. If \((x,y)=(f)\), then \(f\) divides \(x\) and \(y\), so \(f\) is a nonzero constant and \((x,y)=R\), contradicting that every element of \((x,y)\) vanishes at \((0,0)\). By step <1>1, \(M\) is not free.

For \(n\ne0\) and \(q\ne0\), \(nq\ne0\), so \(\QQ\) is torsion-free. Any two nonzero rationals \(p/q\) and \(r/s\) satisfy \((qr)\cdot\frac pq-(ps)\cdot\frac rs=0\) with nonzero integer coefficients, so a basis has at most one element; but \(\QQ\) is not cyclic, since \(\frac{1}{2}\cdot\frac pq\notin\ZZ\frac pq\) for \(p/q\ne0\). Hence \(\QQ\) is not free.

A Euclidean domain is a principal ideal domain, since a nonzero element of minimal Euclidean norm in a nonzero ideal generates it, and a principal ideal domain is a unique factorization domain [@DF04].

By Gauss’s lemma, \(\ZZ[x]\) is a UFD because \(\ZZ\) is [@DF04]. If \((2,x)=(f)\), then \(f\mid2\) forces \(f\in\{\pm1,\pm2\}\); \(f=\pm1\) is impossible because every element of \((2,x)\) has even constant term, and \(f=\pm2\) is impossible because \(x\notin(2)\). So \(\ZZ[x]\) is not a PID.

\(R\) is an integral domain, and the chain \((x_1)\subsetneq(x_1,x_2)\subsetneq(x_1,x_2,x_3)\subsetneq\cdots\) does not stabilize: \(x_{n+1}\notin(x_1,\ldots,x_n)\), since setting \(x_1=\cdots=x_n=0\) sends that ideal to \(0\) and \(x_{n+1}\) to \(x_{n+1}\ne0\).

Steps <1>2 through <1>7 answer parts (a) through (f).

We have \(0 = 0 + 0 \in I+J\). For \(x_1+y_1, x_2+y_2 \in I+J\), \((x_1+y_1) - (x_2+y_2) = (x_1-x_2) + (y_1-y_2) \in I+J\). For any \(r \in R\), \(r(x+y) = rx + ry \in I+J\) and \((x+y)r = xr + yr \in I+J\), because \(I\) and \(J\) are ideals.

For \(x \in I\), \(x = x + 0 \in I+J\) since \(0 \in J\); similarly, for \(y \in J\), \(y = 0 + y \in I+J\).

For \(x \in I \subseteq M\) and \(y \in J \subseteq M\), closure of \(M\) under addition gives \(x + y \in M\).

Steps <2>1, <2>2, and <2>3.

Taking \(n=1\) and \(x_1=y_1=0\) gives \(0\in IJ\). The difference \(\sum_i x_i y_i - \sum_j x'_j y'_j=\sum_i x_iy_i+\sum_j(-x'_j)y'_j\) is again a finite sum of products with first factors in \(I\) and second factors in \(J\).

We have \(rx = \sum_{i=1}^n (rx_i) y_i\) with \(rx_i \in I\), and \(xr = \sum_{i=1}^n x_i (y_i r)\) with \(y_i r \in J\), because \(I\) and \(J\) are two-sided ideals.

Each product \(x_iy_i\) lies in \(I\) because \(x_i\in I\) and \(I\) absorbs right multiplication, and lies in \(J\) because \(y_i\in J\) and \(J\) absorbs left multiplication. The ideal \(I\cap J\) is closed under addition, so it contains \(\sum_i x_iy_i\).

Steps <2>1 and <2>2 show that \(IJ\) is an ideal, and step <2>3 gives \(IJ\subseteq I\cap J\).

In \(R=\ZZ\), let \(I=J=K=2\ZZ\). Then \(IJ\) is the set of finite sums of products \((2a)(2b)=4ab\), which is \(4\ZZ\). The ideal \(K\) is contained in \(I\) and \(J\), but \(2\in K\) and \(2\notin4\ZZ=IJ\).

Step <1>1 answers part (a), step <1>2 proves part (b), and steps <1>3 and <1>4 answer part (c).

The displayed set contains zero, is closed under subtraction, and is closed under multiplication by any \(r\in R\): \(r(ua+vb)=(ru)a+(rv)b\). It is therefore an ideal. It contains both \((a)\) and \((b)\). Any ideal containing these two ideals contains every sum \(ua+vb\), so it contains the displayed ideal. This proves the least-ideal property and hence the asserted identification with the join.

Because \(R\) is a PID, step <1>1 gives \((a)+(b)=(g)\) for some \(g\in R\). This generator is nonzero because the ideal contains \(a\ne0\). Membership of \(a,b\) in \((g)\) says that \(g\) divides both \(a\) and \(b\). Membership of \(g\) in the sum gives \(g=ra+sb\).

