Alternate Proof of Hilbert's Basis Theorem and Gröbner Bases (9.6.43-44)

Dummit and Foote Abstract Algebra, section 9.6, exercises 43-44:

43. Let $\mathcal{S}=\{N~|~N~\text{not a finitely generated monomial ideal}\}$ and assume by way of contradiction that $\mathcal{S}≠\varnothing$.

(a) Prove $\mathcal{S}$ has a maximal element.
(b) Show there exist monomials $x,y∉M$ such that $xy∈M$.
(c) Show there exists $M_0 \subseteq M$ with $M_0$ a finitely generated monomial ideal such that $M_0+(x)=M+(x)$ and $M=M_0+(x)(M : (x))$. Deduce $M$ is finitely generated, forcing $\mathcal{S}=\varnothing$.

44. If $\varnothing≠I\subseteq F[x_1,..,x_n]$ is an ideal, prove $LT(I)$ is finitely generated. Conclude $I$ has a Gröbner basis and deduce Hilbert's Basis Theorem.

Proof: (a) Let $\mathcal{C}$ be a typical chain totally ordered by inclusion. Since $M=\bigcup_{N∈\mathcal{C}} N$ is a monomial ideal by 9.6.39, by Zorn's Lemma it suffices to show it is infinitely generated; assume the contrary, so that $M=(n_1,...,n_r)$. Let $n_i$ be maximal with regard to the ordering $n_i∈N_i \geq n_j∈N_j :⇔ N_j \subseteq N_i$, and observe $N_i \subseteq M = (n_1,...,n_r) \subseteq N_i$ so that $N_i=M$ is finitely generated, a contradiction.

(b) By 9.6.42(a), the only prime monomial ideals are finitely generated. Therefore there exist polynomials $x',y'∈F[x_1,...,x_n]$ such that $x',y'∉M$ and $x'y'∈M$. By 9.6.10, there exist monomial terms in $x',y'$ not divisible by any $m_i$ for $M=(m_1,...)$. Showing $x'y'=m∈M$ and expanding $x'$ to subtract the products of $y'$ and monomial terms of $x'$ divisible by monomials of $M$ (which are elements of $M$), we obtain $xy'∈M$ for $x$ composed of terms not divisible by monomials of $M$. Similarly, $xy∈M$. Now $LT(xy)=LT(x)LT(y)∈LT(M) \subseteq M$ (the latter $\subseteq$ being due again to 9.6.10) so that $LT(x)$ and $LT(y)$ are monomials not within $M$ whose product is within $M$.

(c) Fix a monomial ordering $≤$. First note that $m \mid m' ⇒ md=m'$ so that, since $d \geq 1$, we have $md = m' \geq m$, i.e. $m \mid m' ⇒ m ≤ m'$.

Let $x$ be the minimal monomial by $≤$ not within $M$ such that there exists another monomial $y$ not within $M$ fulfilling $xy∈M$. Revisit $M=(m_1,...)$ and let $M^*=(m_1,...,x)$ as before. Since $M \subset M^*$ by the maximality of $M$ we must have $M^*=(T)$ for some finite subset of monomials $T=\{q_1,...,q_t\}$. Reformat this generating subset so that it is minimal, by excluding a generator $n$ if $M^*=(T \setminus \{n\})$. This process is seen to terminate, and in part we are left with $T=\{n_1,...,n_s\}$ where $n_i \not \mid n_j$ for all $i≠j$. Assume $x∉T$: Then by 9.6.10 since $x∈M^*$ we have $n_i \mid x$ so $n_id=x$ for some $i$, and since $n_i∈M⇒x∈M$, yet also $n_i∉M⇒n_i < x \land n_i(dy)∈M$ with $dy∉M$ since $d < x$ as well, we inevitably violate the minimality of $x$, and we must instead have $x∈T$. Rewrite $T=\{n_1,...,n_{s-1},x\}$.

Now let $M_0=(T \setminus \{x\})$. Assume $n_j∈T \setminus \{x\}$ yet $n_j∉M$: Since the elements of $T$ are from $M^*$ generated by $\{m_1,...,x\}$ we must have either $m_i \mid n_j$ for some $i$ or $x \mid n_j$; the latter is excluded by the minimalization of $T$ and the former implies $n_j∈M$. Therefore $M_0 \subseteq M$, and evidently $M_0+(x)=M+(x)=M^*$.

Show $M=M_0+(x)(M : (x))$: ($\supseteq$) By the definition of $(M : (x))$, the direction is clear. ($\subseteq$) We have $m_i∈M \subseteq M_0+(x)$, so that either $n_k \mid m_i ⇒ m_i∈M_0$ or $x \mid m_i ⇒ m_i ∈ (x)(M : (x))$.

Finally, we see that $M_0$ is finitely generated, and also $y∉M$ with $y∈(M : (x))$ so that $(M : (x))$ is finitely generated by the maximality of $M$, and also $(x)(M : (x))$ and now $M_0+(x)(M : (x))=M$ is finitely generated, a contradiction.$~\square$

44. Since $LT(I)$ is a monomial ideal, by the previous it is finitely generated. In the same way as in 9.6.1, we may generate each of the elements of the finite generating set of $LT(I)$ by a finite number of finite combinations of $LT(g_i)$ for $g_i$ elements of $I$, so that by proposition 24 these $g_i$ form a Gröbner basis and as well a finite generating set for the arbitrary ideal $I$, which is Hilbert's Basis Theorem.$~\square$

Prime and Maximal Monomial Ideals (9.6.42)

Dummit and Foote Abstract Algebra, section 9.6, exercise 42:

(a) Show $M$ is a monomial prime ideal if and only if $M=(S)$ for some $S \subseteq \{x_1,...,x_n\}$.
(b) Show $(x_1,...,x_n)$ is the only maximal monomial ideal.

Proof: (a) ($\Leftarrow$) Letting $I$ be such that $S=\{x_i~|~i∈I\}$, we have $F[x_1,...,x_n]/M ≅ F[x_j~|~j∉I]$ which is again a polynomial ring in several variables over a field and is thus in particular an integral domain. ($⇒$) Since $M$ is a monomial ideal, let $M=(m_1,...)$ with the monomials countably ordered by $m_1 ≤ m_2$ if $\text{deg}(m_2) < \text{deg}(m_1)$ and decided by lexicographic ordering if $\text{deg}(m_2)=\text{deg}(m_1)$. Trim this generating list by a process of eliminating $m_i$ if it is not generatable by linear combinations of previous monomials. Choose $m∉\{x_1,...,x_n\}$ in the generating set. letting $x_i$ be a variable dividing $m$, we must have $x_i∉M$ else by exercise 10 $x_i$ is divisible by a term in the generating list, forcing $x_i$ in the generating list so that by the ordering $x_i$ appears before $m$ and now $m$ could not have been included. Since $x_i$ was arbitrary, we have the product of the variables dividing $m$ violating the primality of $M$. Therefore $m$ does not exist, and this generating subset is a subset of $\{x_1,...,x_n\}$.

(b) Since maximal ideals are prime ideals, if $M=(S)$ with $S≠\{x_1,...,x_n\}$, by part (a) we have $M \subset (x_1,...,x_n) \subset F[x_1,...,x_n]$, a contradiction.$~\square$

Minimal Gröbner Bases' Commonalities (9.6.15)

Dummit and Foote Abstract Algebra, section 9.6, exercise 15:

Fix a monomial ordering on $R=F[x_1,...,x_n]$.
(a) Prove $\{g_1,...,g_n\} \subseteq I \subseteq R$ for $I$ an ideal and monic $g_i$ is a minimal Gröbner basis for $I$ if and only if $\{LT(g_1),...,LT(g_n)\}$ is a minimal generating set for $LT(I)$.
(b) Prove the leading terms and order of a minimal Gröbner basis of $I$ is uniquely determined.

