Put $N=⟨E^{\prime }⟩$ and $P=⟨E^{\prime \prime }⟩$, the submodules generated by the two parts of the basis of a module. Every finite expression in the basis of a module splits into its $E^{\prime }$ terms and its $E^{\prime \prime }$ terms. Thus $M=N+P$.

If $m\in N\cap P$, there are finite expressionsSubtracting gives a finite relation among distinct members of $E$, because $E^{\prime }\cap E^{\prime \prime }=\varnothing$. Their linear independence in a module forces every coefficient to vanish, so $m=0$. HenceThis proves the basis partition gives a module direct sum lemma, also when one part is empty.

In the direct sum of modules $M=N\oplus P$, every $m\in M$ has a unique expression $m=n+p$, with $n\in N$ and $p\in P$. DefineThe uniqueness makes this well-defined. For $m=n+p$, $m^{\prime }=n^{\prime }+p^{\prime }$ and $r\in R$,so $\pi$ is a module homomorphism. Finally, $\pi (n+p)=0$ exactly when $p=0$, equivalently $m\in N$. Therefore $\boxed{\ker \pi =N}$. Its image is $P$, and ${\pi }^2=\pi$, so it is the projection onto that direct summand.

Yes. An $\mathbb{F}$-module over a field is a vector space. Start with a finite generating list. If it is linearly dependent, one vector has a nonzero coefficient in a linear relation; dividing by that coefficient expresses the vector using the others, so it can be removed without changing the span. Repeating terminates with a linearly independent generating list, which is a basis and therefore a basis of a module.

Thus every finitely generated module over a field is a free module. For the zero module, the empty list is the required basis of a module.

A subset $E\subseteq M$ is a basis of a module if every $m\in M$ has a unique expressionUniqueness concerns the entire family of coefficients, including the zero coefficients. Equivalently, $E$ generates $M$ and is linearly independent in a module: every finite relation ${\sum }_{e\in H}{r}_{e}e=0$, with $H\subseteq E$, has all $r_e=0$.

A module is a free module if it has a basis of a module. An infinite basis of a module still uses only finite sums in each expression; arbitrary infinite sums are not part of the module operations.

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.

Choose a generator $u$ of the cyclic group $U(R)$, so $u^p=1$. Part (a)(iii) gives a unital ring homomorphism ${\mathbb{F}}_2\to R$. Its image lies in the centre of $R$, so even if $R$ is not a commutative ring, evaluating a polynomial at $u$ defines a ring homomorphismBecause $\varphi (X^p-1)=0$, it factors through the quotient ring to giveIts image is the subring $S={\mathbb{F}}_2[u]$. Every unit of $S$ is a unit of $R$, because an inverse in $S$ also lies in $R$. Conversely, every member of $U(R)$ is $u^j$ for some $j$, and both $u^j$ and its inverse $u^{p-j}$ lie in $S$. HenceThe inverse argument matters: merely belonging to a subring would not by itself make an ambient unit invertible there.

The polynomial ring ${\mathbb{F}}_2[X]$ is a Euclidean domain under polynomial degree, hence a principal ideal domain. By part (b)(i), eachis a field. By polynomial division, every residue class has exactly one representative of polynomial degree less than $n_j$. Thus $1,\overline{X},\dots ,{\overline{X}}^{n_j-1}$ form a basis over ${\mathbb{F}}_2$, and $|K_j|=2^{n_j}$.

For $j\ne k$, the ideals $(f_j)$ and $(f_k)$ are distinct maximal ideals, so they are comaximal ideals: their sum is an ideal properly containing $(f_j)$ and must be the whole ring. The Chinese remainder theorem for ideals and part (b)(ii) therefore give the ring isomorphismIn particular, this is a product of finite fields of the required sizes. The map preserves multiplication as well as addition and is surjective by the Chinese remainder theorem for ideals.

Every multiple of $x_1\cdots x_r$ is a multiple of each $x_j$, giving one inclusion. For the reverse inclusion, a principal ideal domain is a unique factorization domain, so each irreducible element $x_j$ is a prime element.

