Writing $z=x+iy$ and $f=u+iv$, the Cauchy-Riemann equations areWith continuous second derivatives, mixed derivatives commute. ThusBoth real and imaginary parts are harmonic.

For $g(z)=\mathrm{Re}z$, the real and imaginary parts are $x$ and zero, so $u_x=1\ne 0=v_y$. For $h(z)=\bar{z}$, they are $x$ and $-y$, so $u_x=1\ne -1=v_y$. The necessary Cauchy-Riemann equations fail at every point. Hence both functions are nowhere holomorphic, indeed nowhere complex differentiable.

Self-adjointness means $⟨Tv,w⟩=⟨v,Tw⟩$ for all $v,w\in V$, or $T=T^{\ast }$. Take the inner product linear in its first variable. If $Tv=\lambda v$ with $v\ne 0$, thenTherefore every eigenvalue is real.

If $U$ is invariant and $w\in U^{\perp }$, then for every $u\in U$, $Tu\in U$ and $⟨Tw,u⟩=⟨w,Tu⟩=0$. Hence $Tw\in U^{\perp }$, proving invariance of the orthogonal complement.

Induct on $d=\dim V$. The zero-dimensional case is empty. For $d>0$, choose an eigenvector and normalize it to a unit vector $e$. Its span is invariant, so its orthogonal complement is invariant as just proved. The restriction of $T$ to that complement is again self-adjoint. Its dimension is $d-1$, so induction supplies an orthonormal basis of eigenvectors there. Adjoining $e$ gives an orthonormal eigenbasis of $V$. This proves the finite-dimensional spectral theorem without assuming it in the argument.

${\dim }_{\mathbb{R}}V=4$. For arbitrary matrices,This is bilinear and symmetric. Also $⟨A\mid A⟩={\sum }_{i,j}{a}_{ij}^2\ge 0$, with equality exactly when every entry vanishes. Thus it is an inner product, the Frobenius inner product.

If $S^t=S$ and $K^t=-K$, invariance of trace under transposition and cyclicity giveFurthermore $⟨I\mid A⟩=\mathrm{tr}A$, so every traceless matrix is orthogonal to the identity. One convenient orthonormal basis adapted to these observations isTheir entrywise inner products are zero in distinct pairs and one on the diagonal; four orthonormal vectors in this four-dimensional space form a basis.

Take $I=(2,t)\subseteq \mathbb{Z}[t]$. The ring homomorphismis surjective. Its kernel of a ring homomorphism is exactly $I$: a polynomial maps to zero precisely when its constant coefficient is even, in which case it has the form $2a+th(t)$ with $a\in \mathbb{Z}$ and $h\in \mathbb{Z}[t]$. By the first isomorphism theorem for rings,This quotient ring is a field, so $I$ is a maximal ideal and in particular a prime ideal.

Suppose $I=(h)$ were a principal ideal. Because $2\in I$, the nonzero polynomial $h$ divides the constant $2$, forcing its polynomial degree to be zero: $h=d$ for a nonzero integer $d$. But $t\in I$ also makes $d$ divide the coefficient $1$ of $t$, so $d=\pm 1$. This would give $I=\mathbb{Z}[t]$, contradicting the properness of the kernel of a ring homomorphism $\ker \theta$. ThereforeThere is no conflict with part (c): $I\cap \mathbb{Z}=2\mathbb{Z}\ne \{0\}$. The zero-intersection hypothesis is exactly what excludes this example.

The nonzero ideal $\tilde{P}$ is principal because $\mathbb{Q}[t]$ is a principal ideal domain. Choose a nonzero generator $h\in \mathbb{Q}[t]$. Part (a) givesThe nonzero rational scalar $c(h)$ is a unit in $\mathbb{Q}[t]$, so this rescaling does not change the generated principal ideal:Part (b) now places $f$ in $P$, since $f\in \tilde{P}\cap \mathbb{Z}[t]$. Thus $(f{)}_{\mathbb{Z}[t]}\subseteq P$.

