@Thorgott let me throw you some numbers

actually let me not

lol

@Balarka without knowing how $\phi$ is defined for the irrationals of the subfield $\mathbb{F}$ idk how to argue $r \in \mathbb{F}$ is a root in $f$ iff $\phi(r)$ is a root in $f$. It's clear as day why it is for $r \in \mathbb{Q}$

wait yeah

idk why i deleted that: can $r$ get mapped by $\phi$ if $r\notin\mathbb{Q}$

think of this as similar to the det problem. you applied the det, like degree, on some expression. what happens if you replace det by $\phi$ and the expression by $f(r) = a_n r^n + a_{n-1} r^{n-1} + \cdots + a_1 r + a_0$

what kind of a polynomial is $f(x)$?

@Thorgott yeah, that’s pretty much how I solved it using homogenous coordinates. However, my objective is to apply the definitions of gluing of prevarieties to solve this. Since I’ll have to glue schemes someday, I can’t flee from this forever. 💀

ok so it leaves $f(x)$ unchanged on the right so if $f(r) = 0$ then $\phi(f(r))=0$

write the expression for $\phi(f(r))$ in full like i did for $f(r)$

appeal to amnesia for a moment and forget $f(r)$ is actually $0$

wrong

wait yeah what am I doing

i mean, this is a consequence of the defn.

$\phi(f(r))=\phi(a_n r^n)+\phi(a_{n-1}r^{n-1})+\cdots+\phi(a_1 r)+\phi(a_0)$

ok, simplify more

but i guess you want sth like: a regular function $X\rightarrow U_1\cup_{U_0}U_2$ is the same as a choice of two open subsets $V_1,V_2\subseteq X$ and regular functions $f_i\colon V_i\rightarrow U_i,\,i=1,2$ s.t. $f_1^{-1}(U_0)=f_2^{-1}(U_1)$ and $f_1,f_2$ agree on this set

since $a_n \in \mathbb{Q}$ we have $\phi(f(r))=a_n\phi(r^n)+a_{n-1}\phi(r^{n-1})+\cdots+a_1\phi(r)+a_0$

simplify more

$\phi(f(r)) = \phi(r)(a_n\phi(r^{n-1})+a_{n-1}\phi(r^{n-2})+\cdots+a_1)+a_0$?

simplify more

this is becoming the anakin meme

i.imgflip.com/38hm3e.png?a474288

im serious though

Do I just keep pulling out $\phi(r)$'s lmao Idk how sire

sure why not

at the end what do you have

@EE18 I would just say $A(X-X_0)=0$, so $X-X_0=U$ is a solution of the homogeneous equation.

it doesn't guarantee $\phi(r)$ is a root though

just do it yo

youll see at the end

I'm not doing it right.. I'm getting $\phi(f(r))=\phi(r^2)(a_n\phi(r^{n-2}+a_{n-1}\phi(r^{n-3})+\cdots+a_2)+a_1+a_0$

@Obliv Write out an example explicitly. Suppose $f(x)=x^2-2$. What are $\phi(f(r))$ and $f(\phi(r))$?

@Obliv thats not right

$\phi(f(r))=x^2-2$

you said you'll pull out $\phi(r)$ one by one

@Obliv Where did $r$ go?

$f(r)=0$, presumably.

@BalarkaSen $\phi(f(r))=\phi(r)(\phi(r)(a_n\phi(r^{n-2})+a_{n-1}\phi(r^{n-3})+\cdots+a_2)+a_1)+a_0$?

yes! now tell me what youll get at the end

feels like I'm tying shoelaces

Obliv, I gave you a simple polynomial to do explicitly.

I'M TRYING TO LISTEN TO BOTH OF U HOLD ON

You're lost in the symbols with Balarka's general case.

Do a simple example first so you understand; then do the general case.

i have faith in obliv

OK, I withdraw.

@TedShifrin $\phi(f(r)) = r^2-2$.. i'm lost even in this simple example now

I wanna do the simple case first. $r \in \mathbb{F}$ but $f \in \mathbb{Q}[x]$ so $\phi(f(r))$ for $f(x) = x^2-2$ should fix $f(r)$ if $f(r)$ is a rational?

since $\phi(c) = c$ for $c \in \mathbb{Q}$

oh you're saying $f(r) = 0$ then $\phi(f(r)) = 0 = f(r)$ since $0 \in \mathbb{Q}$

those are all correct observations but doesn't help you prove $\phi(r)$ is a root of $f = 0$ :)

try to do the general expression business we were doing above with this special case

so... $x^2 - 2 = (x+\sqrt{2})(x-\sqrt{2})$ and apply $\phi$ to that where $r = \pm \sqrt{2}$

we didnt factorize above at all

not sure why you're doing that now

$f(r) = r^2 - 2$, written in full. write out $\phi(f(r))$ in full. just like the det problem, apply $\phi$ to this.