Proof: (a) $⇒$ By the definition of a minimal Gröbner basis, we have $(LT(g_1),...,LT(g_n))=LT(I)$ and $LT(g_i) \not \mid LT(g_j)$ for any $i≠j$. We can see that $(\{LT(g_1),...,LT(g_n)\} \setminus \{LT(g_i)\}) \subset I$ since for there to be equality would be, by exercise 10, to presume an element in $\{LT(g_1),...,LT(g_n)\} \setminus \{LT(g_i)\}$ divides $LT(g_i)$. $\Leftarrow$ By proposition 24 we have $\{g_1,...,g_n\}$ is a Gröbner basis of $I$ and since their leading terms is a minimal generating set for $I$ we must have $LT(g_i) \not \mid LT(g_j)$ for $i≠j$.

(b) Since $LT(I)$ is a monomial ideal, by the previous exercise its minimal generating set of monomials is uniquely determined. By part (a) the leading terms of a minimal Gröbner basis's elements must be uniquely determined, as well as the number of leading terms (which is the order of the basis).$~\square$

Uniqueness of Minimal Generating Monomial Subsets of Ideals (9.6.14)

Dummit and Foote Abstract Algebra, section 9.6, exercise 14:

Suppose $I$ is a monomial ideal in $F[x_1,...,x_n]$ minimally generated by $M=\{m_1,...,m_k\}$ for $m_i$ a monomial. Prove this generating subset is unique.

Proof: Note we must necessarily have $m_i \not \mid m_j$ for all $i≠j$ or else it is not a minimally generating set. Now, assume another minimally generating set $N=\{n_1,...,n_l\}$. Since $(M)=(N)=I$ we have by exercise 10 that $LT(n_1)=n_1$ is divisible by, say, $m_1$. Similarly, $m_1$ is divisible by a monomial from $N$. Continue in this fashion so that we must necessarily reach a point in the chain of division such that $m_i$ repeats, so that $m_i \mid ... \mid n_j \mid ... \mid m_i$ and now $m_i=n_j$. Starting again with an element $m$ in $M$ besides $m_i$, we can follow the chain once more which will reveal another equality between elements, which won't be the one between $m_i$ and $n_j$ as then $m \mid m_i$, a contradiction. In this fashion, every element of the minimally-ordered generating set will be paired up with its equal in the opposite set, i.e. will be a subset of the opposite set. By the nature of minimally generated sets, this inclusion cannot be proper, and now $M=N$.$~\square$

Multiplicative and Additive Independence (9.5.7)

Dummit and Foote Abstract Algebra, section 9.5, exercise 7:

For $F$ a field, prove $F^\times \not ≅ F^+$.

Proof: Assume $F$ is finite. Then $|F^\times|=|F^+|-1$, so they cannot be in bijection. Therefore $F$ is infinite: Assume $-1=1$; then $2=0$ implying $a+a=0$ for all $a∈F^+$ so that $u^2=1$ for all $u∈F^\times$, so that there are an infinite number of roots to the equation $x^2-1$, an impossibility. Therefore $1≠-1$ so that $2a=a+a≠0$ for all nonzero $a∈F^+$, so that for all nonidentity $u∈F^\times$ we have $u^2≠1$, a contradiction by $(-1)^2=1$.$~\square$

Polynomial Factorization Practice #2 (9.4.1-4)

Dummit and Foote Abstract Algebra, section 9.4, exercise- 1-4:

1. Factor the following polynomials:
(a) $x^2+x+1∈\mathbb{F}_2[x]$
(b) $x^3+x+1∈\mathbb{F}_3[x]$
(c) $x^4+1∈\mathbb{F}_5[x]$
(d) $x^4+10x+1∈\mathbb{Z}[x]$
2. Prove the following polynomials are irreducible in $\mathbb{Z}[x]$:
(a) $x^4-4x^4+6$
(b) $x^6+30x^5-15x^3+6x-120$
(c) $x^4+4x^3+6x^2+2x+1$
(d) $\dfrac{(x+2)^p-2^p}{x}$ for $p$ an odd prime.
3. Show that $f_n(x)=(x-1)(x-2)...(x-n)-1$ is irreducible over $\mathbb{Z}$ for all $n \geq 1$.
4. Show that $f_n(x)=(x-1)(x-2)...(x-n)+1$ is irreducible over $\mathbb{Z}$ for all $n \geq 1$ and $n≠4$.

Proof: (1) (a) There are no zeros to this equation, so it is irreducible.
(b) By discovering $f(1)=0$, we obtain $x^3+x+1=(x-1)(x^2+x+2)$, where the latter has no zeros.
(c) $x^4+1=x^4-4=(x^2-2)(x^2+2)$.
(d) Reduce modulo 5. If $f(x)=a(x)b(x)$, then due to monicity of $f(x)$ we have $a(x)$ and $b(x)$ are monic and not collapsing to units so that $\overline{a(x)}$ is associate to $\overline{x^2+2}$ or $\overline{x^2-2}$, say the former. Since $\overline{a(x)}$ is monic we have $\overline{a(x)}=\overline{x^2+2}$ which forces the constant term of $a(x)$ to not be $\pm 1$, a contradiction.

Alternatively, since $f(x)$ clearly has no zeros in $\mathbb{Z}$, it would have to factor as the product of two quadratics such that the following hold:$$f(x)=a(x)b(x)$$$$a(x)=x^2+a_1x+a_0$$$$b(x)=x^2+b_1x+b_0$$$$a_0b_0=1$$$$a_1b_0+a_0b_1=0$$$$a_1b_1+a_0+b_0=10$$$$a_1+b_1=0$$Solving for $b_1$ as a typical system of equations reveals$$\dfrac{1}{4}b_1^4+15b_1^2+24=0$$which is impossible.

(2) (a) Eisenstein's criterion for $p=2$ applies.
(b) Eisenstein applies for $p=3$.
(c) We generalize the technique employed in example 2. Lemma 1: Let $R$ be an integral domain with $f(x)=a_0+a_1x+...+a_nx^n∈R[x]$ and $g(x)=f(r(x))$ for $r(x)$ a nonconstant polynomial. If $g(x)$ is irreducible, then $f(x)$ is irreducible. Proof: If $f(x)=a(x)b(x)$ for some nonunit polynomials, then $g(x)=a(r(x))b(r(x))$ where necessarily $\text{deg}(a(r(x)))=\text{deg}(a(x)) \cdot \text{deg}(r(x)) \geq 1$ if $\text{deg}(a(x)) \geq 1$ and $a(r(x))=a(x)$ if $a(x)$ is a constant and similarly for $b(r(x))$, a contradiction.$~\square$

Let $g(x)=f(x-1)=x^4-2x+2$, where Eisenstein applies for $p=2$.
(d) $\dfrac{(x+2)^p-2^p}{x}=x^p+...+2p$, where by the binomial theorem the omitted terms have coefficients of multiples of $p$. Eisenstein applies for $p$.

(3) When $n=1$ we clearly have $f_n(x)$ is irreducible, so assume $n > 1$. Assuming $f_n(x)$ has a zero in $\mathbb{Z}$, then $1=(x-1)(x-2)...(x-n)$ so that in particular $x-1=\pm 1$ and $x-2=\pm 1$, a contradiction. Therefore if $f_n(x)=a(x)b(x)$ then the degrees of $a(x)$ and $b(x)$ are greater than one.

We demonstrate a basic algebraic result, first. If $R$ is an integral domain and $f(x)∈R[x]$ of degree $n \geq 1$, then there are at most $n$ zeros of $f(x)$. If not, then $f(x)=(x-\alpha_1)(x-\alpha_2)...(x-\alpha_{n+1})f'(x)$ entails $\text{deg}(f(x)) > n$, a contradiction.

Since $f_n(m)=-1$ for all integers $1≤m≤n$ we have $a(m)= \pm 1$ and $b(m)= \mp 1$ for at least $n$ values. We thus have $a(x)+b(x)$ has at least $n$ zeros, so that necessarily $a(x)=-b(x)$. But now the largest term's coefficient of $a(x)b(x)$ is $-1$, a contradiction.

