(A Bit of) Abstract Algebra

1 Introduction

This is a deliberately personal and incomplete tour of abstract algebra. It groups course worksheets and homework submissions on divisibility, congruence, rings, polynomials, groups, quotient structures, isomorphism theorems, and a short final encounter with elliptic curves.

The original modern-algebra/ source manifest and migration receipts remain the provenance authority. Visible mistakes, grading feedback, blank answers, and source-authored omissions are retained instead of silently corrected.

2 Integers and division

This chapter transcribes the handwritten work in ‘WorkSheets/412-WS1-Mywork.pdf’, pp. 1–2, and the integer-linear-combination work in ‘WorkSheets/412-WS2-Mywork.pdf’, p. 1. English source wording remains English and Chinese source annotations remain Chinese.

2.1 Divisibility and the division algorithm

Source transcription — WS1, p. 1, Part I, Warm Up’‘.

The source gives the following main outline of the proof of the Division Algorithm Theorem:

1. Existence: $\exists 𝑞,𝑟\in \text{ℤ}$ such that $𝑛=𝑞𝑑+𝑟$ with $0\le 𝑟<𝑑$.
2. Uniqueness: if another expression $𝑛={𝑞}^{\prime }𝑑+{𝑟}^{\prime }$ has $0\le {𝑟}^{\prime }<𝑑$, then ${𝑟}^{\prime }=𝑟$ and ${𝑞}^{\prime }=𝑞$.

It also records the divisibility calculation: if $𝑎,𝑏,𝑐\in \text{ℤ}$, $𝑎|𝑏$, and $𝑏|𝑐$, write $𝑏=𝑎𝑠$ and $𝑐=𝑏𝑡$. Then $𝑐=\left(𝑠𝑡\right)𝑎$, with $𝑠,𝑡\in \text{ℤ}$, hence $𝑠𝑡\in \text{ℤ}$ and $𝑎|𝑐$.

2.2 Linear combinations, gcd, and Bézout

Source transcription — WS2, p. 1, 自主部份, Pf of Thm 2′‘. Define

i.e. $𝑆$ 为 $𝑎,𝑏$ 的所有 linear combination.‘’ The worksheet wants to show:

1. there is $𝑡\in 𝑆$ with $𝑡|𝑎$ and $𝑡|𝑏$;
2. for every $𝑐$ with $𝑐|𝑎$ and $𝑐|𝑏$, one has $𝑐\le 𝑡$.

Let $𝑡$ be the smallest positive element of $𝑆$ (神奇，这里是直接过一个 定理来想到 $\left(𝑎,𝑏\right)$ 是 $𝑆$ 的 smallest positive elem’‘). By well-ordering, $𝑡$ exists, and $𝑡=𝑢𝑎+𝑣𝑏$ for some $𝑢,𝑣\in \text{ℤ}$. Divide $𝑎$ by $𝑡$:

Since $𝑟=𝑎-𝑡𝑞=𝑎-\left(𝑢𝑎+𝑣𝑏\right)𝑞=𝑎\left(1-𝑢𝑞\right)+𝑏\left(-𝑣𝑞\right)$, it is also a linear combination of $𝑎,𝑏$, hence $𝑟\in 𝑆$. Minimality forces $𝑟=0$, so $𝑎=𝑡𝑞$ and $𝑡|𝑎$; similarly $𝑡|𝑏$.‘’ If $𝑐|𝑎$ and $𝑐|𝑏$, write $𝑎=𝑐𝑘$, $𝑏=𝑐𝑠$. Then $𝑡=𝑢𝑎+𝑣𝑏=𝑐\left(𝑢𝑘+𝑣𝑠\right)$, so $𝑐|𝑡$ and $𝑐\le \left|𝑡\right|=𝑡$.

Source transcription — WS2, p. 1, Pf of Corollary 1.3′‘. The sheet records: if $𝑎,𝑏=1$, then by Theorem 2 there are $𝑢,𝑣\in \text{ℤ}$ with $𝑎𝑢+𝑏𝑣=1$. If $𝑎|𝑐$, write $𝑐=𝑏𝑘$ (as written in the source); then $𝑎𝑢+𝑏𝑣=𝑐$ is used to conclude $𝑎|𝑐$. The adjacent Chinese note says: 这个证明的意思是: 如果 $𝑎$ 是 $𝑏,𝑐$ 的因子, 但 $𝑎$ 和 $𝑏$ 互质, 那 $𝑎$ 肯定就是 $𝑐$ 的因子（直观可见）.’‘

2.3 Euclidean algorithm

Source transcription — WS2, p. 1, Worksheet 部分, Pf of Thm 5: Euclidean Algorithm’‘. For $𝑎,𝑏\in \text{ℤ}$, let $𝑑=gcd\left(𝑎,𝑏\right)$ and divide

The source proves both directions of $gcd\left(𝑎,𝑏\right)=gcd\left(𝑏,𝑟\right)$. If $𝑑|𝑏$ and $𝑑|𝑟$, then $𝑑|\left(𝑏𝑞+𝑟\right)=𝑎$; hence every common divisor of $𝑏,𝑟$ is one of $𝑎,𝑏$, and $gcd\left(𝑏,𝑟\right)\le gcd\left(𝑎,𝑏\right)$. Conversely, if $𝑑|𝑎$ and $𝑑|𝑏$, then $𝑏=𝑑{𝑘}_1$ and $𝑎=𝑑{𝑘}_2$ for some integers, so $𝑟=𝑎-𝑏𝑞=𝑑\left({𝑘}_2-{𝑘}_1𝑞\right)$; the same argument gives $gcd\left(𝑏,𝑟\right)\ge gcd\left(𝑎,𝑏\right)$. Therefore $gcd\left(𝑏,𝑟\right)=gcd\left(𝑎,𝑏\right)$.

The Chinese explanation on the page is retained: Worksheet 则介绍了 Euclidean Algorithm（辗转相除）这种方法则证明；当我们知道 $\left(𝑎,𝑏\right)=\left(𝑏,𝑎mod𝑏\right)$ 时，最后会到某时 $𝑢,𝑣$ 使 $𝑢mod𝑣=0$，那么下一步 $𝑣mod0=𝑣$，$\left(𝑢,0\right)=𝑣$，这个 $𝑣$ 就是一连下来最后的 $\left(𝑎,𝑏\right)$ 了。‘’ Each division has the form $𝑐=𝑑𝑞+𝑟$; at the last nonzero remainder one back-substitutes to obtain the promised linear combination.

Worked source calculations — WS2, p. 1.

Thus $gcd\left(524,148\right)=4$, and the page back-substitutes

The second calculation is

so $gcd\left(1103,456\right)=1$ and

3 Gcd, primes, and congruence

This chapter transcribes ‘WorkSheets/412-WS3-Mywork.pdf’, pp. 1–2; ‘412-WS4-Mywork.pdf’, pp. 1–2; and ‘412-WS5-Mywork.pdf’, pp. 1–2.