we're mixing up notation I think

yes, r is meant to be a root. so what?

my bad. ok so $\phi(f(r)) = \phi(r^2)-\phi(2)$

is it really that productive

simplify as much as you can

$= \phi(r)\phi(r)-2$

or, $\phi(r)^2 - 2$

since it's automorphism I can write that (totally meant to do that)

which you can do in the above general expression also, instead of pulling out $\phi(r)$ one by one :)

anyway, so now come out of the amnesia stupor and remember $r$ is a root of $x^2 - 2$

degree has to be 0 since $\text{deg}(f(r))=0=\text{deg}(\phi(r)^2)-\text{deg}(2)$

what is degree? these are elements of a field

maybe you should think about it, without Balarka's help, for a bit. And then come back

you have proved $\phi(r^2 - 2) = \phi(r)^2 - 2$. You know $r$ is a root of $x^2 - 2$. That is, $r^2 - 2 = 0$. What does this say about $\phi(r)$?

Nope. How?

I think that if you ponder it for yourself a bit then you'll get better view of the whole picture

i agree, but I don't wanna let balarka down lol, i'm pretty washed right now

i dont mind if you go and think about it for a bit. its better to be slow and understand better

oh $\phi(x^2-2)$ with $r$ a root means $\phi(r^2-2) = \phi(0)$

and $\phi(r)^2-2=\phi(0)$

continue

$\phi(r)^2-\phi(2)=\phi(0) \implies$ $\phi(x)$ has roots for $f\in\mathbb{Q}[x]$ that are equivalent to roots $x$ in $f(x)$

The polynomial was $f(x) = x^2 - 2$, not $\phi$.

Simplify $\phi(r)^2 - \phi(2) = \phi(0)$

$\phi(r)^2=2 \implies \phi(r) = \pm\sqrt{2}$

Right. $\phi(r)$ is a root of $f$.

You have proved $r$ is a root of $f$ implies $\phi(r)$ is a root of $f$.

Now do the general thing

using the fact that $\phi$ is a hom. we do $\phi(f(r)) = a_n\phi(r)^n+a_{n-1}\phi(r)^{n-1}+\cdots+a_1\phi(r)+a_0$

Yes, thank you, much better :)

since $f(r) = 0$ we have $\phi(0) = 0 = a_n\phi(r)^n+a_{n-1}\phi(r)^{n-1}+\cdots+a_1\phi(r)+a_0$

terrific

so we can just treat $\phi(r)$ as the new indeterminate $x$? :O

cool, thank you @balarka that was illuminating if not embarrassing

it means if you plug $x = \phi(r)$ in place of the indeterminate in $f(x)$, you get zero.

at $r=0$ since $\phi(f(r)) = \phi(0)=0$

not at $r = 0$. $r$ is some root of $f(x)$. $f(r) = 0$, rather.

oh right

Yep! And that's true more generally, $f(\phi(a)) = \phi(f(a))$ even if $a$ is not a root of $f$ or anything.

For all $a \in \Bbb F$

Right, idk why i deleted it i literally just showed it

Just write out the expression on both sides and use $\phi$ is a field homomorphism.

Yes, same proof.

oh right it just has to be homomorphism. but for a polynomial with field coefficients it has to fix the coefficients I think?

yes, correct.

otherwise you don't have the same roots/polynomials

indeed true

proving $\mathbb{Q}[\sqrt{2}]=\{a+b\sqrt{2}\mid a,b \in \mathbb{Q}\}$ is a subfield of $\mathbb{C}$ amounts to just showing closure in subtraction & mult. proving that $\phi:\mathbb{Q}[\sqrt{2}]\to\mathbb{Q}[\sqrt{2}]$ defined by $\phi(a+b\sqrt{2})=a-b\sqrt{2}$ is an automorphism of $\mathbb{Q}[\sqrt{2}]$ just means demonstrating it respects addition&mult. and is bijective. It also mentions one can show using the previous problem (that we just did) the only automorphisms of

$\mathbb{Q}[\sqrt{2}]$ is $\phi$ and the identity map

Idunno how that follows from the last problem when we were doing polynomials and these are just numbers, but it's ok it's not part of the problem to show that anyway

That's all the practice exam problems. Welp i guess time to read & review hw problems till tomorrow

def need to work on the isomorphisms and theorems associated with them

obliv: a homomorphism on Q(sqrt(2)) is determined by its values on 1 and sqrt(2). where 1 goes is determined by the definition of 'homomorphism' (and implies that the homomorphism fixes Q), and [by the exercise you did] there are only two possibilities for where sqrt(2) goes