(4) Similar as in the last problem, $f_1(x)$ is irreducible and $f_n(x)$ has no zeros for $n > 1$, so that necessarily $a(x)=b(x)=\pm 1$ for at least $n$ values, and again $a(x)=b(x)$ so that $f_n(x)$ is a perfect square. However, when $n$ is odd this is an impossibility by degree, so $n$ must be even. For $n < 4$ we can observe $f_n(4)$ is irreducible, and for $n > 4$ and $n$ even we can observe $f_n(3/2)$ is negative, a contradiction since $f_n(x)$ is a perfect square.

When $n=4$, with a process similar to that for (1)(d), we can observe $f_4(x)=(x^2-5x+5)^2$ is its decomposition.$~\square$

Bezout Subring of Q[x] (9.3.5)

Dummit and Foote Abstract Algebra, section 9.3, exercise 5:

Let $R=\mathbb{Z}+x\mathbb{Q}[x] \subset \mathbb{Q}[x]$.

(a) Let $f(x),g(x)∈\mathbb{Q}[x]$ with $x^r$ the maximal power of $x$ dividing both $f(x)$ and $g(x)$, and let $f_r$ and $g_r$ be the coefficients of $x^r$ in $f(x)$ and $g(x)$ respectively. Then $\mathbb{Z}f_r+\mathbb{Z}g_r=\mathbb{Z}d_r$ for some nonzero $d_r\mathbb{Q}$. Prove there is a polynomial $d(x)∈\mathbb{Q}[x]$ such that $d(x)$ is a gcd of $f(x)$ and $g(x)$, and its minimal term is $d_rx^r$.
(b) Prove $f(x)=d(x)q_1(x)$ and $g(x)=d(x)q_2(x)$ for some $q_1(x),q_2(x)∈R$.
(c) Prove $d(x)=f(x)a(x)+g(x)b(x)$ for some $a(x),b(x)∈R$.
(d) Conclude $Rf(x)+Rg(x)=Rd(x)$ in $\mathbb{Q}[x]$, implying in part that $R$ is a Bezout domain.
(e) Show that there must be ideals of $R$ which are not finitely generated, and in fact $x\mathbb{Q}[x]$ is an ideal of $R$ that is not finitely generated.

Proof: (a) We can compute the gcd of $f(x)$ and $g(x)$ in the Euclidean domain $\mathbb{Q}[x]$. By the Euclidean algorithm we have $d(x)=v_1(x)f(x)+v_2g(x)$ for some polynomials $v_1(x),v_2(x)∈\mathbb{Q}$, so that $x^r \mid d(x)$. Furthermore, letting $f_r$ be the one of its pair that is guaranteed nonzero, by $d(x) \mid f(x)$ we must have the coefficient of $x^r$ in $d(x)$ is nonzero. Since this gcd is unique up to multiplication by a nonzero rational number, we can arrange for a gcd $d(x)$ with minimal term $d_rx^r$.

(b) Since $\mathbb{Z}f_r+\mathbb{Z}g_r=\mathbb{Z}d_r$ we have $d_rq_1=f_r$ and $d_rq_2=g_r$ for some $q_1,q_2∈\mathbb{Z}$. By definition of a gcd we have $f(x)=d(x)q_1(x)$ and $g(x)=d(x)q_2(x)$ for some $q_1(x),q_2(x)∈\mathbb{Q}[x]$, and by observation we see the constant terms of $q_1(x)$ and $q_2(x)$ are $q_1$ and $q_2$ respectively, putting these polynomials in $R$.

(c) We can observe $d(x)=f(x)a(x)+g(x)b(x)$ for some $a(x),b(x)∈\mathbb{Q}[x]$, and by 9.2.11 we have a full solution set for $a(x),b(x)$ given by$$a'(x)=a(x)+m(x) \dfrac{g(x)}{d(x)}$$$$b'(x)=b(x)-m(x) \dfrac{f(x)}{d(x)}$$as $m(x)$ ranges over $\mathbb{Q}[x]$.

Assume $f_r$ and $g_r$ are nonzero. First, observe that $d_r=f_r\alpha+g_r\beta$ for some $\alpha,\beta∈\mathbb{Z}$, so that $1=\dfrac{f_r}{d_r}\alpha+\dfrac{g_r}{d_r}\beta$, implying that $v_f=\dfrac{f_r}{d_r}$ is relatively prime to $v_g=\dfrac{g_r}{d_r}$ (which are, foremost, seen to be integers).

Letting $a$ and $b$ be the constant terms of $a(x)$ and $b(x)$ respectively, we can see by our mentioned equation that $(f_rx^r)a+(g_rx^r)b=d_rx^r$, so that $f_ra+g_rb=d_r$ implying $v_fa+v_gb=1$. As such, we can see $v_fz_1+v_gz_2=v_fa+v_gb$ for some $z_1,z_2∈\mathbb{Z}$, so that $v_fz_1-v_fa=-v_gz_2+v_gb$ and we can set $$m=\dfrac{z_1-a}{v_g}=\dfrac{-z_2+b}{v_f}$$so that with some manipulation and replacement we have$$z_1=a+m\dfrac{g_r}{d_r}$$$$z_2=b-m\dfrac{f_r}{d_r}$$Now, by letting $m(x)=m$ and taking the solutions $a'(x)$ and $b'(x)$ as above, by noting that the constant terms of $\dfrac{f(x)}{d(x)}$ and $\dfrac{g(x)}{d(x)}$ must be $\dfrac{f_r}{d_r}$ and $\dfrac{g_r}{d_r}$ respectively we see that the equations above actually describe the constant terms of $a'(x)$ and $b'(x)$, which implies $a'(x),b'(x)∈R$.

Assume one of $f_r$ or $g_r$ is zero, letting $f_r$ be the nonzero. Since now $d_r \mid f_r$ and $f_r \mid d_r$ we have $f_r=\pm d_r$, so that in the equation $d(x)=f(x)a(x)+g(x)b(x)$ we can see the constant term of $a(x)$ must be $\pm 1$ as the latter summand doesn't affect the sum coefficient of $x^r$. Continuing the terminology of the last procession we still have the same complete set of solutions, although observing the constant terms $z_1$ and $z_2$ of $a'(x)$ and $b'(x)$ simplifying to$$z_1=a+m\dfrac{g_r}{d_r}=a=\pm 1$$$$z_2=b-m\dfrac{f_r}{d_r}=b-m$$So choose $m(x)=b$ so that $a'(x),b'(x)∈R$ are valid solutions.

(d) This is evident from parts (a) and (c) as greatest common divisors in $R$ both exist and are expressible as $R$-linear combinations of elements from $R$, so that $(~f(x),g(x)~)=(~d(x)~)$. This implies every finitely generated ideal of $R$ is principle, i.e. $R$ is a Bezout domain.

(e) Since $R$ has been shown in the previous exercise to not be a UFD, it cannot be a PID, so that $R$ must contain infinitely generated nonprinciple ideals. Now, assume $x\mathbb{Q}[x]=(a_1,...,a_n)$ is finitely generated; then $x\mathbb{Q} \subseteq x\mathbb{Q}[x]=a_1R+...+a_nR$. By observing the first degree terms, we have $\mathbb{Q}=a_1\mathbb{Z}+...+a_n\mathbb{Z}$, which translates to saying that $\mathbb{Q}$ as an additive group is finitely generated by the $a_i$, a contradiction by 2.4.14(d).$~\square$

Polynomial Factorization Practice (9.3.4)

Dummit and Foote Abstract Algebra, section 9.3, exercise 4:

Let $R=\mathbb{Z}+x\mathbb{Q}[x] \subset \mathbb{Q}[x]$.

(a) Prove $R$ is an integral domain and $R^\times = \pm 1$.
(b) Show that the irreducibles of $R$ are the irreducibles of $\mathbb{Z}$ together with the irreducible polynomials of $\mathbb{Q}[x]$ with constant term $\pm 1$. Prove these irreducibles are prime.
(c) Show $x$ cannot be written as the product of irreducibles in $R$ (in particular, $x$ is not irreducible) and conclude that $R$ is not a UFD.
(d) Show $x$ is not prime in $R$ and describe $R/(x)$.