If \(h\) divides both \(a\) and \(b\), then it divides every linear combination of them, and in particular divides \(g\). Thus \(g\) is a greatest common divisor in the divisibility sense. Conversely, any other greatest common divisor \(g'\) divides \(g\) and is divisible by \(g\), so \((g')=(g)\). Hence any choice of gcd is a generator of the join.

If \(x\) satisfies the congruences, then \(c-x\in(a)\) and \(x-d\in(b)\). Adding gives \(c-d\in(a)+(b)\), which is the required compatibility.

Conversely, if \(c-d\in(a)+(b)\), choose \(u,v\in R\) with \(c-d=ua+vb\). Set \[ x=c-ua=d+vb. \] Then \(x-c=-ua\in(a)\) and \(x-d=vb\in(b)\), proving both congruences. In terms of step <1>2, write \(c-d=kg\) and \(g=ra+sb\); the explicit solution becomes \(x=c-kra=d+ksb\).

Because \(I\) is two-sided, \(R/I\) is an \((R,R)\)-bimodule by left and right multiplication. Put \(T=(R/I)\otimes_R M\), using the right \(R\)-action on \(R/I\) and the given left action on \(M\). For \(r\in R\), left multiplication on \(R/I\) is a right-linear map: \(r(ab)=(ra)b\). It therefore induces an additive map on \(T\), with \[ r\big((a+I)\otimes m\big)=(ra+I)\otimes m. \] The module identities follow on elementary tensors from the ring identities, and elementary tensors generate \(T\) additively. Thus this defines a left \(R\)-module structure [@DF04].

The set \(IM\) is an additive subgroup, since sums concatenate the finite expressions and negatives can be absorbed in the coefficients. For \(r\in R\), each \(ra_i\) lies in \(I\), so \(r\sum_i a_im_i=\sum_i(ra_i)m_i\in IM\). Hence \(IM\) is a left submodule, and \(M/IM\) is a left quotient module.

Define \(\beta:(R/I)\times M\to M/IM\) by \(\beta(a+I,m)=am+IM\). If \(a\) is replaced by \(a+i\) with \(i\in I\), the value changes by \(im\in IM\), so it is independent of the representative. It is additive in each variable. For \(r\in R\), \[ \beta((a+I)r,m)=(ar)m+IM =a(rm)+IM=\beta(a+I,rm). \] Thus \(\beta\) is balanced, and the universal property of the tensor product gives an additive map \(\Phi:T\to M/IM\) with the asserted rule [@DF04]. It is left \(R\)-linear since \[ \Phi\big(r((a+I)\otimes m)\big) =(ra)m+IM=r(am+IM). \]

Define \(\eta:M\to T\) by \(\eta(m)=(1+I)\otimes m\). It is additive, and balancing gives \[ \eta(rm)=(1+I)\otimes rm=(r+I)\otimes m =r\eta(m). \] For \(i\in I\) and \(m\in M\), \(\eta(im)=(i+I)\otimes m=0\). It follows by additivity that \(\eta\) vanishes on all of \(IM\). It therefore descends to a left-linear map \(\Psi:M/IM\to T\) with the displayed formula.

For every \(m+IM\), one has \(\Phi\Psi(m+IM)=m+IM\). For every elementary tensor, balancing gives \[ \Psi\Phi((a+I)\otimes m) =(1+I)\otimes am=(a+I)\otimes m. \] Since elementary tensors generate \(T\), these identities prove that both composites are the identity. Thus \(\Phi\) is the claimed left \(R\)-module isomorphism.

The homomorphism \(\mathbb Z[x]\to\mathbb F_2\) sending a polynomial to its constant coefficient modulo \(2\) is surjective. Its kernel is \((2,x)\): a polynomial has even constant coefficient exactly when it has the form \(2a+xg(x)\) with \(a\in\mathbb Z\) and \(g(x)\in\mathbb Z[x]\). Thus the quotient is the field \(\mathbb F_2\), so the ideal is maximal.

This is a prime ideal because a product of integers is zero only when one factor is zero. It is not maximal, since \[ (0)\subsetneq 2\mathbb Z\subsetneq\mathbb Z. \] The first containment is strict because \(2\ne0\), and the second because \(1\) is not even.

If \(n\in\mathbb Z\setminus\{0\}\) and \(nq=0\) in \(\mathbb Q\), then \(q=0\). Thus this module is torsion free.

Suppose it had a \(\mathbb Z\)-basis \(B\). Since \(\mathbb Q\ne0\), choose \(b\in B\). The rational number \(b/2\) belongs to the module and has a finite expansion in \(B\) with integer coefficients. Multiplying that expansion by \(2\) expresses \(b\) with an even coefficient on every basis vector. Its coefficient on \(b\) must instead be \(1\) by uniqueness of basis coordinates, a contradiction. This excludes infinite as well as finite free rank.

