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The isomorphism theorems relate homomorphisms, their kernels and images, and quotient constructions in algebra. For groups, the first isomorphism theorem identifies a homomorphism's image with the quotient group by its group kernel. Analogous statements include the first isomorphism theorem for rings and the First isomorphism theorem for vector spaces: identifying precisely the inputs with the same output preserves the remaining algebraic structure.

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

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.

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, $\boxed{J_1\text{ is maximal}}$.

The Chinese remainder theorem for two comaximal two-sided ideals $I,J$ of a ring states that the natural ring homomorphism inducesIndeed, the map $r\mapsto (r+I,r+J)$ has kernel of a ring homomorphism $I\cap J$. Choose $i\in I$, $j\in J$ with $i+j=1$. Given target representatives $a,b\in R$, the element $aj+bi$ is congruent to $a$ modulo $I$ and to $b$ modulo $J$, proving surjectivity. The first isomorphism theorem for rings gives the result. For a commutative ring, $I\cap J=IJ$: if $x\in I\cap J$, then $x=xi+xj\in IJ$, while the reverse inclusion follows from the ideal properties.

Set $X=0$ to define a surjective ring homomorphismIts kernel of a ring homomorphism is exactly $(X)$: the constant coefficient in $X$ must vanish, so every remaining monomial is divisible by $X$. The first isomorphism theorem for rings therefore givesThe polynomial ring $\mathbb{Q}[Y]$ is an integral domain, because the product of two nonzero polynomials has nonzero leading coefficient. It is not a field: $Y$ has no polynomial multiplicative inverse, since a nonzero product $YQ(Y)$ has positive degree of a polynomial. Consequently $(X)$ is prime and is not maximal. Concretely, $(X)⊊(X,Y)⊊\mathbb{Q}[X,Y]$ is a strict chain of ideals.

The first isomorphism theorem for rings states that the mapis a ring isomorphism. It is well defined because two representatives differ by an element of the kernel exactly when their images agree. It preserves addition and multiplication, is onto its stated codomain, and has trivial kernel. In particular, if $f$ is surjective then $R/\ker f\cong S$.

A complex number is an algebraic number over $\mathbb{Q}$ when it is a root of a nonzero rational polynomial. Its minimal polynomial of an algebraic element is the unique monic polynomial in $\mathbb{Q}[X]$ of smallest degree annihilating it; polynomial division shows that it divides every rational polynomial annihilating the element, and it is irreducible.

Consider the surjective evaluation ring homomorphism ${\mathrm{ev}}_{\alpha }:\mathbb{Z}[X]\to \mathbb{Z}[\alpha ]$. Certainly $(f)\subseteq \ker {\mathrm{ev}}_{\alpha }$. Because the nonconstant $f$ is irreducible over $\mathbb{Z}$, its content is a unit: otherwise its content and its nonconstant primitive part give a factorization into nonunits. By the Gauss lemma for polynomials, $f$ is irreducible in $\mathbb{Q}[X]$ and is a scalar multiple of the minimal polynomial.

If $g\in \mathbb{Z}[X]$ vanishes at $\alpha$, then $f$ divides $g$ in $\mathbb{Q}[X]$. Primitive divisibility in Gauss lemma for polynomials implies division in $\mathbb{Z}[X]$ as well. Explicitly, write the rational quotient as $(a/b)h$ with $h$ primitive integral and $\gcd (a,b)=1$; the primitive product $fh$ shows that integrality of $(a/b)fh$ forces $b=1$. Thus the kernel is precisely $(f)$, and the first isomorphism theorem for rings yieldsNo monicity of $f$ is needed; this is an isomorphism of rings, not an assertion that either ring is a field.

Yes. Taking the constant coefficient defines a surjective ring homomorphismIts kernel of a ring homomorphism is $(x)$, so the first isomorphism theorem for rings gives $R[[x]]/(x)\cong R$.

We prove the needed preservation result. If $A$ is a Noetherian ring and $J\subseteq A$ is an ideal, any ascending chain of ideals in $A/J$ lifts under the quotient ring homomorphism to an ascending chain of ideals in $A$. The lifted chain stabilizes; surjectivity makes the original chain stabilize as well. Hence every quotient ring of a Noetherian ring is Noetherian.

