ROUGH NOTES
Lec-21: An ascending chain $I _1 \subseteq I _2 \subseteq \ldots$ of ideals stabilizes if $I _m = I _{m+1} = \ldots$ for some $m$; Ring $R$ is Noetherian (ie every ideal is finitely generated) if and only if every ascending chain of ideals stabilizes
$\implies$ Let ideals $I _1 \subseteq I _2 \subseteq \ldots.$ Now $\cup I _j$ is an ideal, and of the form $(f _1, \ldots, f _r).$ Taking $f _1 \in I _{n _1}, \ldots, f _r \in I _{n _r}$ and $N := \max n _j$ we see $I _N, I _{N+1}, \ldots$ are all $\cup I _j.$
$\impliedby$ Suppose ideal $I$ is not finitely generated. Pick $x _1 \in I.$ As $I \neq (x _1),$ pick $x _2 \in I \setminus \lbrace x _1 \rbrace.$ As $I \neq (x _1, x _2)$, pick $x _3 \in I \setminus \lbrace x _1, x _2 \rbrace,$ and so on. Now $(x _1) \subsetneq (x _1, x _2) \subsetneq (x _1, x _2, x _3) \subsetneq \ldots,$ absurd.
$\mathscr{C}(\mathbb{R}) = \lbrace \text{continuous maps } f : \mathbb{R} \to \mathbb{R} \rbrace$ isnt Noetherian : For any $A \subseteq \mathbb{R}$, let $Z(A) = \lbrace f \in \mathscr{C}(\mathbb{R}) : f \vert _{A} = 0 \rbrace.$ Now ${Z([1, \infty)) \subsetneq Z([2, \infty)) \subsetneq \ldots}$
Subring of a Noetherian ring can be Non-Noetherian. Eg: $\mathbb{R}[x _1, x _2, \ldots ]$ inside its field of fractions; Quotients of Noetherian rings are Noetherian (from correspondence thm);
Lec-22: [Hilbert Basis Thm] If $R$ is Noetherian, so is $R[x]$
Pf: Suppose ideal $I \subseteq R[x]$ isnt finitely generated. As $I \neq (0),$ pick an $f _1$ of least degree (among nonzero elems) in $I.$ As $I \neq (f _1)$, pick an $f _2$ of least degree in $I \setminus (f _1).$ As $I \neq (f _1, f _2),$ pick an $f _3$ of least degree in $I \setminus (f _1, f _2),$ and so on.
Especially $\underbrace{\deg(f _1)} _{d _1} \leq \underbrace{\deg(f _2)} _{d _2} \leq \ldots$
Let $a _i := \text{lc}(f _i)$ be the leading coeffs. From the chain $(a _1) \subseteq (a _1, a _2) \subseteq \ldots$, we have ${(a _1, \ldots, a _N) = \cup _{j \geq 1} (a _1, \ldots, a _j) = (a _1, a _2, \ldots)}$ for some $N.$
Now $a _{N+1} \in (a _1 , \ldots, a _N),$ so $a _{N+1} = \sum _{1} ^{N} b _j a _j.$
Consider ${g := b _1 (f _1 x ^{d _{N+1} - d _1}) + \ldots + b _{N} (f _{N} x ^{d _{N+1} - d _N})}$ : It has degree $d _{N+1},$ leading coeff $a _{N+1},$ and is in $(f _1, \ldots, f _{N}).$
So $g - f _{N+1}$ has degree $\lt d _{N+1},$ and is $\not\in (f _1, \ldots, f _{N}).$ Absurd.
Lec-23: [For integral domains. Also exclude 0 from consideration] Divisors; Proper divisors $(b$ is a proper divisor of $a$ if for some $c,$ $a = bc$ and neither of $b,c$ are units); Associates $(a,b$ are associates if $a = bc$ for some unit $c$); Irreducible elem (nonunit with no proper divisors); $a$ is called prime if $(a)$ is a prime ideal.
In ideal language (after discarding $0$) : “$a$ is a unit” is $(a) = (1)$; “$b$ divides $a$” is $(a) \subseteq (b)$; “$b$ is a proper divisor of $a$” is $(a) \subsetneq (b) \subsetneq (1)$; “$a,b$ are associates” is $(a) = (b)$; “$a$ is irreducible” is “$(a) \subsetneq (1)$ and there is no $b$ with $(a) \subsetneq (b) \subsetneq (1)$”; “$a$ is prime” is “$(a)$ is prime”.
Eg: In $\mathbb{Z}$, units are $\lbrace \pm 1 \rbrace,$ primes and irreducibles are both $\lbrace \pm 2, \pm 3, \pm 5, \ldots \rbrace.$
$2 \in \mathbb{Z}[\sqrt{-5}]$ is irreducible but not prime.