3.1 Prime factorization

Source transcription — WS3, p. 1, Pf of Thm 1′‘. The sheet states

For the forward implication, write $𝑏=𝑘𝑝$ when $𝑝|𝑏$. Since $𝑝$ is prime, the source notes that $gcd\left(𝑝,𝑏\right)$ can only be $1$ or $\pm 𝑝$; $gcd\left(𝑝,𝑏\right)=1$ gives $𝑝|𝑐$ by the preceding corollary, while $gcd\left(𝑝,𝑏\right)=𝑝$ gives $𝑝|𝑏$ by definition. The Chinese margin explanation is: 这一段实际很好想: $𝑝$ is prime => say $𝑝=𝑘𝑑$, $𝑑\in \text{ℤ}$ composite, 就会使 $𝑝|𝑏𝑐$ 但 $𝑝$ 不可能整除 $𝑏$ 或 $𝑐$，可以是整除 $𝑏$ 的某一个 连结的因数，因此 $𝑝|𝑝$，而它不可能整除自己的 factor.‘’

For the converse the source uses the contrapositive: if $𝑝$ is not prime, there are $𝑏,𝑐\in \text{ℤ}$ with $𝑝|𝑏𝑐$, but $𝑝$ does not divide either $𝑏$ or $𝑐$. It then writes $𝑝=𝑘𝑑$, with $𝑘\ne \pm 1$ and $𝑘\ne 𝑝$, so $𝑝$ is composite; because $𝑝$ is not a unit and $|𝑝|>1$, this gives the required nontrivial factors.

Source transcription — WS3, p. 1, Pf of Corollary 1′‘. If $𝑝\in \text{ℤ}$ is prime and $𝑝|\left({𝑎}_1\dots {𝑎}_{𝑛}\right)$, then $𝑝|{𝑎}_{𝑖}$ for some $𝑖$. The source says that this is the same simple argument as the two-factor case: treat ${𝑎}_1\dots {𝑎}_{𝑛-1}$ as one factor and repeat the argument, that is, use induction.

Source transcription — WS3, p. 1, Pf of Thm 2 (FTA), Part I: Existence’‘. Consider

The sheet proves: (4) every element of $𝑆$ is composite (a prime cannot lie in $𝑆$, since $𝑝=𝑝$ is a trivial factorization); (5) if $𝑎,𝑏>1$ and $𝑎𝑏\in 𝑆$, then $𝑎\in 𝑆$ or $𝑏\in 𝑆$ (the contrapositive is written); and (6) $𝑆$ is empty. Indeed, if $𝑆$ were nonempty, let $𝑠$ be its minimum by well-ordering. By (4), $𝑠=𝑎𝑏$ with $𝑎,𝑏>1$; by (5), one of $𝑎,𝑏$ is in $𝑆$ and is smaller than $𝑠$, a contradiction. Thus every nonzero nonunit admits a prime factorization.

Source transcription — WS3, p. 2, Pf of Thm 2 FTA, Part II: Uniqueness’‘. Suppose

are two factorizations into primes. Since ${𝑝}_1|\left({𝑞}_1\dots {𝑞}_{𝑡}\right)$, Euclid’s lemma makes ${𝑝}_1$ divide one ${𝑞}_{𝑖}$; as ${𝑞}_{𝑖}$ is prime, ${𝑝}_1=\pm {𝑞}_{𝑖}$. Eliminating associated prime factors on both sides and repeating, if $𝑠<𝑡$ then the remaining equality would say $1={𝑞}_{𝑠+1}\dots {𝑞}_{𝑡}$, impossible because each ${𝑞}_{𝑖}$ is prime. Hence $𝑠=𝑡$, and after reordering every ${𝑝}_{𝑖}={𝑞}_{𝑖}$ up to associates.

Source transcription — WS3, p. 2, GCD Exercise’‘. If

with $0\le {𝑐}_{𝑖}\le min\left({𝑎}_{𝑖},{𝑏}_{𝑖}\right)$, then every common divisor has the form $𝑑=\pm {𝑝}_1^{{𝑐}_1}\dots {𝑝}_{𝑛}^{{𝑐}_{𝑛}}$. Thus

3.2 Congruence modulo $𝑁$

Source transcription — WS4, p. 1. Congruence 的概念是对 equality relation 的 generalization.‘’ For $𝑎,𝑏\in \text{ℤ}$,

The Congruence modulo N relation is defined by

that is, $𝑁|\left(𝑎-𝑏\right)$. The source explicitly compares the three equality axioms—reflexive, symmetric, transitive—with their congruence counterparts, then defines the congruent class ${\left[𝑎\right]}_{𝑁}$ and lists

For classes the worksheet proves 相等或者是 disjoint.‘’ If $𝑥\in {\left[𝑎\right]}_{𝑁}\cap {\left[𝑏\right]}_{𝑁}$, then $𝑥\equiv 𝑎\left(mod𝑁\right)$ and $𝑥\equiv 𝑏\left(mod𝑁\right)$, hence $𝑎\equiv 𝑏\left(mod𝑁\right)$. Any $𝑦{\in \left[𝑎\right]}_{𝑁}$ is then congruent to $𝑏$ by transitivity, so ${\left[𝑎\right]}_{𝑁}\subset {\left[𝑏\right]}_{𝑁}$; similarly the reverse inclusion holds. It also records ${\left[𝑎\right]}_{10}\subset {\left[𝑎\right]}_{𝑁}$ when $𝑁$ divides $10$.

The source asks whether ${\left[𝑎\right]}_7\mapsto$ round down $𝑎$ to 最近的 $10$ 的倍数’‘ is a function and concludes: 其实不是.’‘ For example, with $𝑥={\left[0\right]}_7=\left\{\dots ,-14,-7,0,7,14,\dots \right\}$, the representatives map to different multiples of $10$, so one class would have multiple images. In contrast, ${\left[𝑎\right]}_7\mapsto {\left[-𝑎\right]}_7$ is a function: replacing one representative by another congruent representative preserves the resulting class.

Source transcription — WS4, pp. 1–2, well-defined operations. The worksheet defines

For multiplication, if $𝑥=𝑎+𝑘𝑎$ and $𝑦=𝑏+𝑙𝑏$ are written as source representatives, then the calculation is retained in its displayed form:

so $𝑎𝑏$ is congruent to $𝑥𝑦$. It says: ${\text{ℤ}}_{𝑁}$ 具有除了 ${𝑥}^{-1}$ 外所有 field 的性质（A/M/D 易证）.‘’ It then proves that if $gcd\left(𝑎,𝑁\right)=1$, then ${\left[𝑎\right]}_{𝑁}𝑥={\left[1\right]}_{𝑁}$ has a solution: Bézout gives $𝑎𝑟+𝑁𝑠=1$, hence ${\left[𝑎\right]}_{{𝑁\left[𝑟\right]}_{𝑁}}={\left[1\right]}_{𝑁}$.