The integers are a unique factorization domain, and the polynomial ring over a unique factorization domain is again one, by Gauss’s lemma [@DF04]. But the proper ideal \((2,x)\) from step <1>1 is not principal. If \((2,x)=(d)\), then \(d\) divides the nonzero constant \(2\). Degree additivity in the integral domain \(\mathbb Z[x]\) forces \(d\) to be a constant integer, necessarily \(\pm1\) or \(\pm2\). Since \(d\) must also divide \(x\), it cannot be \(\pm2\), whose polynomial multiples have even coefficients. Hence \(d=\pm1\), which would make \((d)\) the whole ring, contradicting its properness.

Let \(I=(x_1,x_2,\ldots)\). Suppose \(I\) were generated by finitely many polynomials \(f_1,\ldots,f_s\). Each polynomial involves finitely many indeterminates, so choose \(x_j\) occurring in none of them. Every \(f_i\in I\) has zero constant coefficient. The evaluation homomorphism \[ R\longrightarrow\mathbb Q[t],\qquad x_j\longmapsto t,\qquad x_k\longmapsto0\ (k\ne j) \] therefore sends all \(f_i\) to \(0\), but sends \(x_j\in I\) to the nonzero polynomial \(t\). This contradicts \(I=(f_1,\ldots,f_s)\). Thus \(R\) has an ideal that is not finitely generated and is not Noetherian.

Over \(\CC\), \(x^2+1=(x-i)(x+i)\). Neither factor lies in \((x^2+1)\), since every nonzero element of that ideal has degree at least \(2\), but their product does. So \((x^2+1)\) is not prime, and hence not maximal.

The surjection \(\ZZ[x]\to\ZZ/6\ZZ\), \(f\mapsto f(0)\bmod 6\), has kernel \((6,x)\): a polynomial \(f\) lies in the kernel exactly when \(f=f(0)+xg\) with \(6\mid f(0)\). Hence \(\ZZ[x]/(6,x)\cong\ZZ/6\ZZ\), which has the zero divisors \(2\cdot3=0\) and is not a domain.

View \(y^2-x^3\) as a monic polynomial in \(y\) over \(\CC[x]\). A factorization would be \((y-g)(y+g)\) with \(g\in\CC[x]\) and \(g^2=x^3\), which is impossible since \(x^3\) has odd degree. Hence \(y^2-x^3\) is irreducible in the UFD \(\CC[x,y]\), so it is a prime element and \((y^2-x^3)\) is a prime ideal. It is properly contained in the proper ideal \((x,y)\), since \(y\notin(y^2-x^3)\) by degree, so it is not maximal.

Steps <1>1, <1>2, and <1>3 treat parts (a), (b), and (c).

A unit cannot lie in a proper ideal: if \(u\in I\) and \(u\) is a unit, then \(1=u^{-1}u\in I\). Thus no element of \(M\) is a unit.

Conversely, if \(a\in A\) is not a unit, then \((a)\) is proper. Every proper ideal of a commutative unital ring is contained in a maximal ideal. Order the proper ideals containing \((a)\) by inclusion. The union of any nonempty chain is an ideal containing \((a)\) and remains proper, since membership of \(1\) in that union would put \(1\) in a member of the chain. Zorn’s lemma gives a maximal element, which is a maximal ideal of \(A\) [@DF04]. By uniqueness it is \(M\), so \(a\in M\).

The axioms imply \(|0|=0\) and \(|1|>0\). Multiplicativity gives \(|1|=|1|^2\), hence \(|1|=1\). Also \(|-1|^2=|1|=1\) and \(|-1|\geq0\), so \(|-1|=1\) and \(|-x|=|x|\).

The set \(R\) contains \(0,1\). For \(x,y\in R\), \(|x-y|\leq\max\{|x|,|y|\}\leq1\) and \(|xy|=|x||y|\leq1\). Thus \(R\) is a commutative unital subring of \(F\). The set \(\mathfrak m\) contains zero, is closed under subtraction by the same inequality, and satisfies \(|rx|=|r||x|<1\) for \(r\in R\), \(x\in\mathfrak m\). It is an ideal of \(R\), proper because \(|1|=1\).

For \(x\ne0\), multiplicativity gives \(|x^{-1}|=1/|x|\). Consequently an element \(x\in R\) is a unit in \(R\) exactly when \(|x|=1\): in that case its inverse lies in \(R\), and conversely \(xy=1\) with \(x,y\in R\) forces \(|x|=|y|=1\). Thus every element of \(R\setminus\mathfrak m\) is a unit. An ideal strictly containing \(\mathfrak m\) contains such a unit and hence is all of \(R\), proving maximality. Every proper ideal contains no unit, so is contained in \(\mathfrak m\). This proves uniqueness. The argument also covers the trivial absolute value, when \(R=F\) and \(\mathfrak m=(0)\).

Since \(n\geq1\), one has \(M^n\subseteq M\ne R\). Moreover, \[ (R/M^n)/(M/M^n)\cong R/M \] is a field, so \(M/M^n\) is maximal [@DF04].