Proof: (a) Since this set is clearly closed under subtraction and multiplication, it is a subring, and by its containment in $\mathbb{Q}[x]$ it is an integral domain. Note that the degree norm $N$ has the property $N(\alpha \beta)=N(\alpha)+N(\beta)$ as in a typical polynomial ring, so that the only possible units are within $\mathbb{Z}$, and as such the only units are $\pm 1$.

(b) Should $p=\alpha \beta$ we have $0=N(p)=N(\alpha)+N(\beta)$, so that $\alpha,\beta∈\mathbb{Z}$ and thus by the primality of $p$ we have one of the two is equal to $\pm 1$, i.e. is a unit. Letting $p(x)=q_nx^n + ... + q_1x \pm 1$ be an irreducible element of $\mathbb{Q}[x]$ with constant term $\pm 1$, assuming $p(x)=\alpha \beta$ for nonunit $\alpha,\beta∈R$, by necessity $\alpha$ and $\beta$ have constant term $\pm 1$ so that $N(\alpha),N(\beta) > 0$. But now $\alpha$ and $\beta$ are nonunits in $\mathbb{Q}[x]$, violating the irreducibility of $p(x)$. Finally, assuming $a(x)$ is an irreducible element of $R$ by necessity we have its constant term is $\pm 1$ else for a prime divisor $p$ of the term we have $a(x)=p \dfrac{a(x)}{p}$ is factorization when $a(x)≠\pm p$. Assuming $a(x)=b(x)c(x)$ for some nonunit $b(x),c(x)∈\mathbb{Q}[x]$, we have the constant term of $b(x)$ is $\dfrac{m}{n}$ and necessarily the constant term of $c(x)$ is $\dfrac{n}{m}$ so that $b(x)=\dfrac{m}{n}b'(x)$ and $c(x)=\dfrac{n}{m}c'(x)$ for $b'(x),c'(x)∈R$ and now $a(x)=b'(x)c'(x)$ is a valid factorization of $a(x)$ in $R$ by $0 < N(b(x))=N(b'(x))$ and $0 < N(c(x))=N(c'(x))$ so that $b'(x),c'(x)∉R^\times$.

For prime $p$ the ideal $(p) \subseteq R$ is the kernel of the surjective homomorphism $φ : R → \mathbb{Z}/p\mathbb{Z}$ by $q_nx^n + ... + q_1x + z \mapsto z + p\mathbb{Z}$. For irreducible $q∈\mathbb{Q}[x]$ with constant term $\pm 1$, we have $q$ is prime and thus $R/(q)$ is contained in the integral domain $\mathbb{Q}/(q)$ and is therefore an integral domain itself, proving $q$ is prime.

(c) Note that any $vx∈R$ for $v∈\mathbb{Q}$ is not irreducible, since $vx=(2)(\dfrac{v}{2}x)$ is a valid factorization. Now, if $x=\pi_1...\pi_n$ for irreducible $\pi_i$, then $1=N(\pi_1)+...+N(\pi_n)$ so that exactly one $\pi_k$ is of the form $qx+z$ and for $i≠k$ we have $\pi_i∈\mathbb{Z}$. By our remark above $z≠0$, but now the product $\pi_1...\pi_n$ has a nonzero constant term, a contradiction.

(d) If $x$ were prime in $R$, it would be irreducible, a contradiction by the above. We now claim that $R/(x)=\{\overline{a+bx}~|~a∈\mathbb{Z},b∈\mathbb{Q} \land 0≤b<1\}$. Since clearly $\supseteq$, we must show the reverse: For arbitrary $a(x)∈R$, we have $$a(x)=q_nx^n + ... + q_1x+z=$$$$(q_nx^{n-1}+...q_2x)x+(w+q_1')x+z=(q_nx^{n-1}+...q_2x+w)x+q_1'x+z$$ for some $w∈\mathbb{Z}$ and $0≤q_1'<1$, so that $\overline{a(x)}∈\{\overline{a+bx}~|~a∈\mathbb{Z},b∈\mathbb{Q} \land 0≤b<1\}$. From here, the isomorphism from $R/(x)$ onto $\mathbb{Z}+(\mathbb{Q}/\mathbb{Z})x$ is evident, where $\mathbb{Q}/\mathbb{Z}$ is the ring extension of $\mathbb{Q}/\mathbb{Z}$ as additive groups by standard multiplication.$~\square$

Polynomial Ideals and Primality (9.1.13)

Dummit and Foote Abstract Algebra, section 9.1, exercise 13:

Letting $F$ be a field, prove $F[x,y]/(y^2-x) \not ≅ F[x,y]/(y^2-x^2)$.

Proof: We show something a bit stronger, that $F[x,y]/(y^n-x) \not ≅ F[x,y]/(x^2-y^2)$ for any $n$, with $F$ only an integral domain.

First we show $y-x,y+x∉(y^2-x^2)=(y-x)(y+x)$ so that there are zero divisors in $F[x,y]/(y^2-x^2)$. Assuming the contrary we have for $f$ some polynomial $y-x=(y-x)(y+x)f⇒(y+x)f=1$, an impossibility. Similarly for $y+x∉(y^2-x^2)$.

Now we show $F[x,y]/(y^n-x)=F[y]/(y^n-x)≅F[y]$ is an integral domain, which will complete the proof. To see the first equality, notice that $\supseteq$ is clear and any polynomial of $F[x,y]/(y^n-x)$ can be represented by the image of a polynomial devoid of variables in $x$ since $\overline{y}^n=\overline{x}$. Letting $φ$ be the natural surjective homomorphism of $F[y]$ onto $F[y]/(y^n-x)$, we can observe that if $\text{ker}~φ≠0$ then $0≠(y^n-x)f∈F[y]$ for some $f∈F[x,y]$, an impossibility by $(y^n-x)f=y^nf-xf$ and observing the result of the $xf$ term by choosing the result of the multiplication monomial with maximal degree of $x$ in $f$, which cannot be affected by any terms in $xf$ (else the $f$ monomials' $y$ and $x$ degrees are the same and can be rewritten as one) nor in $y^nf$.$~\square$

Prime Ideal Squared (9.1.12)

Dummit and Foote Abstract Algebra, section 9.1, exercise 12:

Let the overbar denote passage from $\mathbb{Q}[x,y,z]$ into $\mathbb{Q}/(xy-z^2)$. Prove that $\overline{P}=(\overline{x},\overline{z})$ is a prime ideal. Show that $\overline{xy}∈\overline{P}^2$ but that no power of $\overline{y}$ lies in $\overline{P}^2$, showing that $\overline{P}$ is a prime ideal whose square is not a primary ideal.

Proof: It is easy to check that for a surjective homomorphism $φ$, we have $(φ(a_1),φ(a_2),...)=φ(a_1,a_2,...)$, so that $(\overline{x},\overline{z})=(x,z)/(xy-z^2)$. Evidently $(xy-z^2) \subseteq (x,z)$ so that $\overline{\mathbb{Q}[x,y,z]}/\overline{(x,z)}≅\mathbb{Q}[x,y,z]/(x,z)≅\mathbb{Q}[y]$ is an integral domain, so that $\overline{P}$ is a prime ideal. Furthermore, $\overline{xy-z^2}=\overline{0}$ so that $\overline{xy}=\overline{z}^2∈\overline{P}^2$ as claimed. Now, if $\overline{y}^n∈\overline{P}^2 \subseteq \overline{P}$ for some $n∈\mathbb{Z}^+$ then we have the natural image of $\overline{y}^n$ in the previously mentioned quotient isomorphic to $\mathbb{Q}[y]$ is zero, an impossibility by the commutativity of natural projections.$~\square$

Characterization of PIDs (8.3.11)

Dummit and Foote Abstract Algebra, section 8.3, exercise 11:

Prove$$R~\text{a Principle Ideal Domain}~⇔R~\text{a UFD}~\land R~\text{a Bezout Domain}$$Proof: The $⇒$ direction is evident, so it suffices to demonstrate the converse. Observe a typical ideal $I \subseteq R$, and choose arbitrary $x∈I$ with $x=u\pi_1^{\alpha_1}...\pi_n^{\alpha_n}$ being its unique decomposition up to units. For any $e∈\mathbb{Z}^+ \cup \{0\}$ and $i$ with $1≤i≤n$ let $A_{i,e}=\{r∈I ~ | ~ \pi_i^e \mid r\}$. For each $i$ let $\beta_i$ be the largest integer such that $A_{i,\beta_i}=I$, which is guaranteed to exist for arbitrary $i$ as the exponent $\alpha_i$ of $\pi_i$ is finite so that as a bound $x∉A_{i,\alpha_i+1}$. For each $i$, choose $a_i$ such that $a_i ∈ I \setminus A_{i,\beta_i+1}$, so that $a_i=\pi_i^{\beta_i}n_i$ with $\pi_i \not \mid n_i$. Letting $A=\bigcup_{1≤i≤n} a_i$ and taking successive greatest common divisors of the $a_i$ and then taking a greatest common divisor of this and $x$, we have a final common divisor $d=\pi_1^{\beta_1}...\pi_n^{\beta_n}$ (since these are the only irreducibles dividing $x$, and for each $i$ there is an element $a_i$ with its minimal $\pi_i$ exponent $\beta_i$, and all the exponents of the $a_j$ and $x$ of $\pi_i$ is at least $\beta_i$ as $a_j,x∈I=A_{i,\beta_i}$), writable as an $R$-linear combination of elements from $A \cup \{x\} \subseteq I$ by 7.2.7. We claim $(d)=I$: Evidently $(d) \subseteq I$, and for the reverse containment, for any $r∈I$ we have $\pi_i^{\beta_i} \mid r$ for applicable $i$ since $r∈I=A_{i,\beta_i}$, implying $I \subseteq (d)$ and now $I=(d)$ is a principle ideal.$~\square$

Quotients of the Gaussian Integers (8.3.6-7)

Dummit and Foote Abstract Algebra, section 8.3, exercises 6-7:

6. (a) Prove $\mathbb{Z}[i]/(1+i)$ is a field of order 2.
(b) Let $q∈\mathbb{Z}$ be a prime with $q ≡ 3 \mod{4}$. Prove $\mathbb{Z}[i]/(q)$ is a field of order $q^2$.
(c) Let $p∈\mathbb{Z}$ be a prime with $p ≡ 1 \mod{4}$ and let $p=\pi \overline{\pi}$ be its factorization in $\mathbb{Z}[i]$. Prove $\mathbb{Z}[i]/(p)≅\mathbb{Z}[i]/(\pi) \times \mathbb{Z}[i]/(\overline{\pi})$ as rings, $\mathbb{Z}[i]/(p)$ is of order $p^2$ and therefore $\mathbb{Z}[i]/(\pi)$ and $\mathbb{Z}[i]/(\overline{\pi})$ are fields of order $p$.
7. Let $\pi∈\mathbb{Z}[i]$ be irreducible.
(a) For $n \geq 0$ an integer, show that $(\pi^{n+1}) \subseteq (\pi^n)$ and that multiplication by $\pi^n$ induces an isomorphism $\mathbb{Z}[i]/(\pi)≅(\pi^n)/(\pi^{n+1})$ as additive abelian groups.
(b) Prove $|\mathbb{Z}[i]/(\pi^n)|=|\mathbb{Z}[i]/(\pi)|^n$.
(c) Prove for $\alpha≠0$ that $|\mathbb{Z}[i]/(\alpha)|=N(\alpha)$.

Proof: (6) (a) By the division algorithm, every element in $\mathbb{Z}[i]/(1+i)$ can be represented by an element of norm $< 2$. We have $-i=-(1+i)+1$,$i=(1+i)-1$, and $-1=-(1+i)(1-i)+1$, so that the order is $≤2$. Since $1+i$ is not a unit, we have $(1+i) \subset \mathbb{Z}[i]$, so that the order must be 2. Since $\mathbb{Z}[i]/(1+i)$ is now a finite integral domain by the primality of $1+i$, we have it is a field.

(b) Observe arbitrary $a+bi∈\mathbb{Z}[i]$. Let $a=a'+v_aq$ and $b=b'+v_bq$ with $0≤ a',b' < q$. We have $a+bi=(a'+v_aq)+(b'+v_bq)i=a'+b'i+q(v_a+v_bi)$, so that in this sense $a+bi$ can be reduced modulo $q$. This allows a maximum order of $q^2$. Now, assume reduced $\overline{a+bi} = \overline{c+di}$ in $\mathbb{Z}[i]/(q)$, so that $a+bi=c+di+qr$ for some $r∈\mathbb{Z}[i]$. We therefore have $(a-c)+(b-d)i=qr$, i.e. $q$ divides $(a-c)$ and $(b-d)$. Since $a,b,c,d < q$, we have $a=c$ and $b=d$, i.e. $a+bi=c+di$. Therefore there are precisely $q^2$ elements. This order counting holds when $q$ is an arbitrary integer.

Once again let $\overline{a+bi}$ be a typical (nonzero) element. Since $q$ is irreducible thus prime thus $\mathbb{Z}[i]/(q)$ is an integral domain, we have $\overline{(a+bi)(a-bi)}≠\overline{0}$. Letting $v^{-1}$ denote the inverse of $\overline{(a+bi)(a-bi)}$ modulo $q$, we have $\overline{(a+bi)(v^{-1}(a-bi))}=\overline{1}$, so that $\mathbb{Z}[i]/(q)$ is a field.

Alternatively, by a previous exercise $\mathbb{Z}[i]/I$ is finite for any nonzero ideal $I$, so these finite integral domains are clearly fields.

(c) Since $(\pi)$ is prime and thus maximal by a property of PIDs, and since $\pi$ is not associate to $\overline{\pi}$ i.e. $\pi r ≠ \overline{\pi}$ for any $r∈\mathbb{Z}[i]$, we have $(\pi,\overline{\pi})=\mathbb{Z}[i]$ so that the Chinese Remainder Theorem holds by $(\pi)(\overline{\pi})=(\pi\overline{\pi})=(p)$. Parallel to the proof above, $\mathbb{Z}[i]/(p)$ is of order $p^2$ and since $(\pi),(\overline{\pi})≠\mathbb{Z}[i]$ we have they are both finite integral domains (fields) of order $p$.

(7) (a) The containment should be evident, so we must demonstrate the group isomorphism by $φ(x+(\pi))=x\pi^n+(\pi^{n+1})$. Well-definedness:$$x+(\pi)=y+(\pi)⇒x=y+\pi r⇒\pi r = x-y ⇒ \pi^{n+1} r = x\pi^n-y\pi^n⇒$$$$x\pi^n+(\pi^{n+1})=y\pi^n+(\pi^{n+1})⇒φ(x+(\pi))=φ(y+(\pi))$$Homomorphism:$$φ(x+(\pi)+y+(\pi))=φ((x+y)+(\pi))=$$$$x\pi^n+y\pi^n+(\pi^{n+1})=φ(x+(\pi))+φ(y+(\pi))$$ Surjectivity: $$x∈(\pi^n)/(\pi^{n+1})⇒x=\pi^n r + (\pi^{n+1})⇒φ(r+(\pi))=x$$ Injectivity: $$φ(x+(\pi))=(\pi^{n+1})⇒x\pi^n∈(\pi^{n+1})⇒\pi \mid x⇒x+(\pi)=0+(\pi)$$ (b) Assume $\pi$ is not associate to $1+i$. Let $N(\pi)=p$. We claim $\mathbb{Z}[i]/(p^n)≅\mathbb{Z}[i]/(\pi^n) \times \mathbb{Z}[i]/(\overline{\pi^n})$. The greatest common divisor of $\pi^n$ and $\overline{\pi^n}$ according to proposition 13 is $1$, and since $\mathbb{Z}[i]$ is a Euclidean Domain it can written as a linear combination of $\pi^n$ and $\overline{\pi^n}$, implying $(\pi^n,\overline{\pi^n})=\mathbb{Z}[i]$. Since $(\pi^n)(\overline{\pi^n})=(p^n)$, the Chinese Remainder Theorem takes care of the claim.

