I.1. Show that the additive group of the vector space C(ℝ) of continuous functions ℝ → ℝ equipped with pointwise multiplication is a ring, and show that its units are the f ∈ C(ℝ) with f(x) > 0 for all x ∈ ℝ, or f(x) < 0 for all x ∈ ℝ. Is there a ring whose additive group is the additive group of C(ℝ) and whose multiplication is composition of functions?
If $f,g∈C(ℝ)$, then $f+g,fg∈C(ℝ)$, proved in Analysis Ⅱ Theorem 3.7 (C(ℝ), +) is a commutative group with identity 0 the constant function. × is commutative, associative and distributes over +, since ℝ is a field. So C(ℝ) is a commutative ring. f is a unit ⇔ f(x) ≠ 0 ∀x ∈ ℝ ⇔ f(x) > 0 ∀x ∈ ℝ, or f(x) < 0 ∀x ∈ ℝ, by intermediate value theorem Suppose C(ℝ) with composition as multiplication is a ring. ∃ zero $f$ such that $gf=f\,∀g∈C(ℝ)$. Take the constant function $g(x)=1\,∀x∈ℝ$, then $gf=g≠f$, contradiction.I.2. Show that the set of rationals with odd denominator, denoted ℤ⟨2⟩, is a subring of ℚ. What are the units of ℤ⟨2⟩? What are the ideals of ℤ⟨2⟩?
$1=\frac11∈$ℤ⟨2⟩. For any $\frac pq,\frac rs∈$ℤ⟨2⟩, where $p,q,r,s∈ℤ$, and $q,s$ are odd. $\frac pq-\frac rs=\frac{ps-qr}{qs}$ and $\frac pq×\frac rs=\frac{pr}{qs}$ has odd denominator $qs$, so $\frac pq-\frac rs,\frac pq×\frac rs∈$ℤ⟨2⟩. By the subring test, ℤ⟨2⟩ is a subring of ℚ. $\frac pq∈$ℤ⟨2⟩ is a unit, iff p is odd.Let I be an ideal of ℤ⟨2⟩, then I ∩ ℤ is an ideal of ℤ. For all $\frac rs∈$I (r, s ∈ ℤ) we have $r=s×\frac rs∈I$.(1) If I ∩ ℤ = {0}, by (1), numerator of any element of I is 0, so I = {0}. If I ∩ ℤ = nℤ (n ∈ ℕ*). Let m be any odd divisor of n, then $\frac1m∈ℤ_{⟨2⟩}⇒n×\frac1m=\frac nm∈I∩ℤ=nℤ$, so m = 1, so n = $2^k$ (k ∈ ℕ*), by (1), numerator of any element of I is in nℤ, so I ⊂ 2kℤ⟨2⟩. On the other hand, $2^k∈I$ ⇒ 2kℤ⟨2⟩ ⊂ I. So I = 2kℤ⟨2⟩. So the ideals of ℤ⟨2⟩ are {0} and 2kℤ⟨2⟩ (k ∈ ℕ*).
I.3. Show that there is no ring homomorphism ℂ → ℝ, ℝ → ℚ, or ℚ → ℤ. Show that any ring homomorphism ℝ → ℝ is increasing and deduce that there is only one such homomorphism.
Let $f:ℂ→ℝ$ be a ring homomorphism. $-1=f(-1)=f(\mathrm i^2)=f(\mathrm i)^2$ contradicts $f(\mathrm i)∈ℝ$. Let $f:ℝ→ℚ$ be a ring homomorphism. $2=f(2)=f\left((\sqrt2)^2\right)=f(\sqrt2)^2$ contradicts $f(\sqrt2)∈ℚ$. Let $f:ℚ→ℤ$ be a ring homomorphism. $1=f(1)=f\left(2×\frac12\right)=2f\left(\frac12\right)$ contradicts $f\left(\frac12\right)∈ℤ$. Let $f:ℝ→ℝ$ be a ring homomorphism, then f is additive and f(1) = 1, so f|ℚ = idℚ. x ≥ y ⇔ ∃z: x − y = z2 ⇔ ∃z: f(x) − f(y) = f(z)2 ⇔ f(x) ≥ f(y). So f is increasing. ∀x, y ∈ ℝ, if ${|x-y|}≤\frac1n$ then f(x) − f(y) = f(x − y), similarly f(y) − f(x), so f is continuous. Since ℚ is dense in ℝ, f = idℝ.I.4. Show that if R is a ring with ideals I, J, and K then I ∩ J + I ∩ K ⊂ I ∩ (J + K); and if R = 𝔽[X, Y], and I = ⟨Y⟩, J = ⟨X⟩, and K = ⟨X + Y⟩, that I ∩ J + I ∩ K ≠ I ∩ (J + K).