Let \(\mathfrak n\) be any maximal ideal of \(R/M^n\), and let \(N\subset R\) be its inverse image under the quotient map. Then \(M^n\subseteq N\) and \(R/N\cong(R/M^n)/\mathfrak n\) is a field. In particular \(N\) is a proper prime ideal. For each \(m\in M\), one has \(m^n\in M^n\subseteq N\). Primality, applied repeatedly to this product of \(n\) copies of \(m\), gives \(m\in N\). Hence \(M\subseteq N\). Maximality of \(M\) and properness of \(N\) give \(N=M\), so \(\mathfrak n=M/M^n\). Thus no other maximal ideal exists.

The integers form an integral domain. A nonzero ideal \(I\) contains a positive integer; let \(d\) be its least positive element. For \(a\in I\), Euclidean division writes \(a=qd+r\) with \(0\leq r<d\). Since \(r=a-qd\in I\), minimality implies \(r=0\). Hence \(I=(d)\). The zero ideal is also principal, so \(\mathbb Z\) is a PID. It is not a field, because \(2\) has no multiplicative inverse in \(\mathbb Z\).

The integers are a UFD, and the polynomial ring over a UFD is a UFD by Gauss’s lemma [@DF04]. Thus \(D\) is a UFD. The ideal \(I=(2,x)\) is proper: the homomorphism \(\mathbb Z[x]\to\mathbb F_2\), \(f\mapsto f(0)\bmod2\), annihilates both generators but does not annihilate \(1\).

Suppose \(I=(h)\). Since \(h\) divides the nonzero constant \(2\), degree additivity in the domain \(\mathbb Z[x]\) forces \(h\) to be a nonzero integer. Since it also divides \(x\), comparison of the coefficient of \(x\) in \(x=hu(x)\) shows that \(h\) divides \(1\). Thus \(h\) is a unit, which would imply \(I=D\), contradicting properness. Therefore \(D\) is not a PID.

For \(0\ne n\in\mathbb Z\), the equation \(nq=0\) in \(\mathbb Q\) implies \(q=0\), so \(M\) is torsion-free. But \(2M=M\), since \(q=2(q/2)\) for every rational \(q\). In a nonzero free \(\mathbb Z\)-module, a basis vector cannot lie in twice the module: every coefficient in the basis expansion of \(2v\) is even, whereas that basis vector has coefficient one at itself. Since \(\mathbb Q\) is nonzero and equals twice itself, it cannot be free.

Both modules are nonzero. For any elementary tensor \(t=a\otimes b\), the tensor relations give \(2t=(2a)\otimes b=0\) and \(3t=a\otimes(3b)=0\). Therefore \(t=(3-2)t=0\). Every element of the tensor product is a finite sum of elementary tensors, so the whole tensor product is zero, as required.

Let \(I\) be an ideal of \(R\). If \(I=0\), then \(I=(0)\). Otherwise choose \(d\in I\setminus\{0\}\) with \(N(d)\) minimal; then \((d)\subseteq I\). For \(a\in I\), division gives \(q,r\in R\) with \(a=qd+r\) and either \(r=0\) or \(N(r)<N(d)\). Since \(r=a-qd\in I\), minimality of \(N(d)\) excludes \(r\ne0\), so \(a=qd\in(d)\). Hence \(I=(d)\), and \(R\) is a principal ideal domain.

A coset has the form \(\beta+I\) with \(\beta\in\ZZ[i]\). Division by \(\alpha\ne0\) gives \(q,r\in\ZZ[i]\) with \(\beta=q\alpha+r\) and either \(r=0\) or \(N(r)<N(\alpha)\). Then \(r=\beta-q\alpha\in\beta+I\), and in both cases \(N(r)<N(\alpha)\), because \(N(0)=0<N(\alpha)\).

Let \(n=N(\alpha)\). The Gaussian integers \(a+bi\) with \(a^2+b^2<n\) satisfy \(\abs{a},\abs{b}<\sqrt n\), so there are finitely many of them. By step <1>2 every coset of \(I\) contains one of them, so there are finitely many cosets.

Steps <1>1, <1>2, and <1>3 answer parts (a), (b), and (c).

For \(a,b\in\mathbb Z\), the element \(a+b\sqrt2\) satisfies the monic polynomial \[ T^2-2aT+(a^2-2b^2)\in\mathbb Z[T], \] so it is an algebraic integer.

Conversely, write an algebraic integer in \(\mathbb Q(\sqrt2)\) as \(\alpha=a+b\sqrt2\), with \(a,b\in\mathbb Q\). If \(b=0\), the rational root theorem applied to a monic integer polynomial for \(\alpha\) gives \(a\in\mathbb Z\).