Now, we have $\mathbb{Z}/(p^n)$ is of order $p^{2n}$, therefore $|\mathbb{Z}[i]/(\pi^n)| \cdot |\mathbb{Z}[i]/(\overline{\pi^n})| = p^{2n}$ and in particular these former two are orders of $p$ power. We are now in a position to prove our claim by induction; assuming the claim doesn't hold for some particular $n$, we have either $|\mathbb{Z}[i]/(\pi^n)| < p^n$ or $|\mathbb{Z}[i]/(\overline{\pi^n})| < p^n$. Without loss of generality we can assume the former, so that $|\mathbb{Z}[i]/(\pi^n)|=|\mathbb{Z}[i]/(\pi^m)|$ for some $m < n$ by induction. Write $\mathbb{Z}[i]/(\pi^m)≅(\mathbb{Z}[i]/(\pi^n))/((\pi^m)/(\pi^n))$ so that necessarily $(\pi^m) \subseteq (\pi^n)$, a contradiction by $\pi^m \not ∈ (\pi^n)$.

So assume $\pi=1+i$, which by the associativity of $1+i$ to $1-i$ implies this is the last case we have to consider. We see that this is the case when $n=2m$ is even, as $\mathbb{Z}[i]/(\pi^{2m})=\mathbb{Z}/((\pi^2)^m)=\mathbb{Z}/(2^m)$, which has $2^{2m}=2^n=|\mathbb{Z}[i]/(\pi)|^n$ elements as claimed.

So assume $n=2m+1$. We first show that $|\mathbb{Z}[i]/(\pi^{2m+1})|$ is of $2$ power; by way of contradiction, assume there is some prime $p$ dividing the order of this quotient. Since the quotient may be considered a group under addition, where exponentiation of addition is multiplication, by Cauchy's Theorem we have some $\overline{a+bi}∈\mathbb{Z}[i]/(\pi^{2m+1})$ such that $\overline{p}$ is the smallest positive integer for which $\overline{p(a+bi)}=\overline{0}$. In other words, $\pi^{2m+1}$ doesn't divide $a+bi$ so that there are less than $2m+1$ factors of $\pi$ in the decomposition of $a+bi$, yet since $\pi \not \mid p$ for any $p≠2$ we must also have $\pi^{2m+1} \not \mid p(a+bi)$, a contradiction by $\overline{p(a+bi)}=\overline{0}$.

We have $(\pi^{2m+2}) \subset (\pi^{2m+1}) \subset (\pi^{2m})$, so that $$|(\mathbb{Z}[i]/(\pi^{2m+2}))/((\pi^{2m})/(\pi^{2m+2}))| <$$$$|(\mathbb{Z}[i]/(\pi^{2m+2}))/((\pi^{2m+1})/(\pi^{2m+2}))| <$$$$|(\mathbb{Z}[i]/(\pi^{2m+2}))/((\pi^{2m+2})/(\pi^{2m+2}))|$$implying $$|\mathbb{Z}[i]/(\pi^{2m})| = 2^{2m} < |\mathbb{Z}[i]/(\pi^{2m+1})| < |\mathbb{Z}[i]/(\pi^{2m+2})| = 2^{2m+2}$$Since $|\mathbb{Z}[i]/(\pi^{2m+1})|$ is of $2$ power, we must have $|\mathbb{Z}[i]/(\pi^{2m+1})|=2^{2m+1}=|\mathbb{Z}[i]/(\pi)|^{2m+1}$.

(c) Let $\alpha = \pi_1^{a_1}...\pi_n^{a_n}$ be its decomposition, so that $|\mathbb{Z}[i]/(\alpha)|=|\mathbb{Z}[i]/((\pi_1^{a_1})...(\pi_n^{a_n}))|$, the latter of which we claim is equivalent to $|\mathbb{Z}[i]/(\pi_1^{a_1})| ... |\mathbb{Z}[i]/(\pi_n^{a_n})|$ by the Chinese Remainder Theorem; taking any $(\pi_x^{a_x})$ and $(\pi_y^{a_y})$ with $x≠y$ we can see $(\pi_x^{a_x},\pi_y^{a_y})=\mathbb{Z}[i]$ since the greatest common divisor of the pair is $1$. The original claim now follows by$$|\mathbb{Z}[i]/(\alpha)|=|\mathbb{Z}[i]/(\pi_1^{a_1})...(\pi_n^{a_n})|=|\mathbb{Z}[i]/(\pi_1^{a_1})| ... |\mathbb{Z}[i]/(\pi_n^{a_n})| =$$$$|\mathbb{Z}[i]/(\pi_1)|^{a_1} ... |\mathbb{Z}[i]/(\pi_n)|^{a_n}=N(\pi_1)^{a_1} ... N(\pi_n)^{a_n} = N(\alpha)~\square$$

Rings of Fractions of Principle Ideal Domains (8.2.8)

Dummit and Foote Abstract Algebra, section 8.2, exercise 8:

Let $R$ be a principle ideal domain, and let $D$ be a multiplicatively closed subset of $R$. Prove that $D^{-1}R$ is a principle ideal domain.

Proof: Let $I$ be a typical ideal of $D^{-1}R$, and $J$ be the subset of elements of $R$ appearing as numerators of elements of $I$. We claim $J$ is an ideal: For any $a,c∈J$, we have some $\dfrac{a}{b},\dfrac{c}{d}∈I$ by definition, so $\dfrac{a}{b} \cdot \dfrac{b^2}{b} = \dfrac{ae}{e}∈I$ for any $e∈D$ and similarly $\dfrac{ce}{e}∈I$. Now $(\dfrac{ae}{e}-\dfrac{ce}{e}) \cdot \dfrac{e}{e^2}=\dfrac{(a-c)e}{e} \cdot \dfrac{e}{e^2}=\dfrac{a-c}{e}∈I$ so that $a-c∈J$ and $J$ is closed under subtraction. For any $r∈R$, we have $\dfrac{re}{e} \cdot \dfrac{a}{b}=\dfrac{ra}{b}∈I$ so that $ra∈J$ and $J$ is an ideal. Due to the hypothesis, $J=(\alpha)$ for some $\alpha∈R$.

We now claim $I=(\dfrac{\alpha e}{e})$. Since $\alpha∈J$, as before we have $\dfrac{\alpha e}{e}∈I$ so that $(\dfrac{\alpha e}{e}) \subseteq I$. Now, observe arbitrary $\dfrac{x}{y}∈I$, and write $x=z\alpha$. We have $\dfrac{\alpha e}{e} \cdot \dfrac{z}{b} = \dfrac{a}{b}$, so that $I \subseteq (\dfrac{\alpha e}{e})$ and now $I=(\dfrac{\alpha e}{e})$ is a principle ideal, and $D^{-1}R$ is a principle ideal domain.$~\square$

(p-1)th Roots of the p-Adic Integers (7.6.11e)

Dummit and Foote Abstract Algebra, section 7.6, exercise 11e:

Show that if $a_1 \not ≡ 0 \mod{p}$, then there is an element $a=(a_i)$ in the inverse limit $\mathbb{Z}_p$ satisfying $a_j^{p-1} ≡ 1 \mod{p^j}$ and $\mu_{j1}(a_j)=a_1$ for all $j$. Deduce that $\mathbb{Z}_p$ contains $p-1$ distinct $(p-1)$th roots of 1.