MSE MSE ∀j ∈ I ∩ J, k ∈ I ∩ K: j + k ∈ J + K, also j + k ∈ I by closure of I, so j + k ∈ I ∩ (J + K). So I ∩ J + I ∩ K ⊂ I ∩ (J + K). If R = 𝔽[X, Y], and I = ⟨Y⟩, J = ⟨X⟩, and K = ⟨X + Y⟩, then I ∩ J = ⟨YX⟩, I ∩ K = ⟨Y(X + Y)⟩. Y = (X + Y) − X ∈ I ∩ (J + K).Suppose Y ∈ I ∩ J + I ∩ K, then Y = YX p(X, Y) + Y(X + Y) q(X, Y). Dividing by Y, 1 = X p(X, Y) + (X + Y) q(X, Y). Evaluating at (0, 0) we get 1 = 0, contradiction. So I ∩ J + I ∩ K ≠ I ∩ (J + K).
I.5. Let E be the ring of group homomorphisms from the additive group of ℚ[X] to itself. Show that α: ℚ[X] → ℚ[X]; p(X) ↦ Xp(X) is an element of E, and that there is β ∈ E such that βα = 1E, but no γ ∈ E such that αγ = 1E. Hence give an example of A ∉ U(M2(E)) with det A ∈ U(E), and show that$$B=\begin{pmatrix}α&1-αβ\\0&β\end{pmatrix}$$has B ∈ U(M2(E)) and det B ∉ U(E).
For p, q ∈ ℚ[X], α(p + q) = X(p + q) = Xp + Xq = α(p) + α(q), α(−p)= −Xp = −α(p), so α ∈ E. E is injective but not surjective, so ∃β ∈ E such that βα = 1E, but no γ ∈ E such that αγ = 1E. For $A=\begin{pmatrix}β&0\\0&α\end{pmatrix}$, det A = βα = 1E ∈ U(E). If ∃M ∈ M2(E): AM = I2, then α M2,2 = 1E, contradiction. So A ∉ U(M2(E)). Using βα = 1E we calculate\[\pmatrix{α&1-αβ\\0&β}\pmatrix{β&0\\1-αβ&α}=\pmatrix{β&0\\1-αβ&α}\pmatrix{α&1-αβ\\0&β}=\pmatrix{1&0\\0&1}\]So B ∈ U(M2(E)) Suppose ∃δ ∈ E such that (αβ)δ = 1E, then α(βδ) = 1E, contradiction. So det B = αβ ∉ U(E).I.6. Show that if x, y ∈ ℤN have ⟨x⟩ = ⟨y⟩ then there is u ∈ U(ℤN) such that x = uy. Give examples of f, g ∈ C(ℝ) (with pointwise addition and multiplication) such that ⟨f⟩ = ⟨g⟩ but for which there is no h ∈ U(C(ℝ)) with f = hg.