Let $a\in {\bigcap }_{j=1}^r(x_j)$. The case $a=0$ is immediate. Otherwise, write $a=x_{1}b$. Since $x_2\mid a$ and the prime element $x_2$ cannot divide the non-associate irreducible element $x_1$, it divides $b$. Write $a=x_1{x}_2{b}_2$, and repeat. At stage $j$, the prime element $x_j$ divides none of $x_1,\dots ,x_{j-1}$, so it divides the remaining factor. Hence $x_1\cdots x_r\mid a$, proving

Since the irreducible element $x$ is not a unit, the principal ideal $(x)$ is proper. Suppose $(x)\subseteq J\subseteq R$. As $R$ is a principal ideal domain, write $J=(d)$. Then $x=dr$ for some $r\in R$. The definition of an irreducible element says that $d$ or $r$ is a unit.

If $d$ is a unit, then $J=R$. If $r$ is a unit, then $d$ and $x$ are associate elements, so $J=(x)$. Thus $(x)$ is a maximal ideal. By the permitted criterion that a quotient ring by a maximal ideal is a field,

The element $-1_R$ lies in the unit group and satisfies $(-1_R{)}^2=1_R$. Its order of a group element divides two and, by Lagrange's theorem, divides the odd number $|U(R)|$. Its order is therefore one, so $-1_R=1_R$ and $2\cdot 1_R=0$.

It follows thatis a well-defined unital ring homomorphism. For a nonzero ring, this says its characteristic of a ring is two. If the zero ring is permitted, the same map still exists. This is the odd unit group forces characteristic two argument, with that harmless exception to the characteristic wording.

Write $T={\prod }_{j=1}^n{R}_j$, and let ${\rho }_j:T\to R_j$ be the coordinate ring homomorphism. Set $I_j={\rho }_j(I)$. Each $I_j$ is an ideal: it is an additive subgroup, and multiplying a lift in $I$ on either side by a tuple with prescribed $j$th coordinate proves closure under left and right multiplication by $R_j$.

Certainly $I\subseteq {\prod }_j{I}_j$. Conversely, let $a=(a_1,\dots ,a_n)\in {\prod }_j{I}_j$. For each $j$, choose $b^{(j)}\in I$ with $j$th coordinate $a_j$, and let $e_j$ have $1_{R_j}$ in coordinate $j$ and zero elsewhere. Since $I$ is an ideal, $e_j{b}^{(j)}\in I$, andThereforeThe finite sum is essential to this proof of ideals of a finite product of rings.

Multiplication in a direct product of rings is coordinatewise. Thus $(r_1,\dots ,r_n)$ has a two-sided inverse exactly when each $r_j$ has a two-sided inverse in $R_j$, and thenConsequently its unit group is the direct product of groupsThe identification preserves multiplication because both sides use coordinatewise multiplication. This is the unit group of a direct product of rings formula.

Write $q=|\mathbb{F}|$. The unit group $U(\mathbb{F})=\mathbb{F}\setminus \{0\}$ is a finite abelian group of group order $q-1$, since multiplication in a field satisfies commutativity. The structure theorem for finitely generated abelian groups, applied to this finite group, gives invariant factors of a finitely generated moduleafter omitting trivial factors. If there are no factors, the unit group is trivial and hence a cyclic group.

Otherwise put $D=d_r$. Every element $u\in U(\mathbb{F})$ satisfies $u^D=1$, because the order of a group element in each coordinate divides $D$. Hence all $q-1$ distinct nonzero elements of $\mathbb{F}$ are roots of a polynomial $X^D-1$. The root bound for a polynomial over a field givesFor completeness, that bound follows by induction from the factor theorem: if a nonzero polynomial $f$ vanishes at $a$, write $f=(X-a)g$; every other root of a polynomial is a root of $g$, because a field has no zero divisors. Each new distinct root therefore uses one unit of polynomial degree.

On the other hand, the group order in the decomposition isEquality is forced, so $d_1\cdots d_{r-1}=1$. As every retained factor is greater than $1$, there can be only one. ConsequentlyThe same proof establishes the more general result finite multiplicative subgroup of a field is cyclic. The square-free integer result in part (c)(i) alone would not suffice, since $q-1$ need not be a square-free integer; the root bound for a polynomial rules out multiple invariant factors of a finitely generated module even then.

Let $N=|G|$ be a square-free integer. If $N=1$, then $G$ is the trivial cyclic group. Otherwise, finiteness means that a direct sum of modules decomposition into cyclic groups has only finitely many nontrivial factors, each of finite group order. Omit trivial factors and writeThe integers $n_i,n_j$ are coprime integers whenever $i\ne j$. If a prime number $p$ divided both, then $p^2\mid N$, contradicting the square-free integer hypothesis.