For the reverse inclusion, let $g\in P$ be nonzero. The displayed equality supplies $q\in \mathbb{Q}[t]$ with $g=fq$. Multiplicativity of rational polynomial content givesSince $g$ has integer coefficients, $c(g)$ is a positive integer. The normalization from part (a) gives $q/c(q)\in \mathbb{Z}[t]$, so multiplying it by the integer $c(q)=c(g)$ proves $q\in \mathbb{Z}[t]$. Consequently $g\in (f{)}_{\mathbb{Z}[t]}$. The zero polynomial belongs as well, provingThis is the divisibility descent in Gauss lemma for polynomials: dividing an integer-coefficient polynomial by a primitive polynomial cannot introduce a denominator into an exact quotient. Here it yields the description of prime ideals of the integer polynomial ring above zero. The generator has positive polynomial degree, since a primitive constant would be a unit and would put $1$ in $P$.

The set $\tilde{P}$ contains zero. It is closed under negatives because $-f\in P$ whenever $f\in P$. If $f,g\in P$ and $n,m>0$, thensince $P$ is an ideal of $\mathbb{Z}[t]$. Also, any $q\in \mathbb{Q}[t]$ can be written $q=h/d$ with $h\in \mathbb{Z}[t]$ and $d>0$, givingThis proves that $\tilde{P}$ is an ideal of $\mathbb{Q}[t]$.

Clearly $P\subseteq \tilde{P}\cap \mathbb{Z}[t]$, by taking denominator one. Conversely, if $h\in \tilde{P}\cap \mathbb{Z}[t]$, then $nh\in P$ for some positive integer $n$. The condition $P\cap \mathbb{Z}=\{0\}$ implies $n\notin P$. Since $P$ is a prime ideal, $nh\in P$ forces $h\in P$. HenceIn particular, $\tilde{P}$ is nonzero because it contains the nonzero ideal $P$, and it is proper: otherwise its intersection with $\mathbb{Z}[t]$ would contain $1$. This is the concrete extension of an ideal and contraction of an ideal calculation for the localization that inverts nonzero integers.

For a nonzero polynomial $f={\sum }_i{a}_i{t}^i\in \mathbb{Z}[t]$, its polynomial content is the positive greatest common divisor of its coefficients:For nonzero $g\in \mathbb{Q}[t]$, choose a positive integer $n$ with $ng\in \mathbb{Z}[t]$ and define its rational polynomial content byTo check independence of the denominator, write $g=f/n=h/m$ with $f,h\in \mathbb{Z}[t]$ and $n,m>0$. Then $mf=nh$, while multiplication of all coefficients by a positive integer givesDividing by $mn$ proves $c(f)/n=c(h)/m$, as required.

For multiplication, write $g_i=f_i/n_i$. The integer-coefficient content identity permitted in the source givesThusWe will also use the normalization $g/c(g)\in \mathbb{Z}[t]$, with content one: if $g=f/n$, then $g/c(g)=f/c(f)$. It is a primitive polynomial. For a nonzero rational constant $r$, the same definition gives $c(r)=|r|$.

Suppose a finite set of rational numbers $q_1,\dots ,q_m$ generated $\mathbb{Q}$ as a module over $\mathbb{Z}$. A positive common denominator $D$ makes every $D{q}_i$ an integer. For every integer linear combination, therefore,The entire generated submodule is contained in $(1/D)\mathbb{Z}$. But $1/(2D)$ is a rational number outside that set: $1/(2D)=k/D$ would require the noninteger $k=1/2$. This contradicts generation of all of $\mathbb{Q}$. ConsequentlyEach finite set of generators imposes a common denominator; the full rational number system has no such bound.