Irreducibility : Say $a+b\sqrt{-5}$ divides $2$ in $\mathbb{Z}[\sqrt{-5}],$ ie $2 = (a+b\sqrt{-5})x$ for some $x$ in the ring. Taking $\vert \ldots \vert ^{2},$ we see $a ^2 + 5b ^2$ divides $4$ in the usual sense, ie $a ^2 + 5b ^2 = 1 \text{ or } 2 \text{ or } 4.$ Casework gives $(a,b) = (\pm 1, 0), (\pm 2, 0)$ are the only solutions. So $2$ has no proper divisors in $\mathbb{Z}[\sqrt{-5}].$ Its also not a unit (direct check).
Primality : $(1+\sqrt{-5}), (1-\sqrt{-5}) \not\in (2),$ but their product is in $(2).$
Lec-24: In an integral domain primes are irreducible : Let prime $p = ab.$ Then $p \vert a$ or $p \vert b.$
If $p \vert a$ : $a = px$ gives $p = (px)b,$ so $b$ is a unit.
If $p \vert b$ : $b = py$ gives $p = a(py),$ so $a$ is a unit.
So one of the factors must be a unit, ie $p$ is irreducible.
In an integral domain, $\gcd(a,b)$ is a $d$ with : $(d \vert a, d \vert b),$ and $(e \vert a, e \vert b \implies e \vert d).$ This can be rewritten as “$(d) \supseteq (a) + (b),$ and $(e) \supseteq (a)+(b) \implies (e) \supseteq (d)$”.
So $\text{gcd}(a,b)$ is considered only upto the ideal generated by it (ie upto associates).
GCD may not exist : Take $a = x ^5, b = x ^6$ in ${R = \lbrace \text{real polys without }x \text{ term}\rbrace}.$ The divisors of $x ^5$ are $1, x ^2, x ^3, x ^5$ while those of $x ^6$ are $1, x ^2, x ^3, x ^4, x ^6.$ The common divisors are $\lbrace 1, x ^2, x ^3 \rbrace.$ But none of these three can act as the GCD.
Lec-25: PIDs; $\mathbb{Z}[x]$ isnt a PID. Eg ideal $(2,x)$. If it were principal, say $(2,x) = (f),$ $f$ must be a constant (else $2$ isnt a multiple of $f$), nonzero nonunit. But then all elements of $(2,x)$ have coeffs divisible by this constant, absurd; $k[x _1, x _2]$ isnt a PID. Eg ideal $(x _1, x _2);$
In a PID, $\gcd(a,b)$ exists for all $a,b$ : Ideal $(a) + (b) = (d).$ Now $d$ is gcd.
In a PID, irreducibles are primes too : Let $a$ be an irreducible (esp $(a) \neq (1)$). There is no $b$ with $(a) \subsetneq (b) \subsetneq (1),$ and since all ideals are of the form $(p),$ we see $(a)$ is maximal. Hence prime.
Alt: Suppose $xy \in (a).$ Consider gcd $(a,x) = (d).$ Since this divides $a$ and $a$ is irred, $(d) = (1) \text{ or } (a).$
If former : $ra + sx = 1,$ so $y = r{\color{purple}{a}}y + s{\color{purple}{xy}} \in (a).$
If latter : $x \in (a,x) = (a).$
In a PID, nonzero prime ideals are maximal too : Let $\mathfrak{p} \neq 0$ be a prime ideal. Its $(p)$ for some $p.$ As primes are irreducible, there is no $b$ with $(p) \subsetneq (b) \subsetneq (1),$ and as any ideal looks like $(f),$ we see $(p)$ is maximal. That is, $\mathfrak{p}$ is maximal.
Lec-26: [Factoring] Let $R$ be an integral domain, and $a \in R$ (as usual, nonzero nonunit). If $a$ is irred, stop. Else write $a = a _1 b _1$ for nonunits $a _1, b _1.$ If both $a _1, b _1$ are irred, stop, else continue onto them.
This needn’t terminate : Eg In $\mathbb{Z}[2 ^{\frac{1}{2}}, 2 ^{\frac{1}{4}}, 2 ^{\frac{1}{8}}, \ldots],$ we have $2 = 2 ^{\frac{1}{2}} 2 ^{\frac{1}{2}} = (2 ^{\frac{1}{4}}) ^4 = \ldots$ so on.
An elem $a$ has a unique factorisation if it factors into a product of irreducibles, and any two factorisations of $a$ (into product of irreducibles) are same upto associates and ordering. Eg: In $\mathbb{Z}[i],$ $5 = (2+i)(2-i) = (1-2i)(1+2i)$ are same factorisations.
