Fracture Hypothesis (Research)

Hypotheses formulated in mid-May, 2013.

Let $R$ be a non-field UFD, and for $x∈R \setminus \{0\}$ let$$\omega(x)=\omega(up_1^{\alpha_1}...p_n^{\alpha_n})=\sum_{i=1}^n \alpha_i$$For each $i∈\mathbb{N}$ let$$V_i = \{x∈R~|~\omega(x)=i\}∪\{0\}$$

Weak Fracture Hypothesis: For any non-field UFD $R$, there exists $i∈\mathbb{N}$ such that $V_i$ is not a group under addition.

Strong Fracture Hypothesis: For any non-field UFD $R$, for all $i∈\mathbb{N}^+$ necessarily $V_i$ is not a group under addition. Equivalently, $V_1$ is not a group under addition.

What would otherwise be the very "strongest" FH (that $V_0$ is also never an additive group) is false, since taking any field $F$ and letting $R=F[x]$, we have $V_0$ is the subset of constant polynomials, which are seen to be a group under addition.

SFH: Assume $R$ violates the SFH with $V_i$ an additive group.

Proposition: The image of $\mathbb{Z} \setminus \{0\}$ in $R$ is within $V_0$, i.e. $z∈\mathbb{Z} \setminus \{0\}$ is a unit in $R$ if it is nonzero. Proof: For $x∈V_i$, we have $x+...+x \text{(z times)} = zx∈V_i$, so that $z$ doesn't append to the prime factorization of $x$, so is a unit. Since also $(-1)^2=1$, we have $-1$ is a unit as well, and now all nonzero integers are seen to be units.

Proposition: $V_0,V_1,...,V_{i-1}$ are additive groups. Proof: Assume $V_0$ is not a group so that $u_1-u_2$ is divisible by a prime. Now for some prime $p$ we have $u_1p^i-u_2p^i=(u_1-u_2)p^i∉V_i$. Now assume $V_1$ is not a group so that $p_1-p_2∉V_1$. Observe $p_1^i-p_2p_1^{i-1}=(p_1-p_2)p_1^{i-1}∉V_i$. The case follows similarly for $V_k$ for any $k≤i$.

Prime Infinitude: There are an infinite number of distinct primes of $R$. Proof: Assume $P$ is a complete, finite list of primes in $R$. Observe $x=(\prod_{p∈P} p) - 1$. Assume $x$ is a unit, in which case $x+1=\prod_{p∈P} p$ is also a unit, a contradiction. So for some $q∈P$ we have $q$ divides $x$ and by construction $q$ divides $\prod_{p∈P}$ so $q$ divides $-1$, another contradiction.

Proposition: When $x∈V_a$ and $y∈V_b$ for $a≠b$ and $x≠0≠y$, we have $x+y∉V_a \cup V_b$. Proof: Assume $x+y∈V_a$. Now $(x+y)-x=y∈V_a$ and $y=0$. The case for $V_b$ holds similarly.

Algebraic Geometry/Rank of $V_1$ and/or Closedness of $V_0$: (First method due to Professor Li) The rank of $V_1$ as a vector space over $V_0$ is greater than $2$. Proof: Assume the rank is $\leq 2$, so $R \cong V_0[x,y]/P$ for some prime ideal $P$ identifying $x,y$ with primes in $R$. Necessarily $P=(f(x,y),p(x))$ for some $f(x,y) \in V_0[x,y]$, $p(x) \in V_0[x]$, each either prime or zero in their respective rings. We see $p(x)=0$ since each prime in $R$ is transcendental over $V_0$, so simply $P=(f(x,y))$. Now, $V_0$ must be infinite since $V_1$ must be infinite to satisfy infinitude of primes as above, so choose distinct units $u,v$. Write $Q_1=x+u=(\alpha_1x+\beta_1y)...(\alpha_nx+\beta_ny)$ in $R$, so $Q_1-x-u=f(x,y)g(x,y)$ in $V_0[x,y]$. The left hand side demonstrates a polynomial in $V_0[x,y]$ with three homogenous components, so by comparing terms on the right we see the largest homogenous term $m(x,y)$ of $f(x,y)$ divides $Q_1$, so without loss assume $\alpha_1x+\beta_1y$ is a factor of $m(x,y)$. But by taking the same approach to $Q_2=x+v$, we see $Q_2$ as a polynomial in $V_0[x,y]$ has $\alpha_1x+\beta_1y$ as a factor. This is impossible as $Q_1-Q_2 = u-v \in V_0$ (nonzero) while simultaneously $\alpha_1x+\beta_1y$ divides $Q_1-Q_2$ in $V_0[x,y]$ and now too in $R$.