Using the common denominator $70$, the three given generators becomeThus their submodule is contained in $(1/70)\mathbb{Z}$. For the reverse inclusion, the following explicit integer linear combination supplies the missing generator:soTherefore $1/70$ belongs to the generated submodule, and all its integer multiples belong as well. We have established equality in both directions:Finally, $k/70=0$ for an integer $k$ forces $k=0$, so the single generator is linearly independent. Hence the required basis of a module is

Let $x,y\in \mathbb{Q}$. Choose a positive integer $D$ clearing their denominators and write $x=m/D$, $y=n/D$ with $m,n\in \mathbb{Z}$. If $x$ and $y$ are not both zero, thenis a relation with integer coefficients not both zero. If both vanish, use $1\cdot x+0\cdot y=0$. Thus every pair is linearly dependent over $\mathbb{Z}$.

A submodule $M\subseteq \mathbb{Q}$ is a torsion-free module: if $kz=0$ for a nonzero integer $k$ and a rational number $z$, then $z=0$. If $M$ is also a finitely generated module, part (a), applied to the Euclidean domain $\mathbb{Z}$, makes it a free module. Its rank cannot exceed one, since two different members of a basis of a module would be linearly independent, contradicting the relation just proved. Since $M\ne 0$, its rank cannot be zero. ThereforeThe denominator-and-greatest common divisor construction in part (c) gives an explicit version of this result for a finitely generated subgroup of the rational additive group.

A Euclidean domain is a principal ideal domain. The structure theorem for finitely generated modules over a principal ideal domain therefore gives, for every finitely generated module $M$ over $R$,where the $d_i$ are nonzero nonunits and $d_1\mid d_2\mid \cdots \mid d_s$. The factors $d_i$ are unique up to multiplication by units; $r$ is uniquely determined.

The rank of a module is this integer $r$. Equivalently, with $K$ the field of fractions of $R$,The tensor product of modules kills each displayed torsion summand after extending scalars to $K$, while sending $R^r$ to $K^r$, explaining the equality.

If $M$ is a torsion-free module, no summand $R/(d_i)$ can occur: its nonzero element $1+(d_i)$ is killed by the nonzero scalar $d_i$. Hence $s=0$ andThis also covers $M=0$, with $r=0$.

By the first isomorphism theorem for rings, the given surjection identifies $S$ with $R/\ker \varphi$. Since $R$ is a principal ideal domain, write $\ker \varphi =(a)$.

If $a=0$, the quotient ring is $R$, an integral domain, so part (a)(ii) applies. If $a$ is a unit, the quotient ring is the zero ring; its unique element is associate to itself by multiplication by its identity, so it too is an associator ring.

In the remaining case, unique factorization giveswhere $\epsilon$ is a unit, the $p_i$ are pairwise nonassociate prime elements, and $k_i\ge 1$. The principal ideals $(p_i^{k_i})$ are pairwise comaximal: different prime factors have greatest common divisor $1$, and Bézout's identity gives a linear combination equal to $1$. The Chinese remainder theorem therefore gives a ring isomorphismEvery factor is an associator ring by part (c), and their finite direct product of rings is an associator ring by part (b). A ring isomorphism preserves units and principal ideals, so the property transfers to $S$. Hence

Put $A=R/(p^k)$. A principal ideal domain is a unique factorization domain, so every nonzero element of this quotient ring can be writtenIndeed, remove the largest power of the prime element $p$ from a representative that is not divisible by $p^k$. Since $s$ and $p^k$ have greatest common divisor $1$ up to a unit, Bézout's identity in the principal ideal domain gives $xs+y{p}^k=1$. Thus $\overline{s}$ is a unit in $A$.

Consequently every nonzero principal ideal in $A$ is generated by one of $1,\overline{p},\dots ,{\overline{p}}^{k-1}$. Include the zero ideal by assigning it exponent $e=k$. These ideals are distinct: their inverse images in $R$ are $(1),(p),\dots ,(p^k)$, and equality of two different powers would, by cancellation in the integral domain, make the nonunit $p$ a unit.

Now suppose $(\overline{a})=(\overline{b})$. If this is the zero ideal, both generators are zero and are associate elements. Otherwise their exponents agree, so writewith $\overline{s},\overline{t}$ both units. Thenand the multiplier is a unit. ThusThe exponent detects the principal ideal, while its remaining factor is invertible; this is why zero divisors cause no obstruction here.