Suppose \(b\ne0\). Then \(\alpha\) has degree \(2\), with minimal polynomial \(T^2-2aT+(a^2-2b^2)\). A monic factor over \(\mathbb Q\) of a monic integer polynomial has integer coefficients, by Gauss’s lemma [@DF04]. Therefore \[ 2a=m\in\mathbb Z,\qquad a^2-2b^2=n\in\mathbb Z, \qquad 8b^2=m^2-4n\in\mathbb Z. \] Write \(b=u/v\) in lowest terms with \(v>0\). The last integrality condition gives \(v^2\mid8u^2\), hence \(v^2\mid8\) and \(v\in\{1,2\}\). Thus \(b=k/2\) for an integer \(k\), and \[ m^2-2k^2=4n. \] If \(k\) were odd, this would give \(m^2\equiv2\pmod4\), impossible because squares modulo \(4\) are \(0\) and \(1\). Hence \(k\) is even; then the same equality makes \(m\) even. Consequently \(a=m/2\) and \(b=k/2\) are both integers. This proves the asserted equality of rings.

Conjugation \(\tau(a+b\sqrt2)=a-b\sqrt2\) is an automorphism of the field. Thus \[ N(\alpha)=\alpha\tau(\alpha),\qquad N(\alpha\beta)=N(\alpha)N(\beta), \] and absolute values give multiplicativity of \(\delta=|N|\). Step <1>1 makes \(N\) integer-valued on \(O\). If \(\alpha\ne0\), then both \(\alpha\) and \(\tau(\alpha)\) are nonzero field elements, so \(\delta(\alpha)\) is a positive integer.

Let \(\alpha,\beta\in O\) with \(\beta\ne0\), and write \[ \frac{\alpha}{\beta}=u+v\sqrt2,\qquad u,v\in\mathbb Q. \] Choose integers \(m,n\) such that \(|u-m|\leq1/2\) and \(|v-n|\leq1/2\), and put \(q=m+n\sqrt2\in O\) and \(r=\alpha-\beta q\in O\). If \(r=0\), the division has zero remainder. Otherwise step <1>2 gives \[ \begin{aligned} \delta(r) &=\delta(\beta)\left|(u-m)^2-2(v-n)^2\right|\\ &\leq\delta(\beta)\bigl((u-m)^2+2(v-n)^2\bigr)\\ &\leq\frac34\delta(\beta)<\delta(\beta). \end{aligned} \] Thus \(\alpha=\beta q+r\) with zero remainder or strictly smaller positive integer size. Since \(O\) is a subring of a field, it is an integral domain, and this proves that it is Euclidean. Under the definition also requiring \(\delta(\alpha)\leq\delta(\alpha\beta)\) for nonzero \(\alpha,\beta\), that condition follows from multiplicativity and \(\delta(\beta)\geq1\).

Let \(m_1,\ldots,m_k\in V\) be \(R\)-linearly independent and suppose \(\sum_i c_im_i=0\) with \(c_i\in F\). Choose \(d\in R\setminus\{0\}\) with \(dc_i\in R\) for all \(i\). Then \(\sum_i(dc_i)m_i=0\), so \(dc_i=0\) and hence \(c_i=0\) for all \(i\).

If \(r\in R\setminus\{0\}\) and \(v\in M\) satisfy \(rv=0\), then \(v=r^{-1}(rv)=0\) in the \(F\)-vector space \(V\); so \(M\) is torsion-free. By the structure theorem for finitely generated modules over a PID, \(M\cong R^k\) for some \(k\ge0\) [@DF04]. An \(R\)-basis of \(M\) is \(F\)-linearly independent by step <1>1, so \(k\le\dim_FV=n\).

By step <1>1, \(w_1,\ldots,w_n\) are \(n\) linearly independent vectors in the \(n\)-dimensional space \(V\), so they form an \(F\)-basis. Write \(u_i=\sum_j(a_{ij}/b_{ij})w_j\) with \(a_{ij},b_{ij}\in R\) and \(b_{ij}\ne0\), and put \(\alpha=\prod_{i,j}b_{ij}\ne0\). Then each \(\alpha u_i=\sum_j(\alpha a_{ij}/b_{ij})w_j\) has coefficients in \(R\), so \(\alpha u_i\in N\); since the \(u_i\) generate \(M\), \(\alpha M\subseteq N\).

Multiplication by \(\alpha\) is an injective \(R\)-linear map \(M\to\alpha M\), so \(\alpha M\) is a free submodule of rank \(n\) of the free module \(N\) of rank \(n\), by step <1>3. By the stacked-bases theorem for submodules of free modules over a PID [@DF04], there are an \(R\)-basis \(w_1',\ldots,w_n'\) of \(N\) and \(d_1,\ldots,d_n\in R\setminus\{0\}\) such that \(d_1w_1',\ldots,d_nw_n'\) is an \(R\)-basis of \(\alpha M\). Then \(e_i\coloneqq(d_i/\alpha)w_i'\) form an \(R\)-basis of \(M=\alpha^{-1}(\alpha M)\), and with \(\beta_i\coloneqq\alpha/d_i\in F^\times\) we get \(\beta_ie_i=w_i'\), an \(R\)-basis of \(N\).

Step <1>2 answers part (a), and steps <1>3 and <1>4 answer part (b).