Theorem: Assume the rank $n$ of $V_1$ as a vector space over $V_0$ is finite, and that $V_0$ is algebraically closed. Then $R$ is not a PID. Proof: By the above $n \geq 3$. Consider the surjective ring homomorphism $V_0[x_1,...,x_n] → R$ induced by mapping the variables to the generating primes of $V_1$ over $V_0$. Then by Nullstellensatz, there is a common zero $u \in V_0^n$ of the polynomials generating the kernel of this homomorphism. Consider the nontrivial homomorphism $\phi : R → V_0$ induced by this zero. Then since $\phi$ restricts to a linear transformation $V_1$ to $V_0$ over $V_0$, and since $n \geq 3$, by rank-nullity the kernel of this transformation is of rank $\geq 2$, i.e. it contains two nonassociate primes, and hence in $R$ the kernel of the full homomorphism cannot be principal.

WFH: Fix some infinite sequence of primes $p_1,p_2,...$. For each pair of integers $i≤j$ define$$\rho_{ij} : V_i → V_j$$$$x \mapsto (\prod_{k=i+1}^jp_k)x$$These mappings are seen to satisfy the following properties for $i≤j≤k$$$\rho_{ii}=1$$$$\rho_{jk} \circ \rho_{ij}=\rho_{ik}$$As well, they are group homomorphisms (and are in addition injective), so this directed system satisfies all the properties of a direct limit.

Proposition: For any units $u_1≠u_2$ and prime $p$, we have $p+u_1$ is not within the same fracture as $p+u_2$. Proof: Clearly neither are within $V_0$, so assume $p+u_1,p+u_2∈V_i$ for positive $i$. Since $V_i$ is an additive group, we have $(p+u_1)-(p+u_2)=u_1-u_2∈V_0$ as well as $u_1-u_2≠0$ so that $u_1-u_2∉V_i$, a contradiction.

Lemma: $V_0$ is countable. Proof: Fix a prime $p$ and let $φ(u)$ be the index of the fracture $p+u$ is within. By the previous proposition, this mapping is injective.

Lemma: $V_1$ is countable. Proof: Fix a unit $u$ and let $φ(p)$ be the index of the fracture $u+p$ is within. Assume $φ(p_1)=φ(p_2)$, i.e. $p_1+u,p_2+u∈V_i$. Since $V_i$ is additive, we have $(p_1+u)-(p_2+u)∈V_i$ while also $(p_1+u)-(p_2+u)=p_1-p_2∈V_1$. Therefore $(p_1+u)-(p_2+u)=0$ and $p_1=p_2$, and the mapping is injective.

Lemma: $R$ is countable. Proof: Since$$R=\bigcup_{n∈\mathbb{N}} V_n$$it remains to show that $V_n$ is countable. We can establish a surjection$$φ : V_0 \times \prod_{i=1}^nV_1 → V_n$$$$(u,p_1,p_2,...,p_n) \mapsto up_1p_2...p_n$$Since both $V_0$ and $V_1$ are countable, and finite direct products of sets of a given cardinality retain the same cardinality, we have $V_n$ is countable and now $R$ is countable.

Lemma: There are no nontrivial automorphisms of $R$. Proof: Let $φ$ be a nontrivial automorphism of $R$. Then since $φ$ is defined by its action on units and primes, either $φ$ doesn't fix some prime $p$, or $φ$ doesn't fix some unit $u$ in which case either $φ$ doesn't fix any chosen prime $p'$ or $φ$ doesn't fix $p=up'$; hence in either case we may say $φ(p)≠p$ for some prime $p$.

Now, write $1+p∈V_k$, necessarily $k > 1$. For general UFDs, automorphisms send units to units and primes to primes, so also $φ(1+p)∈V_k$. Hence $1+p-φ(1+p)∈V_k$. But also $1+p-φ(1+p)=p-φ(p)∈V_1$, so since $V_1∩V_k=\{0\}$ we must have $p=φ(p)$, a contradiction.