The source proves uniqueness of the solution to ${\left[𝑎\right]}_{𝑁}𝑥={\left[1\right]}_{𝑁}$ under the coprime condition: if ${\left[𝑎\right]}_{{𝑁\left[{𝑥}_1\right]}_{𝑁}}={\left[𝑎\right]}_{{𝑁\left[{𝑥}_2\right]}_{𝑁}}$, then ${\left[𝑎\right]}_{{𝑁\left[{𝑥}_1-{𝑥}_2\right]}_{𝑁}}={\left[0\right]}_{𝑁}$; since ${\left[𝑎\right]}_{𝑁}$ is a unit, ${\left[{𝑥}_1-{𝑥}_2\right]}_{𝑁}={\left[0\right]}_{𝑁}$, hence ${\left[{𝑥}_1\right]}_{𝑁}={\left[{𝑥}_2\right]}_{𝑁}$. It continues that every ${\left[𝑎\right]}_{𝑁}𝑥={\left[𝑏\right]}_{𝑁}$ then has a unique solution by multiplying the solution of ${\left[𝑎\right]}_{{𝑁\left[𝑟\right]}_{𝑁}}={\left[1\right]}_{𝑁}$ by ${\left[𝑏\right]}_{𝑁}$.

3.3 Linear combinations modulo $𝑁$

Source transcription — WS5, pp. 1–2. The source proves

For $𝑄=\left\{𝑘gcd\left(𝑎,𝑛\right):𝑘\in \text{ℤ}\right\}$ and $𝑃=\left\{𝑟𝑎+𝑠𝑛:𝑟,𝑠\in \text{ℤ}\right\}$, Bézout gives $gcd\left(𝑎,𝑛\right)=𝑎𝑢+𝑛𝑣$, hence every $𝑘gcd\left(𝑎,𝑛\right)=𝑎\left(𝑘𝑢\right)+𝑛\left(𝑘𝑣\right)$ lies in $𝑃$. Conversely, $gcd\left(𝑎,𝑛\right)$ divides both $𝑎$ and $𝑛$, so it divides every $𝑟𝑎+𝑠𝑛$ and that expression lies in $𝑄$.

Thus ${\left[𝑎\right]}_{𝑁\left[𝑥\right]}={\left[𝑏\right]}_{𝑁}$ has a solution exactly when $gcd\left(𝑎,𝑁\right)|𝑏$. The sheet writes the example

whose solutions are $\left[3\right],\left[7\right],\left[11\right]$ because $3=gcd\left(9,12\right)$ divides $3$ and the solutions differ by $\frac{12}{3}=4$. The source’s question 有多少 sol?‘’ is answered: if $𝑑=gcd\left(𝑎,𝑁\right)$, the distinct solutions are separated by $\frac{𝑁}{𝑑}$; hence there are $𝑑$ solution classes.

4 Homework 1: Integers and Equivalence Relations (49/50)

Source: ‘Homework/412-Hw-1-graded.pdf’ (30 PDF pages). Canvas grading and navigation pages are interleaved with the submitted handwritten pages. The source trails below identify every page with transcribed work; navigation-only pages are recorded in the migration receipt rather than copied as note content.

4.1 Question 1: square and cubic integers

Source trail: PDF pp. 6 and 8 (handwritten answer); p. 1 (10/10 grading).

For the square claim, the answer begins, “We prove by cases.” It invokes Corollary 2.5 to split the integers into $3𝑘$, $3𝑘+1$, and $3𝑘+2$. For an arbitrary $𝑎$ it records:

• if $𝑎=3𝑘$, then ${𝑎}^2=9{𝑘}^2=3\left(3{𝑘}^2\right)$;
• if $𝑎=3𝑘+1$, then ${𝑎}^2={\left(3𝑘+1\right)}^2=3\left(3{𝑘}^2+2𝑘\right)+1$;
• if $𝑎=3𝑘+2$, it writes ${𝑎}^2={\left(3𝑘+2\right)}^2=3\left(3{𝑘}^2+4𝑘\right)+1$.

The conclusion underneath is that the three cases cover all integers and none gives remainder $2$ upon division by $3$.

For the cubic claim, the answer again uses the same three cases:

• ${𝑎}^3=27{𝑚}^3=9\left(3{𝑚}^3\right)$, so it takes $𝑘=3{𝑚}^3$;
• ${𝑎}^3=27{𝑚}^3+27{𝑚}^2+9𝑚+1=9\left(3{𝑚}^3+3{𝑚}^2+𝑚\right)+1$;
• ${𝑎}^3=27{𝑚}^3+54{𝑚}^2+36𝑚+8=9\left(3{𝑚}^3+6{𝑚}^2+4𝑚+1\right)-1$.

Thus it concludes that a cubic integer has form $9𝑘$, $9𝑘+1$, or $9𝑘-1$. On the first cubic line, the handwriting says ${𝑎}^2=9𝑘$ after a calculation for ${𝑎}^3$; it is retained as a source-writing slip rather than silently treated as a new assertion.

4.2 Question 2: least common multiples

Source trail: PDF pp. 10 and 12 (handwritten answer); p. 2 (10/10 grading).

For part (a), the response lets $𝑎,𝑏$ be arbitrary positive integers and defines the set of positive common multiples

It observes that $𝑎𝑏\in 𝑆$, hence $𝑆$ is nonempty and is a subset of the positive integers. The well-ordering principle supplies a smallest element, which it identifies as the least common multiple of $𝑎$ and $𝑏$.

For part (b), with $𝑑=gcd\left(𝑎,𝑏\right)$, the handwritten calculation is

This is transcribed as written: the displayed factorization omits a factor of $𝑑$ if both preceding equalities are read literally. The written construction targets an integer $𝑚$ satisfying $𝑑𝑚=𝑎𝑏$.

For part (c), it starts from $𝑑𝑚=𝑎𝑏$ and $𝑎=𝑑𝑝$, then writes

and, similarly, $𝑚=𝑞𝑎$. It concludes that both $𝑎$ and $𝑏$ divide $𝑚$.

For part (d), let $𝑀$ be an arbitrary common multiple, so $𝑀=𝑠𝑎=𝑘𝑏$ for integers $𝑠,𝑘$. Bézout is written as

With $𝑚=𝑎\frac{𝑏}{gcd\left(𝑎,𝑏\right)}$, the response calculates

The substitution $𝑘𝑏=𝑠𝑎$ makes the last expression integral, so $𝑚$ divides $𝑀$. Together with the preceding divisibility calculation, it concludes that $𝑚$ is the least common multiple.

4.3 Question 3: greatest common divisors under an integral matrix

Source trail: PDF pp. 14 and 16 (handwritten answer); p. 2 (10/10 grading, with feedback).

Let $𝐴=\left(\begin{array}{cc}𝑎& 𝑏\\ 𝑐& 𝑑\end{array}\right)$ and write its inverse as ${𝐴}^{-1}$. The proof notes that both $det𝐴$ and $det{𝐴}^{-1}=\frac{1}{det𝐴}$ are integers. It concludes that $det𝐴$ is $1$ or $-1$, and writes

It applies this to the column vector with entries $𝑎𝑥+𝑏𝑦$ and $𝑐𝑥+𝑑𝑦$, obtaining $\pm {\left(𝑥,𝑦\right)}^{𝑇}$. In particular,

and

Thus $𝑥$ and $𝑦$ are integer linear combinations of $𝑎𝑥+𝑏𝑦$ and $𝑐𝑥+𝑑𝑦$; conversely those two expressions are integer linear combinations of $𝑥,𝑦$. The response phrases the two divisibility comparisons as mutual greatest-common-divisor inequalities and concludes

Grader feedback (PDF p. 2): “avoid the gcd notation.”

