Topics:
Rings, Fields, The characteristic of a field, Review of polynomial rings, Factoring polynomials, Extension fields, The subring generated by a subset, The subfield generated by a subset, Construction of some extension fields, Stem fields, Algebraic and transcendental elements, Transcendental numbers
Brief review of basic notions (groups, rings, and fields):
(also download from here)

Characteristic of a field, "Freshman's Dream": $\left(a+ b\right)^p$ = $a^p$ + $b^p$

Division Algorithm (with proof); $xc \mid f(x)$ $\iff$ $f \left(a \right)$= 0

This includes items 1.11.4, 1.5, 1.6, 1.7, 1.9, 1.11
$F[x]$ is a PID; precise definition of UFD. If degree $f\left(x\right)\le$ 3 then $f\left(x \right)$ is irreducible $\iff f \left(x \right)$ has no roots; fails for degree 4
Discussion of irreducibility in $\mathbb {Z}[X]$ vs in $\mathbb{Q}[X]$:
(also download from here)

Testing for irreducibility in $\mathbb{Z}[X]$ by checking irreducibility in $\mathbb{F}_p[X]$; proof of why it works, pitfalls of misinterpreting, why $p$ must not divide leading coefficient. Explained what Milne means by "factors nontrivially" (not quite the same, over $\mathbb{Z}$, as being reducible). Examples, of a polynomial that can be shown to be irreducible by going mod 2 and of another polynomial, $X^410X^2+1$ in $\mathbb{Z}[X]$, that factors nontrivially modulo every prime, even though it is irreducible (proof of irreducibility assigned in Problem Set 1).

Finding a polynomial in $\mathbb{Z}[X]$ with $\sqrt 2+\sqrt 3$ as a root (the polynomial above, coincidentally).

Gauss' Lemma 1.13, proved with a more rigorous version of Milne's proof. (Choose appropriate positive rational multiples of $g(x)$ and $h(x)$ so as to minimize their integral product. then show that this minimal product must be 1.)

Vector spaces: Discussed how $F[X]$, and extension fields $K$ of $F$ are vector spaces over $F$.

Proposition 1.8: proved that for $R$ a PID, gcd's exist and can be written as a combination of the generators.
Discussed the Euclidean algorithm method of finding $d=\gcd(a,b)$ and r,s with $ra +sb=d$.

Stated and proved one case of the $\frac{2}{3}$ Lemma ("twoout of three lemma"): If $a,b,c,d\in R$, a commutative ring, with $d\ne 0$ and $a \pm b=c$, and two of a, b, c are multiples of d, then so is the third.
On Tuesday May 17 2016
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Discussed extension fields, notation, as on page 13 of Milne, including ``$F$homomorphism'', and examples.

We added the fact, which we proved later, that, if $\alpha$ and $\beta$ are roots of the same irreducible polynomial there exists
an $F$homomorphism from $F\big[ \alpha \big] $ to $F \big[ \beta \big]$

Proved Proposition 1.20 Multiplication of degrees, part 2. (Will go over part 1 in lecture 4 today.)

Proved Lemma 1.2.3: For $F\subseteq R$, where $F$ is a field and $R$ is an integral domain that has finite dimension over $F$ as a vector space, $R$ is a field.

Discussed constructing an extension field where an irreducible polynomial has a root. Proved it is a field.

Discussed examples such as building the complex numbers from the reals.

Proved the finiteness Theorem: For $F \subseteq E \subseteq K$ fields, these are equivalent:

$E/F$ is finite (vector space dimension).

$E/F$ is finitely generated as a field extension and is algebraic over $F$.

There exist algebraic elements $\alpha_1$, $\alpha_2$, $\dots$, $\alpha_n$ of $E$, such that $E=F(\alpha_1, \alpha_2,$ $\dots, \alpha_n)$.
On Wednesday May 18 2016
(also download from here)

Reviewed the Finitealgebraic Theorem from Lecture 5: For $F$ subseteq $E$ fields, these are equivalent:


$E/F$ is finite (i.e. finite vector space dimension).

$E/F$ is finitely generated as a field extension and $E$ is algebraic over $F$.

There exist algebraic elements $\alpha_1$, $\alpha_2$, $\dots$, $\alpha_n$ of $E$, such that $E=F(\alpha_1, \alpha_2, \dots,$ $\alpha_n)$.

Stated and proved the Corollary:
Let $K/F$ be a field extension and let $\overline F$ be the set of elements of $K$ that are algebraic over $F$. Then $\overline F$ is a subfield of $K$ with $F \subset \overline F \subset K$. The field $\overline F$ is called the algebraic closure of $F \in K$.

Discussed the set of algebraic numbers in the complex numbers $\mathbb {C}$ over the rational numbers $\mathbb {Q}$.

Showed that for $\alpha=\sqrt 2 + \sqrt 3$, $\mathbb{Q}(\alpha)= \mathbb{Q} (\sqrt2, \sqrt 3)$. Later showed that $\sqrt 3 \notin \mathbb{Q} (\sqrt2) $
and consequently that $ \big[ \mathbb{Q}(\alpha): \mathbb{Q} \big]=4$ (By Eisenstein $x^22$, and $x^23$ are irreducible over $\mathbb{Q}$ and so $ \sqrt 2, \sqrt 3 \notin \mathbb {Q}$).

Discussed an example of the Louisville numbers (with big gaps in the decimal expansion):
$\sum_{i=1}^\infty 1/10^{i!}$. These are transcendental and it is easier to show they are transcendental than other transcendental numbers (see book).

Discussed countable and uncountable cardinalities.

Stated and proved the Theorem :
For $K/F$ a field extension with $F$ countable, the set of elements of $K$ that are algebraic over $F$ is countable.

To prove this, stated and outlined/pictureproved $N$ lemmas:
For $X, X_n, Y$ sets

$X,Y$ countable $\implies X \times Y$ countable;

$X$ countable and an injection takes $Y$ into X $\implies Y$ is countable;

$X$ countable and a surjection takes $ X \to Y \implies Y$ countable;

$Y$ a countable union of countable sets $X_n \implies Y$ is countable;

$F$ a countable field and $V$ a vector space over $F$ with a countable basis $\implies V$ is countable;

$\vdots$
 $N$. If $X$ is countable and $Y \to X$ is countabletoone, then $Y$ is countable.

Corollary: In the complex numbers $\mathbb{C}$, the subfield of elements that are algebraic over the rationals $\mathbb{Q}$ is countable. Elements of this set are called algebraic numbers.

Proved part $1$ of Proposition~1.20 Multiplication of degree (part 2 was proved in Lecture 5).

Proved the following tw statements:
Proposition: (algebraic extensions are transitive): If $F \subseteq K \subseteq L$ are fields with $K/F$ algebraic and $L/K$ algebraic, then $L/F$ is algebraic.
Proposition:
If $F\subseteq K \subseteq L$ are fields and $L=F[\alpha]$ where alpha is algebraic, then the minimal polynomial for $\alpha$ over $K$ divides the minimal polynomial
for $\alpha$ over $F$.

Worked two problems:

Show if $[L:F]=p$, a prime integer, then there are no fields $K$ properly between $F$ and $L$.
 If $F \subseteq L$ is algebraic and $T$ is an integral domain with $F \subseteq T \subseteq L$, prove that $T$ is a field.
Another problem the students might like to try for extra practice is Exercise~15 in
Milne.