The constant polynomials identify \(R\) with a subring of the commutative integral domain \(R[x]\), so \(R\) is a commutative integral domain with \(1\ne0\). Fix \(0\ne a\in R\). The ideal \((a,x)\) is generated by some \(f\in R[x]\). As \(a=fg\) for a nonzero \(g\in R[x]\), additivity of degrees over a domain gives \(\deg f=\deg g=0\). Thus \(f\) is a nonzero constant. The identity \(x=fh\) for some \(h\in R[x]\) shows, by comparing coefficients of \(x\), that \(1=fh_1\) for an element \(h_1\in R\). Hence \(f\) is a unit, and \((a,x)=R[x]\).

Write \(1=au(x)+xv(x)\) with \(u,v\in R[x]\). Evaluation at \(x=0\) gives \(1=au(0)\). Every nonzero \(a\in R\) is therefore invertible.

The order of a permutation is the least common multiple of its disjoint-cycle lengths [@DF04]. A permutation of order \(15\) in \(S_8\) must have a cycle with length divisible by \(5\). Every cycle length divides \(15\) and is at most \(8\), so this cycle has length \(5\). Only three letters remain. The order must also be divisible by \(3\), so those letters form a \(3\)-cycle. Thus every generator of a cyclic subgroup of order \(15\) has cycle type \((5)(3)\).

For two such generators, match the five letters in their cyclic orders and then match the three remaining letters in their cyclic orders. The resulting permutation conjugates one generator to the other. It consequently conjugates their generated subgroups. The subgroup generated by \((1\,2\,3\,4\,5)(6\,7\,8)\) shows that such subgroups exist.

For a nonzero integer \(n\) and \(q\in\mathbb Q\), the equation \(nq=0\) implies \(q=0\), so the module is torsion-free. Moreover, \(2\mathbb Q=\mathbb Q\), since \(q=2(q/2)\).

If \(\mathbb Q\) were free over \(\mathbb Z\), its nonzero underlying module would have a nonempty basis. A basis vector \(e\) could not belong to twice the module: in the unique finite basis expansion of \(2m\), every coefficient is even, whereas the coefficient of \(e\) in itself is \(1\). This contradicts \(2\mathbb Q=\mathbb Q\).

Let \(F=\mathbb F_2(t)\) with \(t\) an indeterminate. The polynomial \(X^2-t\) has no root in \(F\). Indeed, a root \(A(t)/B(t)\) with nonzero polynomials \(A,B\) would imply \(A(t)^2=tB(t)^2\). The degrees of the two sides would be respectively even and odd, an impossibility. A quadratic over a field is reducible exactly when it has a root, so \(X^2-t\) is irreducible.

Set \(K=F(u)\) where \(u^2=t\). The minimal polynomial of \(u\) is \(X^2-t\), giving \([K:F]=2\). In characteristic \(2\), \[ X^2-t=(X-u)^2. \] Thus the minimal polynomial has a repeated root, and \(K/F\) is not separable. In particular it is not Galois [@DF04].

The five numbers \(\zeta^j\) are distinct and each has fifth power \(1\). Thus the product of their linear factors divides \(x^5-1\); both polynomials are monic of degree \(5\), so they are equal. A polynomial of degree \(1\) over a field cannot be a product of two positive-degree polynomials, so every displayed factor is irreducible.

Factor \(x^6-1=(x^3-1)(x^3+1)\) and use the sum and difference of cubes. The linear factors are irreducible. The two quadratic factors have discriminant \(-3\), so neither has a rational root. A reducible quadratic over a field has a linear factor and hence a root, proving their irreducibility.

Its leading coefficient is \(1\), all other coefficients are divisible by \(3\), and its constant coefficient is not divisible by \(9\). By Eisenstein’s criterion at \(3\) it is irreducible. In detail, by Gauss’s lemma, a proper factorization over \(\mathbf Q\) would give monic positive-degree factors \(g,h\in\mathbf Z[x]\) [@DF04]. Modulo \(3\), their product is \(x^7\), so each reduction is a positive power of \(x\). Consequently \(3\) divides both \(g(0)\) and \(h(0)\), forcing \(9\mid g(0)h(0)=-15\), a contradiction.

Every nonzero \(a\) belongs to the multiplicative group of order \(10\), so \(a^{10}=1\) and \(a^{11}=a\). The identity also holds at \(a=0\). Thus the eleven distinct field elements give eleven distinct linear factors of the monic degree-\(11\) polynomial. Their product is the polynomial, and each factor is irreducible.

Choose a root \(\alpha\) in an algebraic closure, and let \(d=[\mathbf F_{13}(\alpha):\mathbf F_{13}]\). Its minimal polynomial divides the given polynomial, so \(1\leq d\leq13\). The root equation gives \(\alpha^{13}=\alpha+1\). Induction, using the Frobenius homomorphism in characteristic \(13\), gives \[ \alpha^{13^j}=\alpha+j\cdot1\qquad(j\geq0). \] The field \(\mathbf F_{13}(\alpha)\) has \(13^d\) elements, so every element of it satisfies \(z^{13^d}=z\). Taking \(j=d\) in the displayed identity therefore gives \(d\cdot1=0\), and hence \(13\mid d\). Thus \(d=13\). The degree-\(13\) minimal polynomial divides the monic polynomial \(x^{13}-x-1\) of the same degree, so they are equal. This proves irreducibility and completes all five factorizations.

