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The double dual module is $M^{\ast \ast }={\mathrm{Hom}}_R(M^{\ast },R)$. The evaluation homomorphism sends $m\in M$ to the functional $f\mapsto f(m)$.

An ideal is an additive subgroup of a ring closed under multiplication by arbitrary ring elements; in a potentially noncommutative ring, a two-sided ideal must absorb multiplication on both sides. For a ring homomorphism $\varphi$, its kernel of a ring homomorphism $K=\{r:\varphi (r)=0\}$ contains zero, is closed under subtraction, and satisfies $\varphi (rk)=\varphi (r)\varphi (k)=0$ and $\varphi (kr)=0$. Thus $K$ is an ideal.

Define $\tilde{\varphi }:R/K\to S$ by $r+K\mapsto \varphi (r)$. It is well-defined because two representatives differ by an element of $K$, preserves addition and multiplication, is surjective by hypothesis, and is injective because its zero class is precisely $K$. Therefore $\boxed{R/\ker \varphi \cong S}$, the first isomorphism theorem for rings.

Apply this to the evaluation homomorphism $\mathbb{C}[z]\to \mathbb{C}$, $p\mapsto p(1)$. Constants prove surjectivity, and division by $z-1$ shows that its kernel of a ring homomorphism is $(z-1)$. Hence $\boxed{\mathbb{C}[z]/(z-1)\cong \mathbb{C}}$.

The evaluation homomorphism $\mathbb{R}[X]\to \mathbb{R}$, $p\mapsto p(0)$, is surjective and has kernel of a ring homomorphism $J_1=(X)$. Hence $J_1$ is an ideal, and the first isomorphism theorem for rings gives $\boxed{\mathbb{R}[X]/J_1\cong \mathbb{R}}$. Since this quotient is a field, $\boxed{J_1\text{ is maximal}}$.

Here $J_4=(X^2+1)$ is a principal ideal. The evaluation homomorphism $p\mapsto p(i)$ from $\mathbb{R}[X]$ onto $\mathbb{C}$ has this kernel of a ring homomorphism: division by $X^2+1$ leaves a remainder $a+bX$, and $a+bi=0$ with real $a,b$ forces both to vanish. Surjectivity follows because every complex number is $a+bi$. ConsequentlyEquivalently, $X^2+1$ is an irreducible polynomial over the real numbers.

An algebraic line bundle over $A$ is an invertible module: a finitely generated projective module $L$ such that $L_{\mathfrak{p}}\cong A_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$. Define the Picard group of a ring as the set of its isomorphism classes, withThe tensor product of modules of two finite projective modules is finite projective: write each as a direct summand of a finite free module and tensor the two decompositions. Localization gives $(L{\otimes }_{A}K{)}_{\mathfrak{p}}\cong A_{\mathfrak{p}}$, so this multiplication stays within the set of invertible modules. It is well defined on isomorphism classes. The associativity and symmetry isomorphisms of the tensor product give associativity and commutativity, and $[A]$ is the identity.

Let $L^{\ast }={\mathrm{Hom}}_A(L,A)$ be the dual module. A finite direct-summand presentation shows that $L^{\ast }$ is finite projective and that localization commutes with this dual. Since $L_{\mathfrak{p}}$ is free of rank one, so is $(L^{\ast }{)}_{\mathfrak{p}}$. The evaluation homomorphismis an isomorphism after every prime localization. Its kernel and cokernel are therefore zero, by exactness of localization and localization detects zero elements for modules. Thus $[L{]}^{-1}=[L^{\ast }]$, proving that $\mathrm{Pic}(A)$ is an abelian group.

For a ring homomorphism $f:A\to B$, tensoring a split presentation $L\oplus L^{\prime }\cong A^r$ gives $(B{\otimes }_{A}L)\oplus (B{\otimes }_A{L}^{\prime })\cong B^r$, so $B{\otimes }_{A}L$ is finite projective. For $\mathfrak{q}\in \mathrm{Spec}B$ put $\mathfrak{p}=f^{-1}(\mathfrak{q})$. Every element of $A\setminus \mathfrak{p}$ becomes a unit in $B_{\mathfrak{q}}$, andThis proves base change of invertible modules. Moreover, the canonical mapis an isomorphism, with inverse $b\otimes (\ell \otimes k)\mapsto (b\otimes \ell )\otimes (1\otimes k)$. It shows that $[L]\mapsto [B{\otimes }_{A}L]$ is a group homomorphism. Identity maps induce the identity, and for $A\to B\to C$, the associativity isomorphism $C{\otimes }_B(B{\otimes }_{A}L)\cong C{\otimes }_{A}L$ proves composition compatibility. Therefore $\mathrm{Pic}$ is a covariant functor to abelian groups.

The polynomial ring $\mathbb{C}[x]$ is a principal ideal domain by division with remainder. A submodule $N$ of a finite free module over a principal ideal domain is free: induct on the ambient rank, project onto the first coordinate, and note that the image is a principal ideal, hence either zero or a free rank-one module. This gives a split exact sequence with that image as quotient; the kernel lies in a free module of one smaller rank, completing the induction. A finite projective module is such a submodule, and so is free. An invertible module has rank one after passing to the fraction field, so its free rank is exactly one. Hence every algebraic line bundle over $\mathbb{C}[x]$ is trivial and

No. Take $R=k[t]$ over a field $k$, and inside $R[x]=k[t,x]$ takeThis is a subring containing $R$: its elements have arbitrary constant coefficient in $k[t]$ as polynomials in $x$, and every positive-degree coefficient is divisible by $t$; that condition is preserved by sums and products. The ring $R$ is Noetherian. Indeed, a nonzero ideal of $k[t]$ has an element $g$ of least degree, and division by $g$ shows that every element of the ideal is a multiple of $g$. Thus $k[t]$ is a principal ideal domain, and its ideals are finitely generated.