Module Structure: Note that $V_i$ acts as a vector space over $V_0$. We may define a bilinear map of $V_0$-modules $\prod_{i=1}^n V_1 → V_n$ by $(p_1,...,p_n) \mapsto p_1...p_n$ to induce an appropriate homomorphism of $V_0$-modules $\Phi_i : \otimes_{i=1}^n~V_1 → V_n$.

Exact Sequences: Since $V_0$ is a field and $V_1$ a $V_0$-module, we can see that $V_1$ is injective and by an injective homomorphism $V_1 → V_n$ by $x \mapsto p_1^{n-1}x$ we obtain $V_n ≅ V_1 \oplus V_n'$.

Tensor Algebras: $V_1$ is a vector space over $V_0$, and $R$ is a commutative $V_0$ algebra with a natural inclusion linear transformation $V_1 → R$, inducing an appropriate extension homomorphism of $V_0$-algebras $Φ: S(V_1) → R$. This is nonzero on nonzero 1- and 2-tensor sums, but to this end we claim $\ker{Φ}$ is generated by 3-tensor sums. Proceed by induction on $n$ for an $n$-tensor sum $∑_{i=1}^nt_i$ in $\text{ker }Φ$; for any two simple tensors $s_1$ and $s_2$, we have $Φ(s_1+s_2)=Φ(s_3)$ for some simple tensor $s_3$ due to representation in $R$. Therefore, for any simple tensors $s_1$ and $s_2$ let $f(s_1,s_2)$ be a simple tensor such that $s_1+s_2+f(s_1,s_2)$ is in $\text{ker }Φ$. Back to the claim at hand, we observe$$0=Φ(\sum_{i=1}^nt_i)=Φ(t_1+t_2+\sum_{i=3}^nt_i)=Φ(-f(t_1,t_2)+\sum_{i=3}^nt_i)$$Since $-f(t_1,t_2)+\sum_{i=3}^nt_i$ is an $(n-1)$-tensor sum in $\text{ker }Φ$, by induction it is generated by 3-tensor sums in $\text{ker }Φ$ and we have $\sum_{i=1}^nt_i=-f(t_1,t_2)+\sum_{i=3}^nt_i+(t_1+t_2+f(t_1,t_2))$ to complete the claim.

Furthermore, we can partially predict the behavior of the function $f$. According to a contradiction of WFH, when $s_1$ and $s_2$ are simple tensors in $S^n(V_1)$, then $f(s_1,s_2)$ is in $S^n(V_1)$. Since this function completely defines the 3-tensor sums generating the kernel, we might clearly derive SFH is violated if and only if there is a vector space $V$ over $F$ whose symmetric algebra contains an ideal $A$ such that (i) $A$ is generated by 3-tensor sums (ii) $A$ contains no 1- or 2-tensor sums, and (iii) $A$ for every pair of simple tensors $s_1,s_2$ there is a simple tensor $s_3$ such that $s_1+s_2+s_3∈A$. As well, WFH is violated if and only if there is such $V$, $F$, and $A$ such that (iv) when $s_1,s_2∈S^n(V)$ then $s_3∈S^n(V)$.

Rank of $V_1$: Assume $V_1$ is generated over $V_0$ as a vector space by the elements $p$ and $q$. Then $V_0$ is algebraically closed. Proof: We first show that the elements $p^kq^{n-k}$ for $k∈[0,n]$ are a basis for $V_n$ over $V_0$; this is evident because products of the form $xp+yq$ for $x,y∈V_0$ taken $n$ at a time generate $V_n$ and such products can be expanded to sums in $p^kq^{n-k}$, and also because they are linearly independent: Assume $v_0p^n+v_1qp^{n-1}+...+v_nq^n=0$, and further assume $v_0≠0$; then we have $v_0p^n=qr$ for some $r$, which is a contradiction by unique factorization. So $v_0=0$; now divide the sum out by $q$ and apply the independence statement by induction to $V_{n-1}$ to obtain $v_i=0$ for all $i$.