UFD (ie integral domain where factoring terminates and irred factorisation is unique, for every elem [nonzero nonunit implicit]);
In an integral domain $R$, factoring terminates if and only if ${(a _1) \subsetneq (a _2) \subsetneq \ldots}$ can’t happen
$\implies$ Suppose $(a _1) \subsetneq (a _2) \subsetneq \ldots$ Then $a _1 = a _2 x _2 = a _3 x _3 x _2 = \ldots$ is a factoring process that doesnt terminate, absurd.
$\impliedby$ Suppose $a = a_1 b _1 = a_2 b _2 b _1 = a _2 a _3 b _3 b _1 = \ldots$ is a factoring process which doesnt terminate. Then $(a) \subsetneq (a _1) \subsetneq (a _2) \subsetneq \ldots,$ absurd.
Even if factorisation exists it may not be unique : In $\mathbb{Z}[\sqrt{-5}],$ ${6 = 2 \cdot 3 = (1-\sqrt{-5})(1+\sqrt{-5}).}$ [These are distinct factorisations : All factors are irreducible, and neither is associated to another. Latter is because units in $\mathbb{Z}[\sqrt{-5}]$ are $\lbrace \pm 1 \rbrace$]
So $\mathbb{Z}[\sqrt{-5}]$ isnt a UFD.
Lec-27: An integral domain $R$ where factoring terminates is a UFD if and only if every irred elem is prime.
$\implies$ Suppose $a$ is an irred which isnt prime. So there are $x,y$ with $a\nmid x, a \nmid y$ but $a \mid xy.$ Taking $xy = ar$ and factoring both sides, we see $a$ appears on RHS but not on LHS. Contradicts uniqueness.
$\impliedby$ We need only show uniqueness. Say $a = p _1 \ldots p _k = q _1 \ldots q _l$ are factorisations into irreds. As $p _1$ is prime, it divides some $q _i$ on the right, wlog $q _1.$ Since primes are irred, $p _1 = q _1$ upto association. On cancelling them and continuing onto other terms, we see the two factorisations are the same.
PIDs are UFDs too : PIDs are Noetherian, so factoring terminates. Since irreds are primes, factorisations are unique.
Eg: $k[x], \mathbb{Z}$; $\mathbb{Z}[i]$ is a PID too (Proof is as in $k[x]$, but using Euclidean division with size function $\sigma (a+ib) = a ^2 + b ^2$).
$\mathbb{Z}[x]$ isnt a PID, but turns out is a UFD.
Lec-28: $\mathbb{Z}[x]$ is a UFD (Similar proof shows $R$ UFD $\implies$ $R[x]$ UFD) :
An $f \in \mathbb{Z}[x]$ with positive deg, positive leading coeff and gcd(coeffs) = 1, is called a primitive polynomial.
An $f \in \mathbb{Z}[x]$ with positive deg and positive leading coeff is primitive if and only if it isnt divisible by any prime integer, ie if and only if $\varphi _p (f) \neq 0$ for every prime $p,$ where ${\mathbb{Z}[x] \overset{\varphi _p}{\to} (\mathbb{Z}/{p\mathbb{Z}}) [x]}$ sends ${ a_0 + \ldots + a _n x ^n \mapsto [a _0] + \ldots + [a _n] x ^n.}$
[Gauss Lemma] Product of primitive polynomials $f,g \in \mathbb{Z}[x]$ is primitive.
Pf: Degree and leading coeff of $fg$ is $\gt 0.$ Also for every prime $p,$ ${\varphi _p(fg) = \varphi _p (f) \varphi _p (g) \neq 0},$ as needed.
Let $f \in \mathbb{Q}[x]$ of positive deg. Then $f$ is uniquely written as $c f _0$ where $c \in \mathbb{Q}$ and $f _0$ is primitive. Further ${c \in \mathbb{Z} \iff f \in \mathbb{Z}[x].}$
Pf: [Existence] Clear denominators, to get $df = f _1 (\in \mathbb{Z}[x]).$ Pull out the gcd of coeffs of $f _1$, to get $df = e f _0,$ ie $f = \frac{e}{d} f _0.$
[Uniqueness] Say $c f _0 = d g _0$ are two such expressions. Clearing denominators, $(Nc) f _0 = (Nd) g _0$ with $Nc, Nd \in \mathbb{Z}.$ Since $Nc$ is gcd of coeffs of LHS and $Nd$ is gcd of coeffs RHS, and coeffs are equal, we get $Nc = \pm Nd.$ So $c=d.$ On cancelling, ${\lbrace c = d, f _0 = g _0\rbrace},$ as needed.
Here $c$ is called the content of $f.$
So $\mathbb{Z}[x]$ is just $\mathbb{Z}$ along with polys in $\mathbb{Q}[x]$ with integer content.