4.4 Question 4: prime multiplicities and irrational roots

Source trail: PDF pp. 16, 18, 20, and 22 (handwritten answer); p. 3 (9/10 grading and feedback).

The answer labels the two assertions “1” and “2.” For the first direction of the equivalence in part (1), it writes $𝑛={𝛽}^{𝑑}$, factors

and argues for a prime $𝑝={𝑞}_{𝑖}$ that ${𝑝}^{{𝑎}_{𝑖}𝑑}$ divides $𝑛$; hence the multiplicity is divisible by $𝑑$. For the converse it writes

and, from ${𝑎}_{𝑖}={𝑘}_{𝑖}𝑑$, obtains

This is the stated proof that $𝑛$ is a $𝑑$-th power exactly when all its prime multiplicities are divisible by $𝑑$.

For part (2), it argues by contradiction. Assume $𝑛$ is not a $𝑑$-th power but

with $𝑝,𝑞\in \mathbb{Z}$, $𝑞\ne 0$, and $gcd\left(𝑝,𝑞\right)=1$. It writes $𝑛{𝑞}^{𝑑}={𝑝}^{𝑑}$ and factors $𝑝$ and $𝑞$ by the Fundamental Theorem of Arithmetic. Since their prime factors are disjoint, it concludes that each prime factor of $𝑛$ has multiplicity $𝑑$, so $𝑛$ is a $𝑑$-th power, a contradiction.

Grader feedback (PDF p. 3): “You forgot the powers on the primes.” A second comment says, “Again, if you had written out the powers, you would have gotten that q = 1, and be done with it all.”

4.5 Question 5: equivalence relations and classes

Source trail: PDF pp. 26, 28, and 30 (handwritten answer); p. 4 (10/10 grading).

For part (a), take arbitrary column vectors ${\left(𝑥,𝑦\right)}^{𝑇}$ and ${\left(𝑤,𝑧\right)}^{𝑇}$ with $𝑥-𝑦=𝑤-𝑧$. The response checks reflexivity from $𝑥-𝑦=𝑥-𝑦$, symmetry by reversing the equality, and transitivity by setting $𝑤-𝑧=𝛼-𝛽$ and chaining $𝑥-𝑦=𝑤-𝑧=𝛼-𝛽$. It concludes that this relation on ${ℝ}^2$ is an equivalence relation.

For part (b), it supplies the transitivity counterexample

Then $𝑎𝑏=0$ and $𝑏𝑐=0$, so $𝑎\sim 𝑏$ and $𝑏\sim 𝑐$, while $𝑎𝑐=-1<0$, so $𝑎$ is not related to $𝑐$. Thus the relation $𝑎\sim 𝑏$ when $𝑎𝑏\ge 0$ is not an equivalence relation.

For part (c), if $𝑎\in 𝑋$, reflexivity gives $𝑎\sim 𝑎$, so

Therefore every equivalence class is nonempty.

For part (d), assume the intersection of $\left[𝑎\right]$ and $\left[𝑏\right]$ is nonempty and choose $𝑥$ in both classes. For arbitrary $𝑚\in \left[𝑎\right]$, the response uses $𝑚\sim 𝑎$, $𝑥\sim 𝑎$, symmetry, and transitivity to obtain $𝑚\sim 𝑥$ and hence $𝑚\in \left[𝑏\right]$. Reversing the argument for arbitrary $𝑛\in \left[𝑏\right]$ gives $\left[𝑎\right]=\left[𝑏\right]$. Therefore two classes are either disjoint or equal.

For part (e), every $𝑥\in 𝑋$ has $𝑥\sim 𝑥$, so $\left[𝑥\right]$ is a nonempty class in the set $𝑆$ of all classes and $𝑥$ belongs to that class. The handwritten conclusion is that $𝑋$ is the disjoint union of the classes $𝑌$ with $𝑌\in 𝑆$. Using part (d), the response identifies the union as disjoint, concluding that the equivalence classes partition $𝑋$.

5 Rings and homomorphisms

This chapter transcribes ‘WorkSheets/412-WS7-Ring-Mywork.pdf’, p. 1; ‘412-WS8-Mywork.pdf’, pp. 1–2; and the ring-homomorphism portion of ‘412-WS9-Mywork.pdf’, p. 1.

5.1 Ring structure

Source transcription — WS7, p. 1. An operation on a set $𝑆$ is a function $𝑓:𝑆\times 𝑆\to 𝑆$. A ring is a set $𝑅$ with two operations $+$‘’ and $\times$‘’ such that, for all $𝑎,𝑏,𝑐\in 𝑅$:

1. $\left(𝑅,+\right)$ is an abelian group: closure, associativity, commutativity, $0_{𝑅}$, and additive inverses;
2. multiplication has closure and associativity;
3. there is $1_{𝑅}$ such that $1_{𝑅}𝑎=𝑎1_{𝑅}=𝑎$ (有幺元的环即为环’‘ in the handwritten note); and
4. $\left(𝑎+𝑏\right)𝑐=𝑎𝑐+𝑏𝑐$ and $𝑎\left(𝑏+𝑐\right)=𝑎𝑏+𝑎𝑐$.

The source then proves $0\times 𝑥=0$: from $0\times 𝑥=\left(0+0\right)\times 𝑥=0\times 𝑥+0\times 𝑥$, let $𝑦$ be the additive inverse of $0\times 𝑥$, add $𝑦$ to both sides, and obtain $0=0\times 𝑥$.

Source transcription — WS7, p. 1, D(1). To show a nonempty subset $𝑆$ of a ring $𝑅$ is a subring, the worksheet lists: $1_{𝑅},0_{𝑅}\in 𝑆$; $𝑆$ is closed under $+$ and $\times$; and $𝑆$ is closed under additive inverse. It notes that the inherited $+$ is commutative and associative and $\times$ distributes over it, while $1_{𝑅},0_{𝑅}$ serve as the identities; 所以只要证明 $1_{𝑅},0_{𝑅}\in 𝑆$ 且对 closure 即可.‘’

Source transcription — WS7, p. 1, D(2). The set $\text{Fun}\left(𝑅,𝑅\right)$ of all functions from $𝑅$ to itself, with pointwise operations

is recorded as a ring. The source asks whether there are other subrings: ① 它自己；② 一个 smallest subring: 至少 include $1_{𝑅},0_{𝑅}$. 因而 all elements of $𝑆$: $𝑛\cdot 1_{𝑅}=1_{𝑅}+\dots +1_{𝑅}$, $𝑛\in \text{ℤ}$.‘’ It concludes that $\left\{𝑛\cdot 1_{𝑅}:𝑛\in \text{ℤ}\right\}$ is a subring and is the smallest subring.