MSE Let $R = ℤ/(n)$, and let $n = p_1^{k_1} p_2^{k_2}⋯p_m^{k_m}$ be the prime factorization of $n$. For each $1 \leq i \leq m$, let $I_i = (p_i^{k_i})$, the principal ideal generated by $p_i^{k_i}$. If $i \neq j$, then the gcd of $p_i^{k_i}$ and $p_j^{k_j}$ is $1$, so $I_i + I_j = R$ and the Chinese remainder theorem applies. Note also that $K = I_1 \cap I_2 \cap … \cap I_m = (n)$, which is the "zero" element in $R$, so $R/K$ is simply $R$, and the map $ϕ$ described in the CRT statement is an isomorphism. Claim 1: $(a) = (b)$ in $R$ if and only if $(a) = (b)$ in each $R/I_j$. Proof: Suppose $(a) ⊂ (b)$ in $R$. Then $a = bx$ for some $x∈R$, so $ϕ(a) = ϕ(b)ϕ(x)$ (since $ϕ$ is an isomorphism), which means that $a = bx$ in each $R/I_j$, and therefore $(a) ⊂ (b)$ in $R/I_j$. Conversely, suppose that $(a) ⊂ (b)$ in each $R/I_j$. Then in $R/I_j$ we have $a = bx_j$ for some $x_j$. Therefore, in $(R/I_1) × (R/I_2) ×⋯× (R/I_m)$ we have $(a,a,…,a) = (b,b,…,b)(x_1,x_2,…,x_m)$. Applying $ϕ^{-1}$ to this equation, we get $a = bx$ for some unique $x∈R$. Therefore, $(a) ⊂ (b)$ in $R$. Reversing the roles of $a$ and $b$ gives the proof for the opposite containment. $\square$ Claim 2: $a$ and $b$ are associates in $R$ if and only if $a$ and $b$ are associates in each $R/I_j$. Proof: Suppose that $a = ub$ for some unit $u$ in $R$. Then $ϕ(a) = ϕ(u)ϕ(b)$, so $a = ub$ in each $R/I_j$. Moreover, $ϕ$ maps units to units, so $u$ is a unit in each $R/I_j$. Conversely, suppose that in each $R/I_j$, we have $a = u_jb$, where $u_j$ is a unit. Therefore, in $(R/I_1) × (R/I_2) ×⋯× (R/I_m)$ we have $(a,a,…,a) = (u_1,u_2,…,u_m)(b,b,…,b)$. Applying $ϕ^{-1}$, we get $a = ub$ in $R$, for some unique unit $u∈R$. $\square$ Because of the two claims, it suffices to solve the problem for $R/(p^k)$, where $p^k$ is any prime power. Note that $R = ℤ/(n)$, but $p^k$ divides $n$, so $R/(p^k)$ is really (isomorphic to) $ℤ/(p^k)$. Proof for $R = ℤ/(p^k)$ Let $I = (p^k)$, and suppose that $(a) = (b) = J$ in $R$. Then $J = (x + I) = (x + (p^k))$ for some $x∈ℤ$. As $J$ is an additive subgroup of $R$, $|J|$ must divide ${|R|}=p^k$. Thus ${|J|}=p^{k-j}$ for some $0≤j≤k$. This forces $J = (p^j + (p^k))$. We have established that $(a) = (b) = (p^j + (p^k))$. Thus, working modulo $p^k$, we have:- $a = rp^j$ for some $r$.
- $r$ cannot be divisible by $p$, for if it were, say $r = sp^m$ where $m≥1$, then $a = sp^{m+j}$, so $p^{k-m-j}a = 0$, which means that the size of $(a)$ is strictly smaller than $p^{k-j}={|J|}$.
- $u$ is a unit in $R$ if and only if it is not divisible by $p$.
- Therefore, $r$ is a unit.
- By the same argument, $b = sp^j$ for some unit $s$.
- Therefore, $b = r^{-1}sa$, and $a$ and $b$ are associates.
Consider $f,g∈C(ℝ)$ given by $$ f(x)=\begin{cases} x+1 & x<-1 \\ 0 & -1 \leq x \leq 1 \\ x-1 & x>1 \end{cases}, \qquad g(x)=\begin{cases} -x-1 & x<-1 \\ 0 & -1\leq x \leq 1 \\ x-1 & x>1. \end{cases} $$Let $$ h(x)=\begin{cases} -1 & x<-1 \\ x & -1 \leq x≤1 \\ 1 & x>1 \end{cases} $$Note that $g=fh$ and $f=gh$, and so $⟨f⟩=⟨g⟩$. Let $w∈C(ℝ)$ such that $g=fw$, then $w(-2)=-1<0<w(2)=1$, by Q1, $w$ is not a unit.
I.7. Show that if R is an integral domain of non-zero characteristic p then p is prime, and R can be given the structure of an 𝔽p-vector space in such a way that left (and right) multiplication is linear. Deduce that every finite integral domain is a field and every finite field has size pn for some prime p.
Suppose $p = rs$ for $r,s<p$, then $r1_R⋅s1_R=0$ in $R$, so $s1_R$ is a zero divisor, contradiction.For R a finite integral domain, to show R is a field, we need to show that any a ∈ R* is a unit. Consider a, a2, a3, ... Since R finite, for some m < n, am = an. Then 0 = am − an = am(1 − an−m). Since there are no zero divisors and am ≠ 0, hence 1 − an−m = 0 and so 1 = a(an−m−1) = (an−m−1)a, so a is a unit. Alternatively, $∀r∈R$, since there are no zero divisors, the map f: R→ R; x ↦ rx is injective, and R is finite, so f is surjective, so $∃s∈R$ such that $rs=f(s)=1$, and thus r is a unit.