Applying this with $A=R[[x]]$ and $J=(x)$ gives

By the prime ideal quotient criterion,For a principal ideal, this is precisely the assertion that $p$ is a prime element. This closes the cycle and proves that (i)--(v) are equivalent.

Now let $\phi :R\to S$ be a surjective ring homomorphism, where $R$ is a principal ideal domain and $S$ is an integral domain. WriteIf $a=0$, the first isomorphism theorem for rings makes $\phi$ an isomorphism. Otherwiseis an integral domain, so the equivalences just proved show that $(a)$ is maximal and $S$ is a field.

Next suppose that the polynomial ring $R[X]$ is a principal ideal domain. It follows first that $R$ is an integral domain. For nonzero $a\in R$, the idealis principal. Since $f$ divides the nonzero constant $a$, it is constant; since it also divides $X$, that constant divides the coefficient $1$, so $f$ is a unit. Therefore $(a,X)=R[X]$, and there are polynomials $g,h$ such thatSetting $X=0$ gives $1=ag(0)$, so every nonzero $a\in R$ is a unit. Hence

Finally, let $R$ be an integral domain in which every two nonzero elements have a greatest common divisor. If an irreducible $p$ divides $ab$, then $\gcd (p,a)$ is either a unit or an associate of $p$. In the second case $p\mid a$. In the first case, the Euclid lemma in a greatest-common-divisor domain gives $p\mid b$. Thus every irreducible element of $R$ is prime.

The first isomorphism theorem for rings states that a ring homomorphism $\varphi :R\to S$ induces an isomorphismBecause $J$ is an ideal, sums and products of elements of $R+J$ remain in $R+J$, so it is a subring. The surjective homomorphismhas kernel $R\cap J$. The theorem givesEvaluation at $-2$ givesso this quotient is a field of characteristic zero. The other quotient isIt has characteristic $3$ but is not a field, since the nonzero class of $X-1$ is nilpotent.

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.

Using the assumed square-free polynomial property, factorinto distinct monic polynomials that are irreducible polynomials, with $n_j=\mathrm{deg}{f}_j$. Part (b)(iii) identifies $T={\mathbb{F}}_2[X]/(X^p-1)$ with ${\prod }_j{K}_j$, where $|K_j|=2^{n_j}$.

By the first isomorphism theorem for rings, the image $S$ from part (c)(i) is $T/J$, where $J$ is the kernel of a ring homomorphism. By part (a)(ii), $J={\prod }_j{J}_j$ for ideals $J_j\subseteq K_j$. A field has only the zero ideal and the whole field as ideals, so each $K_j/J_j$ is either $K_j$ or the zero ring. Consequently, for a subset $D\subseteq \{1,\dots ,r\}$,This is the quotient of a finite product of fields principle. The subset is nonempty because $|U(S)|=p>2$, whereas the zero ring has only one unit. By part (a)(i),Each factor is a positive integer. Since $p$ is a prime number, exactly one factor is $p$ and all remaining factors are one. Thusso $p$ is a Mersenne prime. If $n=ab$ with $a,b>1$, then $2^a-1$ would be a proper divisor of $2^{ab}-1$; thus $n$ is itself a prime number, although only the displayed form is needed.

The assumed square-free polynomial property can also be checked directly: in characteristic of a ring two, the formal derivative of $X^p-1$ is $X^{p-1}$ because $p$ is odd. These two polynomials have greatest common divisor one, since $X$ does not divide $X^p-1$, so no repeated irreducible polynomial factor occurs.

A quotient ring of a finite direct product of rings whose factors are fields is a product of a subset of those factors. By ideals of a finite product of rings, write $J={\prod }_j{J}_j$. Since a field has only the zero ideal and the whole field as ideals, every $J_j$ is one of these. The ring homomorphism $(x_j)\mapsto (x_j+J_j)$ is surjective with kernel of a ring homomorphism $J$, so the first isomorphism theorem for rings gives $({\prod }_j{K}_j)/J\cong {\prod }_j(K_j/J_j)$. Discard the zero ring factors. If no factors survive, the result is the zero ring.