5.2 Ring homomorphisms

Source transcription — WS8, p. 1, A. The page lists seven maps and their status:

1. the inclusion $𝜑:\text{ℤ}\to \text{ℚ}$, $𝑧\mapsto \frac{𝑧}{1}$, is a hom but not an isomorphism (for example $\frac{2}{3}$ is not $𝜑\left(𝑧\right)$);
2. the doubling map $𝜑:\text{ℤ}\to \text{ℤ}$, $𝑧\mapsto 2𝑧$, is not a hom because $1\mapsto 2$ and it does not preserve $1_{\text{ℤ}}$;
3. the residue map $𝜑:\text{ℤ}\to {\text{ℤ}}_{𝑁}$, $𝑧\mapsto {\left[𝑧\right]}_{𝑁}$, is a hom by modular arithmetic, is surjective, but is not an isomorphism because it is not one-to-one;
4. the evaluation at $0$′‘ map $𝜑:\text{ℝ}\left[𝑋\right]\to \text{ℝ}$, $𝑓\left(𝑋\right)\mapsto 𝑓\left(0\right)$, is a hom: the page writes $\text{eval}\left(𝑓\left(𝑋\right)+𝑔\left(𝑋\right)\right)=\text{eval}\left(𝑓\left(𝑋\right)\right)+\text{eval}\left(𝑔\left(𝑋\right)\right)$ and similarly for products;
5. $𝜑:\text{ℝ}\left[𝑋\right]\to \text{ℝ}\left[𝑋\right]$, $𝑓\left(𝑋\right)\mapsto {𝑓}^{\prime }\left(𝑋\right)$, is not a hom because $1\mapsto 0$;
6. $𝜑:\text{ℝ}\to {𝑀}_2\left(\text{ℝ}\right)$, $𝜆\mapsto \left(\begin{array}{cc}𝜆& 0\\ 0& 𝜆\end{array}\right)$, is a hom, with the addition and product of diagonal matrices written out; and
7. $𝜑:{𝑀}_2\left(\text{ℤ}\right)\to \text{ℝ}$, $𝐴\mapsto det\left(𝐴\right)$, is not a hom, as a displayed pair of matrices shows $det\left(𝐴+𝐵\right)\ne det\left(𝐴\right)+det\left(𝐵\right)$.

Source transcription — WS8, p. 1, B(1)–(3). Every hom preserves $0_{𝑅}$: from $0_{𝑆}+0_{𝑆}=0_{𝑆}$ one gets $𝜑\left(0_{𝑆}\right)=𝜑\left(0_{𝑆}+0_{𝑆}\right)=𝜑\left(0_{𝑆}\right)+𝜑\left(0_{𝑆}\right)$ and cancels an additive inverse. It preserves additive inverse because $𝜑\left(𝑥\right)+𝜑\left(-𝑥\right)=𝜑\left(0_{𝑆}\right)=0_{𝑇}$, hence $-𝜑\left(𝑥\right)=𝜑\left(-𝑥\right)$. It preserves units: if $𝑢{𝑢}^{-1}=1_{𝑆}$, then $𝜑\left(𝑢\right)𝜑\left({𝑢}^{-1}\right)=𝜑\left(1_{𝑆}\right)=1_{𝑇}$, so $𝜑\left(𝑢\right)$ and $𝜑\left({𝑢}^{-1}\right)$ are units.

The same page gives the kernel definition and an example:

for $𝜓:𝑅\times 𝑆\to 𝑅$, $\left(𝑟,𝑠\right)\mapsto 𝑟$. It also writes the informal summary isomorphism preserves 基本 everything（而 hom 只需要 surjective 也 preserve 所有的单位元）‘’, followed by the counter-cue 不是所有 field, domain …‘’.

Source transcription — WS8, pp. 1–2, C–E. A homomorphism kernel is nonempty because $𝜑\left(0_{𝑆}\right)=0_{𝑅}$; in particular $0_{𝑆}\in ker𝜑$. The source proves

If $𝜑$ is injective and $𝑥\in ker𝜑$, then $𝜑\left(𝑥\right)=𝜑\left(0_{𝑆}\right)$, so $𝑥=0_{𝑆}$. Conversely, if $ker𝜑=\left\{0_{𝑆}\right\}$ and $𝜑\left(𝑥\right)=𝜑\left(𝑦\right)$, then $𝜑\left(𝑥\right)+\left(-𝜑\left(𝑦\right)\right)=0_{𝑅}=𝜑\left(𝑥+\left(-𝑦\right)\right)$, hence $𝑥-𝑦\in ker𝜑$, $𝑥=𝑦$.

The Chinese/English note continues: 如何都有一个 unique 的从 $\text{ℤ}$ 到 $𝑅$ 之间的 hom $𝜓:\text{ℤ}\to 𝑅$，这个 hom 叫做 canonical ring homomorphism.‘’ If such a $𝜓$ exists, $𝜓\left(1\right)=1_{𝑅}$ and $𝜓\left(0\right)=0_{𝑅}$; for $𝑛\ge 1$, $𝜓\left(𝑛\right)=𝜓\left(1+\dots +1\right)=𝑛\cdot 1_{𝑅}$, while for $𝑛\le -1$, $𝜓\left(𝑛\right)=-𝑛\cdot 1_{𝑅}$. Thus the possible map is unique, and this calculation also verifies it is a hom: $𝜓\left(𝑛+𝑚\right)=𝜓\left(𝑛\right)+𝜓\left(𝑚\right)$ and $𝜓\left(𝑛𝑚\right)=𝜓\left(𝑛\right)𝜓\left(𝑚\right)$.

5.3 Domains and fields

Source transcription — WS8, p. 2, D–E. The worksheet proves that $0_{𝑅}=1_{𝑅}$ if and only if $𝑅=\left\{0_{𝑅}\right\}$: for $𝑟\in 𝑅$, $𝑟=𝑟1_{𝑅}=𝑟0_{𝑅}=0_{𝑅}$. It then records Thm 3.8: every field 一定 domain’‘: if $𝑎,𝑏\in 𝐹$, $𝑎𝑏=0$, and $𝑎\ne 0$, multiply by ${𝑎}^{-1}$ to get $𝑏=0_{𝐹}$. The red Chinese explanation adds: 任何自乘的去，$+$ 中的所有非 0 元不能乘起就 0；因为 $𝐹$ 上 $+-\times$ 是 well-defined，如果有元被中 $+-\times$ 就失去唯一性了.’‘

A subring of a domain is a domain (去除了不必要了，本来所有非 0 元不能乘 到 0，$+\times$ 也肯定一样’‘). For $𝑆\subset 𝑅$, the inclusion map $𝜑:𝑆\to 𝑅$ is a ring hom exactly when $𝑆$ is a subring of $𝑅$; the source explains that the issue is the map is the inclusion and therefore one must retain the same $0,1,+,\times$.

6 Homework 2: Congruence Classes and Functions (35/40)

Source: ‘Homework/412-Hw-2-graded.pdf’ (33 PDF pages). The source interleaves Canvas grading and navigation pages with the handwritten submission. Source trails identify all transcribed pages; navigation-only and empty-shell pages are recorded in the migration receipt.

6.1 Grading record and feedback

Source trail: PDF p. 1.