Proposition 5.14. If char(R) = p, then {0, 1, …, p − 1} is closed under subtraction and multiplication. Thus it is a subring of R. Hence it is a finite integral domain and hence a field isomorphic to 𝔽p. If R is finite, let $e_1,…,e_n$ be a basis of R over 𝔽p, then $$R=\left\{a_1 e_1+⋯+a_n e_n \middle| a_i∈𝔽_p\right\}$$Hence ${|R|}=p^n$.
I.8. Show that if α0, …, αd ∈ 𝔽 are distinct then for any c0, …, cd ∈ 𝔽 there is p ∈ 𝔽[X] such that p(α0) = c0, …, p(αd) = cd. Show that there are α0, α1 ∈ ℤ with α0 ≠ α1 and c0, c1 ∈ ℤ such that there is no p ∈ ℤ[X] with p(α0) = c0 and p(α1) = c1.
Check the condition for CRT: $α_i-α_j∈⟨x-α_i⟩+⟨x-α_j⟩⇒⟨x-α_i⟩+⟨x-α_j⟩=𝔽[X]$ Using the construction in the proof of CRT: Lagrange interpolation formula: $f(x)=∑_{j=0}^dc_jδ_j(x)$ where $δ_j(x)=\frac{\prod_{i=0;i≠j}^N\left(x-α_i\right)}{\prod_{i=0;i≠j}^N\left(α_j-α_i\right)}∈𝔽[X]$. Take α0 = 0, α1 = 2, c0 = 1, c1 = 0. For p = a0 +a1X + ⋯ + anXn ∈ ℤ[X], p(0) = 1 ⇒ a0 = 1 p(2) = 0 ⇒ a0 +a1⋅2 + ⋯ + an⋅2n = 0, so a0 is even, contradiction. So there is no such p.I.9. Show that if there is u ∈ R with 2u = 1R then R has no element of order 2 in its additive group. Hence deduce that the quotient group ℝ/ℤ is not the additive group of a ring. Is the quotient group ℝ/ℚ the additive group of a ring?
If 2x = 0, then x = x1 = x(2u) = 2xu = 0u = 0, so (R, +) has no element of order 2. $(\frac12+ℤ)+(\frac12+ℤ)=ℤ⇒\frac12+ℤ$ has order 2 in ℝ/ℤ. Let $R$ be a ring with additive group ℝ/ℤ and 1R = e + ℤ, then 2($\frac e2$+ ℤ) = e + ℤ, contradiction. ℝ and ℝ/ℚ are ℚ-vector spaces of same cardinality so ℝ ≅ ℝ/ℚ is a ring.I.10. Suppose that p ∈ ℚ[X] maps integers to integers and is such that for all z ∈ ℤ either 2 divides p(z) or 3 divides p(z). Show that the quantifiers may be reversed i.e. that either for all z ∈ ℤ we have 2 divides p(z), or for all z ∈ ℤ we have 3 divides p(z).
MSE Let $n=\deg p(x)$, we know $p(x)$ is a linear combination of $\binom xn,…,\binom x0$, so $g(x)=n!p(x)∈ℤ[X]$. Let $ν_2(n!)=r,ν_3(n!)=s$. Assume that there exists $a,b∈ℤ$ such that $2∤f(a)$ and $3∤f(b)$. So $2^{r+1}∤g(a)$ and $3^{s+1}∤g(b)$. By Chinese remainder theorem, there exists $c∈ℤ$ such that\begin{align*}c&≡a\pmod{2^{r+1}}\\c&≡b\pmod{3^{s+1}}\end{align*}hence\begin{align*} g(c)&≡g(a)≢0\pmod{2^{r+1}}\\ g(c)&≡g(b)≢0\pmod{3^{s+1}} \end{align*}consequently, $2∤f(c)$ and $3∤f(c)$—a contradiction.Alternative method: Take $D>1$ such that $\deg p<2^D$ then $\deg p<3^D$. Write p-adic valuation of n as νp(n). Let $a_1$ be any integer. Consider $a_1 + k2^D$. Let $a = a_1$ and $b = a_1 + k2^D$ in the lemma. We get :$$ν_2(f(a_1+k2^D) - f(a_1))≥ν_2(k2^D)-D+1≥1$$Therefore, $f(a_1+k2^D)-f(a_1)$ is a multiple of $2$ for all integers $k$. Similarly, for all $a_2$, we know that $f(a_2 + l3^D) - f(a_2)$ is a multiple of $3$ for all integers $l$.