In $S$, let $J_n=(tx,t{x}^2,\dots ,t{x}^n{)}_S$. These form an ascending chain. We show that $t{x}^{n+1}\notin J_n$. If it belonged, we could writeCancel $t$ in the integral domain $k[t,x]$, then apply the evaluation homomorphism $t\mapsto 0$. Every $s_j\in S$ becomes a scalar in $k$, so this would givewhich is impossible by comparison of degrees. Hence $J_n⊊J_{n+1}$ for every $n$, andThis is a constant-plus-ideal non-Noetherian subring construction: the restricted coefficients prevent a finite list of positive powers of $x$ from generating all the required powers.

Letbe a primitive polynomial. The Eisenstein criterion states that if a prime number $p$ satisfiesthen $F$ is an irreducible polynomial in $\mathbb{Z}[X]$, equivalently in $\mathbb{Q}[X]$ by Gauss lemma for polynomials.

To prove it, suppose that $F=GH$ with $G,H\in \mathbb{Z}[X]$ of positive degree. Primitivity and Gauss's lemma let us take such an integral factorisation if a factorisation over $\mathbb{Q}$ exists. Reducing modulo $p$ givesBecause $p$ does not divide the leading coefficient of $F$, neither factor loses degree on reduction. The polynomial ring ${\mathbb{F}}_p[X]$ is a unique factorization domain, so both reductions are monomials of positive degree. In particular, $p$ divides both constant terms $G(0)$ and $H(0)$. It follows that $p^2$ dividesa contradiction. This proves the criterion.

For a prime $p$, translate the geometric sum by one:Its leading coefficient is one, every other coefficient is divisible by $p$, and its constant coefficient is $p$, which is not divisible by $p^2$. It is therefore Eisenstein at $p$. Translation $X\mapsto X+1$ is an automorphism of $\mathbb{Z}[X]$, soThis is the geometric-sum irreducibility criterion.

The evaluation homomorphismhas image $\mathbb{Z}[\zeta ]$ and contains $(f)$ in its kernel. Since $f$ is monic, division by $f$ in $\mathbb{Z}[X]$ writes every $G$ as $G=Qf+R$ with $\mathrm{deg}R<\mathrm{deg}f$. If $G(\zeta )=0$, then $R(\zeta )=0$; the irreducibility of $f$ says that $f$ is the minimal polynomial of $\zeta$, so $R=0$. Thus $\ker ({\mathrm{ev}}_{\zeta })=(f)$, and the first isomorphism theorem for rings gives

Now take $p=3$. Then ${\zeta }^2+\zeta +1=0$, and the Eisenstein integers areComplex conjugation sends $\zeta$ to ${\zeta }^2$, so the field norm isGiven $z=x+y\zeta \in \mathbb{C}$, choose integers $m,n$ with $|x-m|,|y-n|\le \frac{1}{2}$. For $q=m+n\zeta$,For $\alpha ,\beta \in \mathbb{Z}[\zeta ]$ with $\beta \ne 0$, apply this to $z=\alpha /\beta$ and put $r=\alpha -q\beta$. ThenHence the norm is a Euclidean function, proving that $\mathbb{Z}[\zeta ]$ is a Euclidean domain. This is the Euclidean norm on the Eisenstein integers.

Finally suppose $A\in {\mathrm{GL}}_n(\mathbb{Z})$ satisfies $A^2+A+I=0$. Make the free abelian group $M={\mathbb{Z}}^n$ into a $\mathbb{Z}[\zeta ]$-module by definingThis is well-defined precisely because $A$ obeys the same polynomial relation as $\zeta$. The module is finitely generated. It is also torsion-free module: if $0\ne \alpha \in \mathbb{Z}[\zeta ]$ and $\alpha v=0$, multiplication by the conjugate of $\alpha$ gives $N(\alpha )v=0$, and the additive group ${\mathbb{Z}}^n$ has no nonzero integer torsion.

A Euclidean domain is a principal ideal domain, and the structure theorem for finitely generated modules over a principal ideal domain says that a finitely generated torsion-free module over one is free. Thusfor some $m$. Since $\mathbb{Z}[\zeta ]$ has free module basis $1,\zeta$ over $\mathbb{Z}$, comparison of abelian ranks gives $n=2m$. There can consequently be no such matrix when $n$ is odd.

If $n=2m$, choose a $\mathbb{Z}[\zeta ]$-basis $v_1,\dots ,v_m$. Thenis a $\mathbb{Z}$-basis, and $A^2{v}_i=-v_i-A{v}_i$. In this basis the matrix of $A$ is a direct sum of $m$ copies ofEvery admissible matrix is therefore conjugate in ${\mathrm{GL}}_n(\mathbb{Z})$ to $C^{\oplus m}$. Hence there is exactly one conjugacy class for even $n$, and none for odd $n$. This is the classification of integral matrices satisfying the third cyclotomic polynomial.

A finitely generated module $M$ over a commutative ring $A$ is reflexive when its evaluation homomorphism $M\to M^{\ast \ast }$ to the double dual module is an isomorphism.