The graded total is 35/40. Question scores are 8/10, 10/10, 10/10, and 7/10. The substantive comments are retained verbatim:

• Question 1: “I didn’t really understand anything, try to always point out what you are accomplishing with each step.”
• Question 4: “This is somewhat philosophically unsatisfactory for then, there may be an empty eq. class. And all that follows would be inaccurate. I’m just going to subtract one point for it, but remember to think of eq. classes as partitions first.”
• Question 4: “To define Z/nZ you need the equivalence relation you are trying to show exists as of this form. The logic is somewhat circular.”

A separate one-character Question 2 comment reads “n”.

6.2 Question 1: simultaneous congruences

Source trail: PDF pp. 3-4, 6, 8, 10, and 11 (prompt and handwritten answer).

The system is $𝑥\equiv 𝑎$ modulo $𝑚$ and $𝑥\equiv 𝑏$ modulo $𝑛$, where $gcd\left(𝑚,𝑛\right)=1$.

For part (a), assuming $𝑟𝑚+𝑠𝑛=1$, the response takes

Modulo $𝑚$, it rewrites $𝑎𝑠𝑛$ as $𝑎\left(1-𝑟𝑚\right)=𝑎-𝑎𝑟𝑚$, while $𝑏𝑟𝑚$ is divisible by $𝑚$, so $𝑥\equiv 𝑎$. Similarly, modulo $𝑛$, it rewrites $𝑏𝑟𝑚$ as $𝑏\left(1-𝑠𝑛\right)=𝑏-𝑏𝑠𝑛$, while $𝑎𝑠𝑛$ is divisible by $𝑛$, giving $𝑥\equiv 𝑏$. It therefore states that this $𝑥$ solves the system.

For part (b), $gcd\left(𝑚,𝑛\right)=1$ and Bézout give integers $𝑟,𝑠$ with $𝑟𝑚+𝑠𝑛=1$. Part (a) then supplies $𝑥=𝑎𝑠𝑛+𝑏𝑟𝑚$ for every choice of $𝑎,𝑏$.

For part (c), fix a solution ${𝑥}_1$ and take an arbitrary $𝑥\in {\left[{𝑥}_1\right]}_{𝑚𝑛}$. The answer writes $𝑥={𝑥}_1+𝑘𝑚𝑛$ for some integer $𝑘$. Consequently $𝑥\equiv {𝑥}_1\equiv 𝑎$ modulo $𝑚$ and $𝑥\equiv {𝑥}_1\equiv 𝑏$ modulo $𝑛$, so every element of the class is a solution.

For part (d), it lets ${𝑥}_1=𝑎+𝑔𝑚=𝑏+𝑓𝑛$ and an arbitrary solution $𝑥=𝑎+𝑝𝑚=𝑏+𝑞𝑛$. Hence both $𝑚$ and $𝑛$ divide $𝑥-{𝑥}_1$. The response invokes the Fundamental Theorem of Arithmetic and the relative primality of $𝑚,𝑛$ to conclude that $𝑚𝑛$ divides $𝑥-{𝑥}_1$, so $𝑥\in {\left[{𝑥}_1\right]}_{𝑚𝑛}$. Together with part (c), this proves that the set of solutions is exactly ${\left[{𝑥}_1\right]}_{𝑚𝑛}$.

For part (e), the Euclidean algorithm in the answer is

Back-substitution gives

For the system $𝑥\equiv 11$ modulo $72$ and $𝑥\equiv 30$ modulo $169$, it takes

The response checks ${𝑥}_1=11-594\cdot 72$, hence ${𝑥}_1\equiv 11$ modulo $72$, and states similarly that ${𝑥}_1\equiv 30$ modulo $169$. Its full answer is

6.3 Question 2: maps between congruence classes

Source trail: PDF pp. 11, 13, 15, and 17 (prompt and handwritten answer).

For part (a), the proposed map ${\mathbb{Z}}_3\to {\mathbb{Z}}_6$, ${\left[𝑎\right]}_3\to {\left[𝑎\right]}_6$, is declared not well-defined. The counterexample is $𝑎=1$, $𝑏=4$: ${\left[𝑎\right]}_3={\left[𝑏\right]}_3$, but ${\left[1\right]}_6$ and ${\left[4\right]}_6$ are distinct.

For part (b), the map ${\mathbb{Z}}_6\to {\mathbb{Z}}_3$, ${\left[𝑎\right]}_6\to {\left[𝑎\right]}_3$, is declared well-defined. If ${\left[𝑎\right]}_6={\left[𝑏\right]}_6$, then $𝑏=𝑎+6𝑘=𝑎+3\left(2𝑘\right)$, so $𝑏\equiv 𝑎$ modulo $3$ and ${\left[𝑏\right]}_3={\left[𝑎\right]}_3$.

For part (c), assume $𝑛$ divides $𝑚$ and write $𝑚=𝑛𝑝$. If ${\left[𝑎\right]}_{𝑚}={\left[𝑏\right]}_{𝑚}$, then $𝑏=𝑎+𝑚𝑘=𝑎+𝑛\left(𝑝𝑘\right)$ for some integer $𝑘$. Thus $𝑏\equiv 𝑎$ modulo $𝑛$, proving that ${\left[𝑎\right]}_{𝑚}\to {\left[𝑎\right]}_{𝑛}$ is well-defined.

For part (d), assume $𝑛$ does not divide $𝑚$. The two source representatives ${\left[𝑎\right]}_{𝑚}$ and ${\left[𝑎+𝑚\right]}_{𝑚}$ are the same class. If their targets in ${\mathbb{Z}}_{𝑛}$ were equal, then $𝑎+𝑚=𝑎+𝑘𝑛$ for some integer $𝑘$, so $𝑛$ would divide $𝑚$, a contradiction. The response concludes that the rule is not well-defined.

6.4 Question 3: solutions of a congruence-class equation

Source trail: PDF pp. 19-21, 23, 25, and 27-28 (prompt and handwritten answer).

Let $𝑑=gcd\left(𝑎,𝑛\right)$ and consider ${\left[𝑎\right]}_{𝑛}𝑦={\left[𝑏\right]}_{𝑛}$.

For part (a), the proof is by contraposition. If $𝑦={\left[𝑟\right]}_{𝑛}$ is a solution, then ${\left[𝑎𝑟\right]}_{𝑛}={\left[𝑏\right]}_{𝑛}$, hence $𝑎𝑟=𝑏+𝑝𝑛$ for some integer $𝑝$. Thus

is an integer linear combination of $𝑎,𝑛$. The answer invokes the description of all such combinations as the multiples of $gcd\left(𝑎,𝑛\right)$, and obtains $𝑑$ dividing $𝑏$. Hence if $𝑑$ does not divide $𝑏$, there is no solution.