Now, observe the polynomial $f(x)=x^n+a_{n-1}x^{n-1}+...+a_0$ in $V_0[x]$. Then observe the element $p^n+a_{n-1}qp^{n-1}+a_{n-2}q^2p^{n-2}+...+a_0q^n∈V_n$. We must be able to write this according to its prime factorization as $(p-α_0q)(p-α_1q)...(p-α_nq)$ for some $α_i∈V_0$ (the negative signs making the notational argument simpler; note that we may ignore the primes of the form $yq$ for $y∈V_0$ as the expansion requires the coefficient of $p$ to be nonzero in all multiplicands). We see that in fact $α_0,...,α_n$ behaves exactly as would the solutions $β_0,...,β_n$ in writing $f(x)=(x-β_0)...(x-β_n)=(x-α_0)...(x-α_n)$ so that $f(x)$ splits over $V_0$.

Open Problems: When is $R$ generated as an additive group by $V_0$ and $V_1$? If $R$ is generated as such, then any sum $r_1+r_2$ in $R$ can be evaluated with knowledge of the generation of $r_1$ and $r_2$, knowledge of addition in $V_0$ and $V_1$ (to group the sums again) and knowledge of sums of the form $v_0+v_1$. In other words, since $Φ∘f$ above is a homomorphism on $V_0 × V_1$, we see $R$ is generated by primes and units iff $\text{img }Φ∘f=R$. Anyway, we see when $R$ is prime unit generated that it is additively isomorphic to $V_0 × V_1$, and when we have knowledge about the generation of $V_2$ from $V_0$ and $V_1$ we can make $V_0 × V_1$ into a ring isomorphic to $R$ via $(v_0,v_1)(v_0',v_1')=(v_0v_0',v_0'v_1+v_0v_1')+S$, where $S$ is the element of $V_0 × V_1$ corresponding to $v_1v_1'$. This may be to say that the additive structure depends only on $V_0$ and $V_1$ in this case, but the multiplicative structure requires knowledge about $V_2$, i.e. the map $V_1^2↦V_0 × V_1$. Keep in mind the additive automorphism group of $V_0 × V_1$ is $\text{Aut}(V_0) × \text{Aut}(V_1)$.

A UFD not generated as an additive group by its primes and units would be $\mathbb{C}[x]$, and some UFDs that are would be $\mathbb{Z}$, $\mathbb{Z}[x]$, $\mathbb{Q}[x]$, and $\mathbb{F}_p[x]$.

If $V_0,V_1$ generate $V_2$ as an additive group, then $R$ is generated by $V_0,V_1$. To see this, assume an ungenerated element $x$ with $n=\omega (x)$ minimal. Then observe the product of the generation of the product of its first $n-1$ primes, with its last prime, which is a sum of elements in $V_1$ and $V_2$, which by the hypothesis can be represented by a sum of elements in $V_0$ and $V_1$.

$R$ is not generated as an additive group by $V_0$ and $V_1$, and in fact $V_2$ isn't either. This would presume the existence of nonassociate $u_1+p_1,u_2+p_2∈V_2$, where necessarily $u_1,u_2≠0$, so we can write $u_1+p_1-(u_1/u_2)(u_2+p_2)=p_1-(u_1/u_2)p_2∈V_1∩V_2=\{0\}$ but we see $u_1+p_1=u_1+(u_1/u_2)p_2$ is associate to $u_2+p_2$ by $u_2/u_1$.

Vandermonde and Discriminants (Graph Theoretic Proof) (14.6.27)

Dummit and Foote Abstract Algebra, section 14.6, exercise 27:

Let $f(x)$ be a monic polynomial with roots $α_1,...,α_n$.
(a) Show that the discriminant $D$ of $f(x)$ is equal to the square of the Vandermonde determinant. $$\begin{vmatrix} 1&α_1&α_1^2&\cdots&α_1^{n-1} \\ 1&α_2&α_2^2&\cdots&α_2^{n-1} \\ 1&α_3&α_3^2&\cdots&α_3^{n-1} \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ 1&α_n&α_n^2&\cdots&α_n^{n-1} \end{vmatrix} = \prod_{i > j}(α_i-α_j)$$ (b) Taking the Vandermonde matrix above, multiplying on the left by its transpose, and taking the determinant, show that one obtains $$D = \begin{vmatrix} p_0&p_1&p_2&\cdots&p_{n-1} \\ p_1&p_2&p_3&\cdots&p_n \\ p_2&p_3&p_4&\cdots&p_{n+1} \\ \vdots&\vdots&\vdots&\ddots&\vdots \\ p_{n-1}&p_n&p_{n+1}&\cdots&p_{2n-2} \end{vmatrix}$$ where $p_k=∑α_i^k$, which can be computed in terms of the coefficients of $f(x)$ using Newton's Formulas from exercise 22. This gives an efficient procedure for calculating the discriminant of a polynomial.