Moreover, $n_1\cdots n_{j-1}$ and $n_j$ are coprime integers: any common prime divisor would divide one of the earlier factors as well as $n_j$. Thus part (b)(ii) can be applied repeatedly to combine factors, givingIn particular $G$ is a cyclic group. Under the displayed identifications, the element whose coordinate in every factor is the class of $1$ is a generator of a group.

Define the group homomorphismBy part (b)(i), its kernel of a group homomorphism isIt is surjective. Indeed, choose $r,s\in \mathbb{Z}$ with $rd+sh=1$ by Bézout's identity. For any prescribed classes $u+d\mathbb{Z}$ and $v+h\mathbb{Z}$, the integersatisfies $t\equiv u(\mathrm{mod}d)$ and $t\equiv v(\mathrm{mod}h)$. The first isomorphism theorem for groups therefore givesThis proves the required isomorphism and is the Chinese remainder theorem for coprime integers. It also shows that $(1+d\mathbb{Z},1+h\mathbb{Z})$ generates the direct sum of modules when these additive groups are viewed as $\mathbb{Z}$-modules.

Every multiple of $dh$ is a multiple of both $d$ and $h$, so $dh\mathbb{Z}\subseteq d\mathbb{Z}\cap h\mathbb{Z}$.

For the reverse inclusion, let $z=du=hv$ lie in this intersection of ideals. Since $d,h$ are coprime integers, Bézout's identity supplies $r,s\in \mathbb{Z}$ with $rd+sh=1$. Multiplying by $u$ givesThus $h\mid u$, and $dh\mid du=z$. ConsequentlyHere Bézout's identity follows from the allowed principal ideal domain property of $\mathbb{Z}$: the ideal $d\mathbb{Z}+h\mathbb{Z}$ is generated by their greatest common divisor, which is $1$.

Let $n_1,\dots ,n_t$ generate the submodule $N$, and let $m_1+N,\dots ,m_s+N$ generate the quotient module $M/N$. Choose each $m_i\in M$ as a representative of its class. For any $m\in M$, there are $r_1,\dots ,r_s\in R$ such thatTherefore $m-{\sum }_i{r}_i{m}_i\in N$. Expressing that difference using the generating set of a module $n_1,\dots ,n_t$ givesHenceis a finitely generated module. This proves that finite generation is preserved by module extensions.

Choose a finite generating set of a module $n_1,\dots ,n_t$ for the finitely generated module $N$. For any $m\in M$, the surjectivity of the module homomorphism $\varphi$ gives $n\in N$ with $\varphi (n)=m$. Write $n={\sum }_{j=1}^t{r}_j{n}_j$. Since $\varphi$ is a module homomorphism,Thusis a finitely generated module. The argument includes $N=0$, in which case surjectivity forces $M=0$ and the empty generating set of a module suffices.

The required property isFor a nonzero element $x$ of an integral domain $R$, this follows fromIndeed, $\overline{a}\overline{b}=0$ in the quotient ring exactly when $x\mid ab$. The prime element condition says this forces $x\mid a$ or $x\mid b$, exactly the absence of zero divisors in the quotient ring. The nonunit condition makes the quotient ring nonzero. Being an integral domain is preserved by every ring isomorphism.

With the usual definition of a prime element, $x\ne 0$ is understood. The prime ideal formulation works for every ideal of any commutative ring, including the zero ideal; it avoids this convention about elements.

No such property exists. The ring isomorphism in part (b)(ii) identifies quotient rings by elements having different irreducible element behaviour.

The element $X^3$ is an irreducible element of $T$. It is nonzero and a nonunit. Every nonzero nonunit of $T$ has polynomial degree at least $2$: the only possible polynomials of smaller polynomial degree are constant polynomials, and every nonzero constant polynomial is a unit. If $X^3=uv$ with $u,v\in T$ both nonunits, additivity of polynomial degree in the integral domain $\mathbb{R}[X]$ would givewhich is impossible. Thus one factor must be a unit.

In $\mathbb{R}[Y]$, however, $Y^3=Y\cdot Y^2$ is a product of two nonunits, so $Y^3$ is not an irreducible element. If a ring isomorphism invariant $P$ characterized irreducible elements by the indicated quotient ring, the irreducible element $X^3\in T$ would imply that $T/(X^3{)}_T$ has $P$. Its isomorphic quotient ring $\mathbb{R}[Y]/(Y^3)$ would then have $P$, incorrectly making $Y^3\in \mathbb{R}[Y]$ an irreducible element.

Consequently,