In a field \(\FF_{p^2}\) with \(p^2\) elements, every element satisfies \(\alpha^{p^2}=\alpha\), so \(x^{p^2}-x=\prod_{\alpha\in\FF_{p^2}}(x-\alpha)\) has distinct roots. The minimal polynomial over \(\FF_p\) of \(\alpha\in\FF_{p^2}\) has degree \([\FF_p(\alpha):\FF_p]\), which divides \(2\). Conversely, a monic irreducible \(g\) of degree \(d\in\{1,2\}\) has a root \(\beta\) with \(\FF_p(\beta)\) of order \(p^d\), so \(\beta^{p^d}=\beta\) and hence \(\beta^{p^2}=\beta\); thus \(g\), the minimal polynomial of \(\beta\), divides \(x^{p^2}-x\).

If \(f\) is reducible, it has a monic irreducible factor of degree \(1\) or \(2\), since \(\deg f=4\); by step <1>1 that factor divides \(x^{p^2}-x\), so the gcd is not \(1\). If \(f\) is irreducible, a common monic irreducible factor with \(x^{p^2}-x\) would be \(f\) itself, of degree \(4\), contradicting step <1>1; so the gcd is \(1\).

Adjoining one root \(\alpha\) of \(f\) gives \(\FF_p(\alpha)\) of degree \(4\), a field \(\FF_{p^4}\) with \(p^4\) elements. The extension \(\FF_{p^4}/\FF_p\) is Galois with cyclic group of order \(4\) generated by \(\sigma\) [@DF04], so it contains every root of the irreducible polynomial \(f\) and is its splitting field.

Steps <1>2 and <1>3 answer parts (a) and (b).

Follows directly from <1>1 and <1>2. Q.E.D.

The roots are the \(p\) distinct numbers \(\alpha\zeta^j\) for \(j\in\mathbb Z/p\mathbb Z\). They lie in \(K\), and the field they generate contains \(\alpha\) and \(\zeta=(\alpha\zeta)/\alpha\). Thus \(K\) is exactly the splitting field.

Let \(F=\mathbb Q(\zeta)\). The polynomial \[ \Phi_p(X)=1+X+\cdots+X^{p-1} \] vanishes at \(\zeta\). Its translate is \[ \Phi_p(Y+1)=\frac{(Y+1)^p-1}{Y} =\sum_{j=1}^p\binom pj Y^{j-1}. \] Every coefficient other than the leading coefficient is divisible by \(p\), and the constant coefficient is \(p\), not divisible by \(p^2\). Eisenstein’s criterion [@DF04] and invertibility of translation show that \(\Phi_p\) is irreducible. Hence \([F:\mathbb Q]=p-1\).

Put \(d=[K:F]\). Since \(K=F(\alpha)\) and \(\alpha\) satisfies a degree-\(p\) polynomial over \(F\), we have \(1\leq d\leq p\). The original irreducibility gives \([\mathbb Q(\alpha):\mathbb Q]=p\). Applying the tower law through this subfield and through \(F\) yields \[ p\mid[K:\mathbb Q]=(p-1)d. \] As \(p\) is relatively prime to \(p-1\), it divides \(d\). The bounds on \(d\) force \(d=p\), giving the asserted total degree. For \(p=2\) the cyclotomic field is \(\mathbb Q\) and the same argument applies.

Let \(\sigma\in\operatorname{Gal}(K/\mathbb Q)\). It preserves the multiplicative order of \(\zeta\), so \(\sigma(\zeta)=\zeta^k\) for a unique \(k\in(\mathbb Z/p\mathbb Z)^\times\). It sends \(\alpha\) to a root of \(X^p-a\), so \(\sigma(\alpha)=\alpha\zeta^l\) for a unique \(l\in\mathbb Z/p\mathbb Z\). Consequently \[ \sigma(\alpha\zeta^j)=\alpha\zeta^{kj+l}. \] Thus the permutation of the labels is \(j\mapsto kj+l\). The root action is a homomorphism into the permutation group, and is faithful because the roots generate \(K\).

The affine maps form a group: composition and inversion are \[ f_{k,l}\circ f_{u,v}=f_{ku,kv+l},\qquad f_{k,l}^{-1}=f_{k^{-1},-k^{-1}l}. \] Their number is \(p(p-1)\), because \(l=f(0)\) and \(k=f(1)-f(0)\) determine the map uniquely, with \(p\) choices for \(l\) and \(p-1\) nonzero choices for \(k\).

The splitting field \(K/\mathbb Q\) is Galois in characteristic zero, so its automorphism group has order \([K:\mathbb Q]\) [@DF04], which is \(p(p-1)\) by step <1>1. The injective homomorphism in step <1>2 is therefore a bijection onto this affine group, as required.