Proof: (a) Expanding $\prod_{i > j}(α_i-α_j)$ into sums, one finds that the summands can be represented by the collection of graphs on $n$ points where for every point $α_i$ and $α_j$, a "choice" is made according to whether the element from the term $(α_i-α_j)$ is $α_i$ or (negative) $α_j$. Visually, this may be depicted by a regular $n$-gon with exactly one arrow from each vertex to another, with the vertices successively labeled $α_1,...,α_n$. Conversely, a sum may be obtained from a diagram of this type by examining how many arrows point to the vertex $α_i$ (notated $d(i)$) and retrieving $\pm ∏α_i^{d(i)}$, where the sign is negative if the number of arrows pointing from higher-ordered vertices to lower-ordered vertices is odd.

First, a lemma.

Lemma: Given such a diagram on $n$-points, if $d(i)=d(j)$ for any $i≠j$, then there exist points $a,b,c$ such that $a→b→c→a$.

Proof: The lemma is vacuously true for $n=2$. We shall prove it inductively for $n$: Given a counterexample, assume there does not exist $z$ such that $d(z)=n-1$. Then for every point, there is an arrow from that point to some other point. Starting from any point $x_1$, one may follow arrows without dead-ends to obtain a path $x_1→x_2→...→x_i→...→x_i$, so that the existence of a circuit $x_i→...→x_i$ is guaranteed. Observe a circuit $y_1→y_2→...→y_k→y_1$ of minimum distance $k$: if $k≠3$, then either $y_1$ points to $y_3$ or $y_3$ points to $y_1$. In the former case $y_1→y_3→...→y_k→y_1$ is a shorter circuit violating minimality, and in the latter case so too is $y_1→y_2→y_3→y_1$. Hence $k=3$ and the lemma's hypothesis is contradicted.

Therefore there is $z$ such that $d(z)=n-1$, i.e. all other points point to $α_z$. Note that in the equality $d(i)=d(j)$ assumed by a counterexample necessarily $i,j≠z$ as there cannot be two points toward which every point yields an arrow (since one of the two must yield an arrow to the other). As such, examine the diagram on $n-1$ points given by removing $α_z$ from the diagram and all the arrows pointing toward it ($α_z$ itself yields no arrows) to obtain a counterexample of the lemma on a smaller diagram, an inductive contradiction. This establishes the lemma.

Returning to the view of the diagrams as individual summands, establish a map between positive and negative diagrams of the type in the lemma ($d(i)=d(j)$ for some $i≠j$) given by reversing the directions of the arrows on the "first" triangle $a→b→c→a$ appearing in the diagram (triangle in order of labels $a_1,b_1,c_1$ is before triangle $a_2,b_2,c_2$ if the label of $a_1$ is smaller than the label of $a_2$, then compare $b_1$ with $b_2$ then $c_1$ with $c_2$). This map is a bijection as it is its own inverse. This ultimately implies the summands of the form $±α_1^{β_1}α_2^{β_2}...α_n^{β_n}$ where $β_i=β_j$ for some $i≠j$ cancel out in $∏_{i > j}(α_i-α_j)$, and the only ones remaining are those corresponding to diagrams where one point receives $0$ arrows, another $1$ arrow, $...$, and the last $n-1$ arrows, of which there are exactly $n!$; in fact, these are all point-permutations of the graph on $n$ points where point $α_i$ receives $i-1$ arrows. These permutations and negations correspond precisely to the definition of the determinant of the Vandermonde matrix above.

(b) If the $i,j$ entry of the Vandermonde matrix $V$ is $v_{ij}=α_i^{j-1}$, then we see the $i,j$ entry of $V^tV$ is, $$\sum_{k=1}^n v_{ki}v_{kj}=\sum_{k=1}^n α_k^{(i-1)+(j-1)}$$ which is the matrix written above. As described above, Newton's Formulas provide an efficient means of inductively calculating $p_i$ in terms of the symmetric functions which appear in the coefficients of $f(x)$, thus giving an efficient means of calculating the discriminant of $f(x)$.$~\square$

Newton's Formula and Applications (14.6.22-26)

Dummit and Foote Abstract Algebra, section 14.6, exercises 22-26:

22. Let $f(x)$ be a monic polynomial with roots $α_1,...,α_n$. Let $s_k$ be the elementary symmetric function of degree $k$ in the roots and define $s_k=0$ for $k > n$. Let $p_k=∑α_i^k$ be the sum of the $k^\text{th}$ powers of the roots for $k ≥ 0$. Derive Newton's Formulas for all $j∈ℕ$:$$p_1-s_1=0$$$$p_2-s_1p_1+2s_2=0$$$$p_3-s_1p_2+s_2p_1-3s_3$$$$...$$$$p_j-s_1p_{j-1}+s_2p_{j-2}-...+(-1)^{j-1}s_{j-1}p_1+(-1)^jjs_j=0$$ 23. (a) If $x+y+z=1$, $x^2+y^2+z^2=2$, and $x^3+y^3+z^3=3$, determine $x^4+y^4+z^4$.
(b) Prove $x,y,z∉ℚ$ but $x^n+y^n+z^n∈ℚ$ for all $n∈ℕ$.
24. Prove that an $n×n$ matrix $A$ over a field $F$ of characteristic $0$ is nilpotent iff $\text{Tr}(A^k)=0$ for all $1≤k≤n$.
25. Prove than two $n×n$ matrices $A$ and $B$ over a field $F$ of characteristic $0$ have the same characteristic polynomial iff $\text{Tr}(A^k)=\text{Tr}(B^k)$ for all $1≤k≤n$.
26. When $A$ and $B$ are two $n×n$ matrices over a field $F$ of characteristic $0$, show the characteristic polynomials of $AB$ and $BA$ are the same.

Proof: (22) When $I_n=\{1,2,...,n\}$, let $A_i=\{A⊆I_n~|~|A|=i\}$ so that we may say $$s_i=\sum_{A∈A_i}\prod_{a∈A}α_a$$ Now, define $$q_1=s_{j-1}p_1-js_j$$$$q_i=s_{j-i}p_i-q_{i-1}$$ By rearranging Newton's $j^\text{th}$ Formula we see that we must prove $$p_j=s_1p_{j-1}-(s_2p_{j-2}-(...-(s_{j-1}p_1-js_j)...)=q_{j-1}$$ To this end, we shall inductively prove, $$q_i=\sum_{A∈A_{j-i}} \sum_{a∈A} α_a^{i+1} \prod_{\substack{b∈A \\ b≠a}} α_b$$ which will establish the formula when $i=j-1$. For the base case $i=1$, $q_1=s_{j-1}p_1-js_j$ is seen to be the sum of all combinations of roots multiplied $j-1$ at a time with a squaring of one of those roots, which agrees with the inductive pattern stated above.

Now, for the inductive step, we see $q_i=s_{j-i}p_i-q_{i-1}$. Note $s_{j-i}p_i$ is the sum of all roots multiplied $j-i$ at a time with one root to the $(i+1)^\text{th}$ power, plus the sum of all roots multiplied $j-i+1=j-(i-1)$ at a time with one root to the $i^\text{th}=((i-1)+1)^{th}$ power. In fact, that second sum mentioned is inductively $q_{i-1}$, so that in fact $q_i$ accords with the pattern above and the induction is complete.

(23) (a) Letting $f(X)=(X-x)(X-y)(X-z)$, we see $p_1=1$, $p_2=2$, and $p_3=3$. Newton's Formulas provides a sufficient system: $$p_1-s_1=0⇒s_1=1$$ $$p_2-s_1p_1+2s_2=0⇒s_2=-\dfrac{1}{2}$$ $$p_3-s_1p_2+s_2p_1-3s_3=0⇒s_3=\dfrac{1}{6}$$ Since $s_4=0$, we may proceed to calculate $p_4$: $$p_4-s_1p_3+s_2p_2-s_3p_1=0⇒p_4=\dfrac{25}{6}$$ (b) Since we have $s_1$,$s_2$, and $s_3$, we may explicitly observe the polynomial $f(X)=x^3-x^2-\dfrac{1}{2}x-\dfrac{1}{6}$. Multiplying by $6$ one may prove the polynomial $6x^3-6x^2-3x-1$ has no rational roots by checking it against the rational root theorem. Meanwhile, $p_n$ may be deduced for all $n∈ℕ$ by the inductive process shown at the end of (a) which makes all of its calculations in $ℚ$, hence $p_n∈ℚ$.