Proof: We claim that for any $a_1 \not ≡ 0 \mod{p}$, the coordinates of $a$ are uniquely determined and satisfy the given criteria. We proceed by induction on the $n$th coordinate: The case holds true for $n=1$, so observe the $n+1$th coordinate. $a_{n+1}^{p-1} ≡ 1 \mod{p^{n+1}}$ if and only if its associated polynomial representation $(b_0+b_1p+...+b_np^n)^{p-1} ≡ 1 \mod{p^{n+1}}$ holds true, so this is equivalent to solving for $b_n$ since $a_n=b_0+b_1p+...+b_{n-1}p^{n-1}$ has been uniquely determined.$$(b_0+b_1p+...+b_np^n)^{p-1} ≡ 1 \mod{p^{n+1}}⇔$$$$((b_0+b_1p+...+b_{n-1}p^{n-1})+(b_np^n))^{p-1} ≡ 1 \mod{p^{n+1}}⇔$$$$\sum_{k=0}^{p-1} {p-1\choose k}(b_0+b_1p+...+b_{n-1}p^{n-1})^k(b_np^{n})^{p-1-k}≡1 \mod{p^{n+1}}⇔$$$$(b_0+b_1p+...+b_{n-1}p^{n-1})^{p-1}+$$$$(p-1)(b_0+b_1p+...+b_{n-1}p^{n-1})^{p-2}(b_np^n) ≡ 1 \mod{p^{n+1}}$$Since $(b_0+b_1p+...+b_{n-1}p^{n-1})^{p-1}≡ 1 \mod{p^n}$ has been uniquely determined, we have $(b_0+b_1p+...+b_{n-1}p^{n-1})^{p-1}=1+vp^n$ for some integer $v$.$$1+vp^n+(p-1)(b_0+b_1p+...+b_{n-1}p^{n-1})^{p-2}(b_np^n) ≡ 1 \mod{p^{n+1}}⇔$$$$1+vp^n-b_0^{p-2}b_np^n ≡ 1 \mod{p^{n+1}}⇔$$$$1+(v-a_1^{p-2}b_n)p^n ≡ 1 \mod{p^{n+1}}⇔$$$$v-a_1^{p-2}b_n ≡ 0 \mod{p}⇔$$$$b_n≡a_1^{2-p}v \mod{p}~\square$$

Direct Limits (7.6.8)

Dummit and Foote Abstract Algebra, section 7.6, exercise 8-9:

8. Let $I$ be an indexing set with partial order $≤$, with $A_i$ an additive abelian group for all $i∈I$. Furthermore, $I$ is a directed set: For any $i,j∈I$, there exists $k∈I$ such that $i,j≤k$. Suppose that for every pair of indices $i,j$ with $i≤j$ there exists a map $\rho_{ij} : A_i → A_j$ such that $\rho_{jk} \circ \rho_{ij}$ for all $i≤j≤k$ and $\rho_{ii} = 1$. Letting $B$ be the disjoint union of all the $A_i$, define a relation $\sim$ on $B$ by$$a \sim b ⇔ ∃k∈I (i,j≤k \land \rho_{ik}(a)=\rho_{jk}(b))$$for $a∈A_i$ and $b∈A_j$.
(a) Show that $\sim$ is an equivalence relation on $B$. The set of equivalence classes is called the direct limit of the direct system $\{A_i\}$ and is denoted $\varinjlim A_i$. For the rest of the exercise let $A = \varinjlim A_i$.
(b) Let $\overline{x}$ denote the class of $x$ and define $\rho_i : A_i → A$ by $\rho_i(a)=\overline{a}$. Show that if $\rho_{ij}$ is injective for all $i,j$ then $\rho_i$ is injective for all $i$ (so that each $A_i$ can be viewed as a subset of $A$).
(c) Assume $\rho_{ij}$ are all group homomorphisms. For $a∈A_i$,$b∈A_j$ show that the operation$$\overline{a}+\overline{b}=\overline{\rho_{ik}(a)+\rho_{jk}(b)}$$ is well defined, for any $k$ such that $i,j≤k$. Show that this makes $A$ into an abelian group, and that $\rho_i$ in part (b) are group homomorphisms.
(d) Show that if $A_i$ are commutative rings with 1 and $\rho_{ij}$ are ring homomorphisms that send 1 to 1, then $A$ may likewise be given the structure of a commutative ring with 1 such that $\rho_i$ are all ring homomorphisms.
(e) Under the hypotheses in (c), prove that the direct limit has the following universal property: if $C$ is any abelian group such that for each $i∈I$ there is a homomorphism $φ_i : A_i → C$ with $φ_i = φ_j \circ \rho_{ij}$ whenever $i≤j$, then there is a unique homomorphism $φ : A → C$ such that $φ \circ \rho_i = φ_i$ for all $i$.

Proof: (a) Reflexivity: $\rho_{ii}(a)=a=\rho_{ii}(a)⇒a \sim a$. Symmetry: $a \sim b$$⇒ \rho_{ik}(a)=\rho_{jk}(b) ⇒ \rho_{jk}(b)=\rho_{ik}(a) ⇒ b \sim a$ for some $k$. Transitivity: Let $c∈A_l$. $a \sim b \land b \sim c ⇒ \rho_{ik}(a)=\rho_{jk}(b) \land \rho_{jm}(b)=\rho_{lm}(c)$ for some $k$ and $m$ with $i,j≤k$ and $j,l≤m$. Let $k,m≤n$. We have $$\rho_{in}(a)=\rho_{kn} \circ \rho_{ik}(a)=\rho_{kn} \circ \rho_{jk}(b)=\rho_{jn}(b)=$$$$\rho_{mn} \circ \rho_{jm}(b)=\rho_{mn} \circ \rho_{lm}(c)=\rho_{ln}(c)$$ so that $a \sim c$.
(b) Assume $a,b∈A_i$ and $\rho_{i}(a)=\rho_{i}(b)$, so that $\overline{a}=\overline{b} ⇒ a \sim b$. Then $\rho_{ij}(a)=\rho_{ij}(b)$ for some $j$, so that $a=b$.
(c) Start with well-definedness. Let $\overline{a'}=\overline{a}$ by $\rho_{i'n}(a')=\rho_{in}(a)$ for some $i,i'≤n$, and denote $a'∈A_{i'}$, and likewise for $b'$ with $m$ in place of $n$. It suffices to show $\rho_{i'k'}(a')+\rho_{j'k'}(b') \sim \rho_{ik}(a) + \rho_{jk}(b)$ for arbitrary applicable $k'$. Choose $k^o$ such that $n,k,k',m≤k^o$ (by choosing a value greater than or equal to $n$ and $k$, and another greater than or equal to $m$ and $k'$, and letting $k^o$ be a value greater than or equal to both). We have$$\rho_{k'k^o}(\rho_{i'k'}(a')+\rho_{j'k'}(b'))=\rho_{k'k^o} \circ \rho_{i'k'}(a')+\rho_{k'k^o} \circ \rho_{j'k'}(b')=\rho_{i'k^o}(a')+\rho_{j'k^o}(b')=$$$$\rho_{nk^o} \circ \rho_{i'n}(a')+\rho_{mk^o} \circ \rho_{j'm}(b')=\rho_{nk^o} \circ \rho_{in}(a) + \rho_{mk^o} \circ \rho_{jm}(b)=$$$$\rho_{ik^o}(a)+\rho_{jk^o}(b)=\rho_{kk^o} \circ \rho_{ik}(a)+\rho_{kk^o} \circ \rho_{jk}(b)=\rho_{kk^o}(\rho_{ik}(a)+\rho_{jk}(b))$$Due to the homomorpicity of $\rho_{ij}$, we have $\overline{e}=0$ in $A$ for any $e∈A_i$ a zero and $-\overline{a}=\overline{-a}$. Associativity, closure, and abelianness are assured by their demonstration in $A_i$. Furthermore, for $a,b∈A_i$, we have $\rho_{i}(a)+\rho_{i}(b)=\overline{a}+\overline{b}=\overline{\rho_{ii}(a)+\rho_{ii}(b)}=\rho_{i}(a+b)$.
(d) This is simply part (c) with the operation of addition replaced by multiplication, together with the evident fact that $\overline{e}=1$ in $A$ for any $e∈A_i$ an identity.
(e) Any such homomorphism $φ$ is uniquely defined by $φ(\overline{a})=φ \circ \rho_{i}(a)=φ_i(a)$, so now well definedness and homomorphicity must be demonstrated: $φ(\overline{a'})=φ_{i'}(a)=φ_n \circ \rho_{i'n}(a')=φ_n \circ \rho_{in}(a)=φ_i(a)=φ(\overline{a})$, and $φ(\overline{a}+\overline{b})=φ(\overline{\rho_{ik}(a)+\rho_{jk}(b)})=φ_k(\rho_{ik}(a)+\rho_{jk}(b))=φ_i(a)+φ_j(b)=φ(\overline{a})+φ(\overline{b})$where addition is interchangeable with multiplication.$~\square$

Collapsing Units Between Modular Rings (7.6.7)

Dummit and Foote Abstract Algebra, section 7.6, exercise 7:

Let $n \mid m$, and prove that the natural projection map $(\mathbb{Z}/m\mathbb{Z})^{\times} → (\mathbb{Z}/n\mathbb{Z})^{\times}$ is surjective.