Let $f,g \in \mathbb{Z}[x]$ and $f$ primitive. If $g \in f\mathbb{Q}[x],$ then $g \in f\mathbb{Z}[x].$
Pf: [For $g = 0,$ trivial, so we take $g \neq 0$] Say $g = fh$ with $h \in \mathbb{Q}[x].$ Pulling out the contents, $c _g g _0 = f(c _h h _0)$ with $c _g \in \mathbb{Z}.$ As $f$ is primitive, $f h _0$ is the primitive part of RHS. So contents $c _h = c _g (\in \mathbb{Z}),$ ie $h \in \mathbb{Z}[x],$ as needed.
Let $f \in \mathbb{Z}[x]$ be irred with positive leading coeff. If $\deg(f) = 0,$ $f$ is a prime integer, and if $\deg(f) \gt 0,$ its a primitive poly irred over $\mathbb{Q}.$
Pf: Degree $0$ case is clear, so let $\deg(f) \gt 0.$ Pulling out the content, $f = c _0 f _0.$ Irreducibility over $\mathbb{Z}$ ensures $c _0 = 1,$ ie $f$ is primitive.
Suppose $f$ isnt irreducible over $\mathbb{Q},$ ie $f = gh$ for positive deg polys $g,h \in \mathbb{Q}[x].$ Pulling out the contents, $f = (c _g g _0)(c _h h _0).$ As $g _0 h _0$ is primitive part of RHS, $c _g c _h = 1$ and $f = g _0 h _0.$ The latter contradicts irreducibility over $\mathbb{Z},$ absurd.
$2+2x$ is irred in $\mathbb{Q}[x]$ but reduces as $2 \cdot (1+x)$ in $\mathbb{Z}[x].$
[Every primitive poly $f$ irred in $\mathbb{Q}[x]$ is irred in $\mathbb{Z}[x]$ : Primitivity ensures it cant be break as (deg 0 non-unit) times (positive deg non-unit). Irreducibility in $\mathbb{Q}[x]$ ensures it cant break as product of two positive deg non-units]
$\mathbb{Z}[x]$ is a UFD : Since its Noetherian, factoring terminates. To show factorisations are unique, suffices to show irreducibles are prime.
Degree $0$ irreds are prime integers, so prime in $\mathbb{Z}[x].$ Positive degree irreds are (taking leading coeffs positive) primitives $f \in \mathbb{Z}[x]$ irred over $\mathbb{Q}.$ These $f$ are prime in $\mathbb{Z}[x].$
Say $gh \in f\mathbb{Z}[x]$ for some $g, h \in \mathbb{Z}[x].$ So $gh \in f\mathbb{Q}[x].$ As $f$ is irred (hence prime) in $\mathbb{Q}[x],$ we have $g \in f\mathbb{Q}[x]$ or $h \in f\mathbb{Q}[x].$ As $f$ is primitive, $g \in f\mathbb{Z}[x]$ or $h \in f\mathbb{Z}[x],$ as needed.
Lec-29: [Eisenstein Criterion] Let $f = a _n x ^n + \ldots + a _0 \in \mathbb{Z}[x]$ with $a _n \neq 0,$ and $p$ a prime such that $p \nmid a _n,$ $p$ divides all of $a _0, \ldots, a _{n-1},$ and $p ^2 \nmid a _0.$ Then $f$ is irred in $\mathbb{Q}[x].$ (If it’s also primitive, it’s irred in $\mathbb{Z}[x]$)
Pf: Suppose $f = gh$ is a reduction in $\mathbb{Q}[x].$ Pulling out contents, $c _f f _0 = c _g g _0 c _h h _0,$ and equating primitive parts $f _0 = g _0 h _0.$ So $f = {\color{red}{c _f g _0}} {\color{purple}{h _0}}$ is a reduction in $\mathbb{Z}[x]$ with positive deg factors.
From the mod $p$ homom $\mathbb{Z}[x] \twoheadrightarrow (\mathbb{Z}/{p\mathbb{Z}})[x],$ we see $[a _n]x ^n = [c _f g _0][h _0]$ & $[a _n] \neq 0.$
Using unique factorisation in $(\mathbb{Z}/{p\mathbb{Z}})[x],$ we must have ${\lbrace [c _f g _0] = [b _1] x ^k ; [h _0] = [b _2] x ^l\rbrace}$ with $k,l \geq 0.$
In fact $k,l \gt 0$ (Because $l \leq \deg(h _0) \lt n.$ Similar for $k$).
So constant terms of both $c _f g _0$ and $h _0$ are divisible by $p,$ giving $p ^2 \mid a _0,$ absurd.