Using \((\mathbb Z/m)\otimes_{\mathbb Z}B\cong B/mB\), \[ (\mathbb Z/2)\otimes_{\mathbb Z}(\mathbb Z/3) \cong (\mathbb Z/3)/2(\mathbb Z/3)=0, \] because multiplication by \(2\) is an automorphism of \(\mathbb Z/3\).

A \(\mathbb Q\)-module is a vector space. If \(0\ne v\in V\) and \(0\ne w\in W\), choose linear functionals \(\lambda:V\to\mathbb Q\) and \(\mu:W\to\mathbb Q\) with \(\lambda(v)=\mu(w)=1\). The induced map \[ \lambda\otimes\mu:V\otimes_{\mathbb Q}W\to\mathbb Q \] sends \(v\otimes w\) to \(1\), so \(V\otimes_{\mathbb Q}W\ne0\).

The \(\mathbb Z\)-module \(\mathbb Q\) is torsion-free. It is not free: \(\mathbb Q\) is divisible, whereas a nonzero free abelian group is not divisible.

In a ring with identity, a maximal two-sided ideal \(I\) is prime: if \(a,b\notin I\) and \(aRb\subseteq I\), then \(R=I+RaR\) and \(R=I+RbR\), so \(R=R\cdot R\subseteq I+RaR\,RbR\subseteq I+R(aRb)R\subseteq I\), a contradiction. In \(R=2\mathbb Z\), the ideal \(4\mathbb Z\) has index \(2\), so no ideal lies strictly between it and \(R\), and it is maximal. It is not prime, since \(2\cdot2=4\in I\) while \(2\notin I\).

The ascending chain \[ (x_1)\subsetneq(x_1,x_2)\subsetneq(x_1,x_2,x_3)\subsetneq\cdots \] does not stabilize, so \(R\) is not Noetherian.

Its center is \[ Z(R)=\{aI_2:a\in\mathbb Q\}. \] This is a proper subset of \(R\) but contains the identity matrix. Any two-sided ideal containing \(1\) is all of \(R\), so \(Z(R)\) is not an ideal.

Suppose \(M=N\oplus N'\) and let \(\pi:M\to N\) be the projection onto \(N\). It is \(D\)-linear and restricts to the identity on \(N\). If \(y\in N\), \(a\in D\), and \(ax=y\) has a solution \(x\in M\), then \[ a\pi(x)=\pi(ax)=\pi(y)=y. \] Thus \(\pi(x)\in N\) solves the same equation, as required.

Since \(d\in\operatorname{ann}_D(z+N)\), the element \(dz\) lies in \(N\). The equation \(dx=dz\) has the solution \(z\in M\). Purity supplies a solution \(n\in N\). Put \(w=z-n\). Then \(w+N=z+N\) and \(dw=dz-dn=0\), so \((d)\subseteq\operatorname{ann}_D(w)\).

Conversely, if \(aw=0\), then \(a(z+N)=a(w+N)=0\) in \(M/N\). Thus \(a\in\operatorname{ann}_D(z+N)=(d)\). This proves the reverse inclusion and hence \(\operatorname{ann}_D(w)=(d)\). No assumption that \(d\) is nonzero or a nonunit is needed.

The quotient \(Q=M/N\) is finitely generated, by the images of a finite generating set of \(M\). It is torsion: a nonzero scalar annihilating a representative also annihilates its coset. The structure theorem over the PID \(D\) therefore gives \[ Q\cong\bigoplus_{i=1}^r D/(d_i) \] for nonzero nonunits \(d_i\in D\) [@DF04]. The empty sum is allowed when \(Q=0\). Let \(u_i\in Q\) correspond to \(1\) in the \(i\)th cyclic summand, so \(\operatorname{ann}_D(u_i)=(d_i)\). Choose \(z_i\in M\) with \(q(z_i)=u_i\). By step <1>2, there is \(w_i\in M\) with \(q(w_i)=u_i\) and \(\operatorname{ann}_D(w_i)=(d_i)\).

Define \[ s:Q\longrightarrow M,\qquad s\left(\sum_{i=1}^r a_i u_i\right)=\sum_{i=1}^r a_i w_i. \] In the displayed direct sum, two coefficient lists represent the same element exactly when their \(i\)th coefficients differ by a multiple of \(d_i\) for every \(i\). Since \(d_iw_i=0\), those changes do not change the proposed value of \(s\). Thus \(s\) is well-defined and \(D\)-linear. Moreover \(q(s(u_i))=u_i\) for all generators, so \(q\circ s=\operatorname{id}_Q\). For \(Q=0\), the zero map is the required section.

For every \(m\in M\), \[ m=(m-s(q(m)))+s(q(m)), \] where the first term lies in \(\ker q=N\) and the second in \(s(Q)\). If \(s(u)\in N\), applying \(q\) gives \(u=q(s(u))=0\), and hence \(s(u)=0\). Therefore \(N\cap s(Q)=0\) and \(M=N\oplus s(Q)\).