For part (b), with $𝑏=0$, it first takes $𝑥=𝑘\frac{𝑛}{𝑑}$ and computes

so ${\left[𝑎\right]}_{𝑛}{\left[𝑥\right]}_{𝑛}={\left[0\right]}_{𝑛}$. Conversely, if ${\left[𝑥\right]}_{𝑛}$ is a solution, then $𝑎𝑥=𝑝𝑛$. After division by $𝑑$,

The response proves $gcd\left(\frac{𝑎}{𝑑},\frac{𝑛}{𝑑}\right)=1$ by contradiction: a common divisor greater than $1$ would make a common divisor of $𝑎,𝑛$ greater than $𝑑$. It then applies the Fundamental Theorem of Arithmetic to conclude that $\frac{𝑛}{𝑑}$ divides $𝑥$. Thus the displayed solution set is

For part (c), Bézout gives $𝑟𝑎+𝑠𝑛=𝑑$. Since $𝑑-𝑟𝑎=𝑠𝑛$, the response writes ${\left[𝑑\right]}_{𝑛}={\left[𝑟𝑎\right]}_{𝑛}$. If $𝑏=𝑘𝑑$, then

Thus ${\left[𝑟\frac{𝑏}{𝑑}\right]}_{𝑛}$ is a solution.

For part (d), fix a solution ${𝑦}_1={\left[{𝑟}_1\right]}_{𝑛}$. If $𝑦={\left[𝑟\right]}_{𝑛}$ is another solution, then $𝑎𝑟=𝑏+𝑝𝑛$ and $𝑎{𝑟}_1=𝑏+{𝑝}_1𝑛$, so

Thus $𝑧=𝑦-{𝑦}_1$ solves ${\left[𝑎\right]}_{𝑛}𝑧={\left[0\right]}_{𝑛}$. Conversely, if $𝑧=𝑦-{𝑦}_1$ solves the zero equation, the response uses distributivity in congruence classes to add ${\left[𝑎\right]}_{𝑛}{𝑦}_1={\left[𝑏\right]}_{𝑛}$ and obtain ${\left[𝑎\right]}_{𝑛}𝑦={\left[𝑏\right]}_{𝑛}$. Therefore the number of solutions to the original equation is the same as for the zero equation, namely exactly $𝑑$.

6.5 Question 4: equivalence relations induced by functions

Source trail: PDF pp. 28, 30-31, and 33 (prompt and handwritten answer).

For part (a), let $𝑥,𝑦,𝑧\in 𝑋$. Since $𝑓\left(𝑥\right)=𝑓\left(𝑥\right)$, the relation defined by $𝑓\left(𝑥\right)=𝑓\left({𝑥}^{\prime }\right)$ is reflexive. If $𝑥\sim 𝑦$, then $𝑓\left(𝑥\right)=𝑓\left(𝑦\right)$ and therefore $𝑓\left(𝑦\right)=𝑓\left(𝑥\right)$, proving symmetry. If $𝑥\sim 𝑦$ and $𝑦\sim 𝑧$, then $𝑓\left(𝑥\right)=𝑓\left(𝑦\right)=𝑓\left(𝑧\right)$, proving transitivity. The response concludes that it is an equivalence relation.

For part (b), it defines the class indexed by an image value as

and the set of all such classes as

It then defines $𝜑:{𝑋}_{𝑓}\to \Im \left(𝑓\right)$ by $\left[𝑦\right]\to 𝑦$. The answer argues that every member of $\left[𝑦\right]$ has image $𝑦$, so the map is well-defined; it argues injectivity by contradiction from unequal classes allegedly mapping to the same image; and it proves surjectivity because every $𝑦\in \Im \left(𝑓\right)$ is $𝑓\left(𝑥\right)$ for some $𝑥\in 𝑋$, whose class maps to $𝑦$. It concludes that $𝜑$ is bijective.

For part (c), it takes

Then $𝑓\left(𝑥\right)=𝑓\left({𝑥}^{\prime }\right)$ exactly when $𝑥\equiv {𝑥}^{\prime }$ modulo $𝑛$, so congruence modulo a fixed $𝑛$ is the preceding function-induced relation. The response concludes that this gives a partition of $\mathbb{Z}$ whose equivalence classes are

7 Polynomials and quotient rings

This chapter transcribes the polynomial and quotient material in ‘WorkSheets/412-WS9-Mywork.pdf’, p. 2 and ‘WorkSheets/412-WS10-Mywork.pdf’, pp. 1–3.

7.1 Domains, polynomial units, and division

Source transcription — WS9, p. 2. The worksheet records: if $𝑅$ is a domain, then $𝑅\left[𝑋\right]$ is a domain. Its explanation is that the degree of the product of two nonzero polynomials is the sum of their degrees, so the product cannot be zero. It then notes that the units of $𝑅\left[𝑋\right]$ are exactly the units of $𝑅$; a nonconstant polynomial cannot have a polynomial inverse. It gives the special example that in ${\text{ℤ}}_{𝑝\left[𝑋\right]}$, the units are the nonzero elements of ${\text{ℤ}}_{𝑝}$, because ${\text{ℤ}}_{𝑝}$ is a field.

Source transcription — WS10, p. 1, Part 1(A). Long division gives

The page labels the quotient $𝑞={𝑋}^3+1$ and remainder $𝑟=0$. Its red note states: Division algorithm 只能在 field 上有用，因为只有 field 上才对 division 有 well-definedness.‘’ It then records the failure over $\text{ℤ}\left[𝑋\right]$: when $deg𝑓<deg𝑔$, a putative quotient can be $𝑞\left(𝑋\right)=\frac{1}{2}𝑋+\frac{1}{2}$, which is not in $\text{ℤ}\left[𝑋\right]$, so the source’s division-algorithm hypothesis fails.

Source transcription — WS10, p. 1, C(1). Fix $𝑓\in 𝐹\left[𝑋\right]$. Divide $𝑓$ by $𝑋-𝜆$:

Thus $𝑟$ is constant. Substituting $𝑋=𝜆$ gives $𝑓\left(𝜆\right)=𝑟$. The source calls this the Pf of Remainder Thm’‘ and writes $𝑓\left(𝜆\right)$ 是 $\left(𝑋-𝜆\right)$ 的 remainder.’‘

Source transcription — WS10, p. 1, C(2). The factor theorem is recorded in both directions:

If $𝑓\left(𝜆\right)=0$, division gives $𝑓=𝑞\left(𝑋-𝜆\right)+0$; conversely, substitute $𝜆$ in a multiple of $𝑋-𝜆$.

7.2 Factorisation and irreducibility

Source transcription — WS10, p. 1, B. For the polynomial gcd exercises,

so the source writes $gcd=𝑋-3$. It also records in ${\text{ℤ}}_2\left[𝑋\right]$:

then identifies ${𝑋}^2\left({𝑋}^2+1\right)$ as the gcd. The handwritten explanation says: official def: 一直有定义的 ring 下使只要是 subring，且 $1$ 和 $0$ 也在 （事 $𝑋$ 的 multiplication 是 well-defined 的）$0_{𝑅}={\left[0\right]}_2=0_{𝑇}$.‘’

For the Bézout prompt the source writes that there must be $𝑓,𝑔\in \text{ℚ}\left[𝑋\right]$ with

$𝑓\left(2{𝑋}^2-10𝑋+12\right)+𝑔\left({𝑋}^2-3𝑋+2\right)=gcd\left(𝑓,𝑔\right)=𝑋-3.$

It also notes that $1,2,3,4$ are the only units of ${\text{ℤ}}_5\left[𝑋\right]$ (plug in 就好’‘) and factors

in ${\text{ℤ}}_5\left[𝑋\right]$ by checking roots $0,1,2,3,4$.