Proof: Firstly, if $u$ is a unit of $\mathbb{Z}/m\mathbb{Z}$, then $(u,m)=1$, so that $(u,n)=1$, so that $u$ is a unit of $\mathbb{Z}/n\mathbb{Z}$, proving the mapping is as claimed. To show surjectivity, let $p_i$ be a prime not dividing $n$ but dividing $m$ for all applicable $i$. For a typical unit $x∈(\mathbb{Z}/n\mathbb{Z})^{\times}$, solve the simultaneous system of equations$$y ≡ x \mod{n}$$$$y ≡ 1 \mod{\prod p_i}$$which is always possible given that $n$ and $\prod p_i$ are coprime. For any prime $q$ dividing $m$, we have $q$ divides either $n$ or $\prod p_i$. In the former case, since $y$ is coprime to $n$, we have $y$ coprime to $q$. In the latter case, since $y$ is coprime to $\prod p_i$, we have $y$ coprime to $q$. Since $y$ is coprime to all the prime divisors of $m$, it is coprime to $m$ itself, and is thus a unit of $(\mathbb{Z}/m\mathbb{Z})^{\times}$ which naturally projects to $x$ in $(\mathbb{Z}/n\mathbb{Z})^{\times}$.$~\square$

Classification of Finite Boolean Rings with Identity (7.6.1-2)

Dummit and Foote Abstract Algebra, section 7.6, exercises 1-2:

1. An element $e∈R$ is idempotent if $e^2=e$. Assume $e$ is idempotent and within the center of $R$. Prove $Re$ and $R(1-e)$ are two-sided ideals of $R$ and $R≅Re \times R(1-e)$. Show $e$ and $1-e$ are identities for $Re$ and $R(1-e)$ respectively.
2. Let $R$ be a finite Boolean ring with identity. Prove $R≅\mathbb{Z}/2\mathbb{Z} \times ... \times \mathbb{Z}/2\mathbb{Z}$.

Proof: (1) $Re$ and $R(1-e)$ are already left ideals. For any $r_1e∈Re$ and $r∈R$, we have $r_1er=(r_1r)e∈Re$ so that $Re$ is a two-sided ideal. As well, we have $(1-e)^2=1-2e+e^2=1-2e+e=1-e$ and $r(1-e)=r-re=r-er=(1-e)r$ so that $(1-e)$ is idempotent and central and similarly $R(1-e)$ is a two-sided ideal.

Define a mapping $φ : R → Re \times R(1-e)$ by $r \mapsto (re,r(1-e))$. This is clearly a homomorphism of rings, and we can prove surjectivity by observing $φ(e)=(e^2,e(1-e))=(e,0)$ and $φ(1-e)=((1-e)e,(1-e)^2)=(0,1-e)$ and now for any $(r_1e,r_2(1-e)∈Re \times R(1-e)$ we have $φ(r_1e+r_2(1-e))=φ(r_1)(e,0)+φ(r_2)(0,1-e)=(r_1e,r_2(1-e))$. Finally, this mapping is injective since $$φ(r)=(0,0)⇒re=0 \land r(1-e)=0 ⇒$$$$r-re=r-0=0 ⇒ r=0$$ As we've already seen, $e$ and $(1-e)$ are central and idempotent, so that their identity qualities are deduced straightforwardly.$~\square$

(2) Lemma 1: Let $φ_i : A_i → B_i$ be an isomorphism of rings for all applicable $i$. We have$$A_1 \times A_2 \times ... ≅ B_1 \times B_2 \times ...$$Proof: Define a function that maps the $i$th coordinate of $A_1 \times A_2 \times ...$ to the $i$th coordinate of $B_1 \times B_2 \times ...$ as its image under $φ_i$. Due to the nature of the isomorphisms $φ_i$, we have this map is an isomorphism.$~\square$

Definition 1: Let $R$ be a ring with $r∈R$. Define the rank of $R$ from $r$ as the order of the smallest subset $A \subseteq R$ such that $r∈A$ and $A$ generates $R$.

We will proceed by induction on the rank of $R$ from $1$, where $R$ is a finite Boolean ring with identity. When the rank from $1$ is 1, since $R$ is of characteristic 2 we have $R ≅ \mathbb{Z}/2\mathbb{Z}$. Assume that for some integer $n$ we can deduce that a finite Boolean ring with rank from $1$ of $m≤n$ is necessarily isomorphic to a direct product of copies of $\mathbb{Z}/2\mathbb{Z}$, and now observe $R$ with rank from $1$ of $n+1$. Let $A=\{1,a,r_3,...,r_{n+1}\}$ be its fulfilling generating subset, with $1≠a$. Since $a$ is necessarily central and idempotent, we have $R ≅ Ra \times R(1-a)$ by the first exercise.

We claim that $\{1 \cdot a,a \cdot a,r_3 \cdot a,...,r_{n+1} \cdot a\}=\{a,r_3a,...,r_{n+1}a\}$ is a generating subset of $Ra$ containing the identity and of order $≤n$ so that $Ra$ has a rank from $a$ (the identity of $Ra$) of $≤n$ and so is isomorphic to some number of direct products of $\mathbb{Z}/2\mathbb{Z}$ by induction. This is immediate because any series of operations taking elements from this subset is divisible by $a$, and for any $ra∈Ra$ we have $ra=Na$ for some series of operations $N$ from $A$ and evidently $Na$ is replicable by a series of operations from the subset. Similarly the case holds for $R(1-a)$ where a potential generating subset containing its identity is $\{1 \cdot (1-a),a \cdot (1-a),r_3 \cdot (1-a),...,r_{n+1} \cdot (1-a)\}=$$\{1-a,0,r_3-r_3a,...,r_{n+1}-r_{n+1}a\}$. We may remove the superfluous zero from this generating subset. Therefore by induction and the lemma$$R≅Ra \times R(1-a) ≅(\mathbb{Z}/2\mathbb{Z} \times ... \times \mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z} \times ... \times \mathbb{Z}/2\mathbb{Z}) ≅$$$$\mathbb{Z}/2\mathbb{Z} \times ... \times \mathbb{Z}/2\mathbb{Z}~\square$$ In addition, this provides a unique construction of the finite Boolean ring $R$ with identity of rank $n$ for any $n∈\mathbb{Z}^+$.

Zorn's Lemma on the Real Numbers (7.5.6)

Dummit and Foote Abstract Algebra, section 7.5, exercise 6:

Prove that $\mathbb{R}$ contains a subring $A$ with identity and maximal under inclusion such that $\dfrac{1}{2} \not ∈ A$.

Proof: Let $\mathcal{F}$ be the set of all subrings of $\mathbb{R}$ not containing $\dfrac{1}{2}$, and let $\mathcal{C}$ be a typical chain of subrings $R_0 \subseteq R_1 \subseteq ...$. Admit$$R=\bigcup_{n∈N}R_n$$and prove it is an upper bound of $\mathcal{C}$. For any $a,b∈R$, we have $a∈R_x$ and $b∈R_y$ for some $x,y$ implying $a-b∈R$ and $ab∈R$ so that $R$ is a subring. Furthermore, by definition of the union, $1∈R$ and $\dfrac{1}{2} \not ∈ R$. By Zorn's Lemma, $\mathcal{F}$ has a maximal element.$~\square$