(24) ($⇒$) We've seen the negative of the trace of a matrix manifests as the coefficient of $x^{n-1}$ in its characteristic polynomial. Hence, since a nilpotent matrix has a minimal polynomial of the form $x^m$, we see the characteristic polynomial must be of the form $x^{m'}$ so that $\text{Tr}(A)=0$. Since $A^k$ is also nilpotent for all $k$, the same argument applies. ($⇐$) This shall follow from the more general argument in (25).

(25) ($⇒$) Put $A$ and $B$ in Jordan form over the algebraic closure $K$ of $F$. We see that their primary diagonals' sum is the same given that their characteristic polynomials are the same and hence have the same roots in the same multiplicities. In general, given an upper triangular matrix $C$ with primary diagonal elements $c_{ii}=γ_i$, it is simple to show by induction that $C^k$ is an upper triangular matrix with primary diagonal elements $γ_i^k$, i.e. the strict upper triangular elements have no effect on the primary diagonal, ultimately implying through the Jordan forms of $A$ and $B$ that $\text{Tr}(A^k)=\text{Tr}(B^k)$. ($⇐$) Once again given the Jordan forms $A'$ and $B'$ of $A$ and $B$ over $K$, we may deduce from the Newton's Formula system implied by the diagonal sums entailed by $\text{Tr}(A'^k)=\text{Tr}(B'^k)$ for $k∈I_n$ that the elementary symmetric functions in the roots of the characteristic polynomials for $A$ and $B$ are identical (here characteristic $0$ or at least $> n$ is needed so that the term $(-1)^jjs_j$ doesn't vanish), hence their characteristic polynomials are identical.

(26) For any two matrices $A$ and $B$ with $AB=C$ we see $$\text{Tr}(AB)=\sum_{m=1}^n c_{mm}=\sum_{m=1}^n \sum_{k=1}^n a_{mk}b_{km} = \sum_{k=1}^n \sum_{m=1}^n b_{km}a_{mk} = \text{Tr}(BA)$$ Thus for all $k$ we have $\text{Tr}((AB)^k)=\text{Tr}(A(BA)^{k-1}B)=\text{Tr}((BA)^{k-1}BA)=\text{Tr}((BA)^k)$ hence by (25) $AB$ and $BA$ have the same characteristic polynomial.$~\square$

Noncyclicity of Negative-Discriminant Quartic Fields over Q (14.6.19)

Dummit and Foote Abstract Algebra, section 14.6, exercise 19:

Let $f(x)$ be an irreducible polynomial of degree $4$ in $ℚ[x]$ with discriminant $D$. Let $K$ denote the splitting field of $f(x)$, viewed as a subfield of $ℂ$.
(a) Prove $ℚ(\sqrt{D})⊂K$.
(b) Let $τ$ denote complex conjugation and let $τ_K$ denote complex conjugation restricted to $K$. Prove $τ_K$ is an element of $\text{Gal}(K/ℚ)$ of order $1$ or $2$ depending on whether $K⊆ℝ$.
(c) Prove that if $D < 0$ then $K$ cannot be cyclic of degree $4$ over $ℚ$.
(d) Prove generally that $ℚ(\sqrt{D})⊈K$ for any $D < 0$ when $K$ is a cyclic quartic field.

Proof: (a) $\sqrt{D}$ is an expression in the roots of $f(x)$ and hence is clearly within $K$. Since $f(x)$ is irreducible we have $[K~:~ℚ]≥4$ and so the proper containment is evident.

(b) When $K⊆ℝ$, $τ_K$ is merely the identity. So assume $K⊈ℝ$: Then $[K~:~K∩ℝ]=2$ as $≥$ clearly holds, and $≤$ must hold as $K$ is Galois over $K∩ℝ$ and the only nonidentity automorphism must be $φ(a+bi)=a+bφ(i)=a-bi$, i.e. $τ_K$.

(c,d) Assume $D < 0$ so that $\sqrt{D}∉ℝ$. Then $τ_K≠1$ and it must be the sole automorphism of degree 2 generated by either of the two other nonidentity automorphisms. But now $ℚ(\sqrt{D})$ is not fixed by anything but the identity, a contradiction as $ℚ(\sqrt{D})≠K$. Nothing more was needed than the fact that $D < 0$.$~\square$