Source transcription — WS10, pp. 1–2, D. If $𝑓\in 𝐹\left[𝑋\right]$ has degree $2$ or $3$, then $𝑓$ is irreducible iff it has no root. The forward implication uses the factor theorem: irreducibility forbids a factor $𝑋-𝜆$ and so forbids $𝑓\left(𝜆\right)=0$. Conversely, if $𝑓=𝑔ℎ$ is nontrivial, degrees add in a field/domain. For degree $2$ or $3$, one factor must have degree $1$; writing that factor as $𝑎𝑋+𝑏$ yields the root $-\frac{𝑏}{𝑎}$.

The source then factors ${𝑋}^4-1$ in ${\text{ℤ}}_2\left[𝑋\right]$. It explicitly says that one must check whether ${𝑋}^2+1$ is irreducible; by the degree-$2$ criterion it has no root in ${\text{ℤ}}_2$, so

is the recorded factorization.

7.3 Congruence modulo a polynomial and quotient rings

Source transcription — WS10, p. 2, Part 3. For $𝑔,ℎ\in 𝐹\left[𝑋\right]$ define

The source calls ${\left[𝑔\right]}_{𝑓}$ the collection of all polynomials congruent to $𝑔$ modulo $𝑓$ and writes

It explicitly notes: 这里有证锣了，我们易证’‘ that congruence modulo $𝑓$ is an equivalence relation, $ℎ\in {\left[𝑔\right]}_{𝑓}\Rightarrow {\left[𝑔\right]}_{𝑓}={\left[ℎ\right]}_{𝑓}$, and distinct congruence classes are disjoint.

Source transcription — WS10, p. 2, F. Every class ${\left[𝑔\right]}_{𝑓}$ has a unique $ℎ\left(𝑋\right)\in 𝐹\left[𝑋\right]$ with $degℎ<deg𝑓$. Existence is by division. For uniqueness, if $𝑚=𝑟+𝑘𝑓$ with $𝑟$ the remainder, then $𝑘=1$ would make the degree of $𝑚$ equal to $deg𝑓>deg𝑟$, while $𝑘=-1$ gives the same degree obstruction; no other degree can make two different low-degree representatives congruent.

Source transcription — WS10, p. 3, G. Let $𝑓\in 𝐹\left[𝑋\right]$ have positive degree and put

$𝑅=\left\{{\left[𝑔\right]}_{𝑓}:𝑔\in 𝐹\left[𝑋\right]\right\}.$

The source defines ${\left[𝑔\right]}_{𝑓}+{\left[ℎ\right]}_{𝑓}={\left[𝑔+ℎ\right]}_{𝑓}$, ${\left[𝑔\right]}_{{𝑓\left[ℎ\right]}_{𝑓}}={\left[𝑔ℎ\right]}_{𝑓}$, $0_{𝑅}={\left[0\right]}_{𝑓}$, and $1_{𝑅}={\left[1\right]}_{𝑓}$, marking the operations well-defined’‘ and calling $𝑅$ a ring. For the example

$𝑅=\left\{{\left[𝑔\right]}_{{𝑋}^2}:𝑔\in {\text{ℤ}}_2\left[𝑋\right]\right\}$,

the page maps the four classes to ${\text{ℤ}}_2\times {\text{ℤ}}_2$: $\left[0\right]$ to $\left(0,0\right)$, $\left[1\right]$ to $\left(1,1\right)$, $\left[𝑋\right]$ to $\left(0,1\right)$, and $\left[1+𝑋\right]$ to $\left(1,0\right)$, and labels the map isomorphic to ${\text{ℤ}}_2\times {\text{ℤ}}_2$′‘.

8 Homework 3: rings, nilpotents, and homomorphisms

Personal finished homework transcription from 412-Hw-3-finished.pdf.

8.1 1. Subsets of $𝐹𝑢𝑛\left(ℝ,ℝ\right)$

Let $𝑅=𝐹𝑢𝑛\left(ℝ,ℝ\right)$ be the ring in exercise D2 of the “Ring Basics” adventure sheet. $0_{𝑅}$ and $1_{𝑅}$ are the constant functions zero and one. Show which of the following subsets of $𝑅$ are subrings of $𝑅$. If they are not subrings, show whether they are rings (with a different multiplicative identity than $1_{𝑅}$, but endowed with the same operations as in $𝑅$) or not.

(a) The set $𝐶$ of constant functions.

(b) The set $𝑆$ of those functions $𝑓$ such that $𝑓\left(𝑞\right)=0$ for any $𝑞\in \mathbb{Q}$.

(c) The set $𝑇$ consisting of $0_{𝑅}$, together with those functions with no zeros, or only a finite number of zeros. (A zero of a function $𝑓\in 𝑅$ is an element $𝑥\in ℝ$ such that $𝑓\left(𝑥\right)=0$.)

(a) $𝐶$ is a subring of $𝑅$.

Pf. Since $0_{𝑅}$ is $𝑓\left(𝑥\right)=0$ and $1_{𝑅}$ is $𝑓\left(𝑥\right)=1$, $0_{𝑅},1_{𝑅}\in 𝐶$. Let $𝑓,𝑔$ be two elements in $𝐶$, and suppose $𝑓\left(𝑥\right)=𝑎$, $𝑔\left(𝑥\right)=𝑏$ for some $𝑎,𝑏\in ℝ$. Then

so $𝐶$ is closed under addition. Also,

so $𝐶$ is closed under multiplication. Since $-𝑓\left(𝑥\right)=-𝑎$ is also a constant function, $-𝑓\in 𝐶$, so $𝐶$ is closed under additive inverse. Since $𝐶\subset 𝑅$ and $𝑅$ is a ring, by worksheet 3 it suffices to show these four facts. So $𝐶$ is a subring of $𝑅$.

(b) $𝑆$ is not a subring of $𝑅$. Since $1_{𝑅}$, the constant function $𝑓\left(𝑥\right)=1$, is not in $𝑆$ (for $𝑥\in \mathbb{Q}$, $𝑓\left(𝑥\right)\ne 0$), $𝑆$ violates the definition of subring. And $𝑆$ is not even a ring because it does not have a multiplicative identity.

To show this, assume there is a function $𝑓\in 𝑆$ such that for all $𝑔\in 𝑆$, $𝑓𝑔=𝑔𝑓=𝑔$. Take $𝑔\left(𝑥\right)=2$. Then, for any $𝑥\in ℝ$, $2𝑓\left(𝑥\right)=2$, so $𝑓\left(𝑥\right)=1$, which is not in $𝑆$. Thus $𝑆$ does not have a multiplicative identity; therefore it is not a ring.

(c) $𝑇$ is not a subring of $𝑅$, and not a ring. Consider $𝑓$ defined by $𝑓\left(𝑥\right)=1$ for $𝑥\ge 0$ and $𝑓\left(𝑥\right)=𝑥$ for $𝑥<0$; and $𝑔$ defined by $𝑔\left(𝑥\right)=-1$ for $𝑥\ge 0$ and $𝑔\left(𝑥\right)=𝑥$ for $𝑥<0$.