ring

[logistic_guy]

here is the question

Let \(\displaystyle K/F\) be an algebraic extension and let \(\displaystyle R\) be a ring contained in \(\displaystyle K\) and containing \(\displaystyle F\). Show that \(\displaystyle R\) is a subfield of \(\displaystyle K\) containing \(\displaystyle F\).


my attemb
do i have to assume \(\displaystyle R\) is subring of \(\displaystyle K\)
or it's by default a subring
my problem is i don't know how to show \(\displaystyle R\) is closed on inverses

 

[fresh_42]

You have to use that it is an algebraic extension. Select an element [imath] r\in R\setminus F. [/imath] What does it mean that [imath] r [/imath] is algebraic over [imath] F\,? [/imath]

 

[logistic_guy]

You have to use that it is an algebraic extension. Select an element [imath] r\in R\setminus F. [/imath] What does it mean that [imath] r [/imath] is algebraic over [imath] F\,? [/imath]

i think it mean there's polynomial with coeffcient in \(\displaystyle F\) which have \(\displaystyle r\) as a root
and \(\displaystyle r\) isn't necessary in \(\displaystyle F\) as this expression say \(\displaystyle r\in R\setminus F. \)
\(\displaystyle r\) is an element of \(\displaystyle R\), not \(\displaystyle F\)

 

[fresh_42]

i think it mean there's polynomial with coeffcient in \(\displaystyle F\) which have \(\displaystyle r\) as a root
and \(\displaystyle r\) isn't necessary in \(\displaystyle F\) as this expression say \(\displaystyle r\in R\setminus F. \)
\(\displaystyle r\) is an element of \(\displaystyle R\), not \(\displaystyle F\)

Correct. But let us move forward step by step. We need every single condition that is given to us.

We now know that there is a polynomial [imath] p(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots +a_2x^2+a_1x+a_0\in F[x] [/imath] such that [imath] p(r)=0. [/imath] We may also assume that this polynomial is one of the lowest degree among all polynomials [imath] \overline{p}(x) [/imath] with [imath] \overline{p}(r)=0. [/imath] One of them must have a lowest degree since the degrees of polynomials are bounded from below, by [imath] 0, [/imath] or [imath] 1 [/imath] in our case.

Step 2: If [imath] a_0=0 [/imath] then [imath] p(r)=r\cdot \underbrace{(a_nr^{n-1}+a_{n-1}r^{n-2}+\ldots +a_2r+a_1)}_{q(r)}=0 .[/imath]

Question: Does [imath] R [/imath] have zero divisors other than [imath] 0 [/imath], and if not, then why?
A zero divisor is a number [imath] a [/imath] such that there is a number [imath] b\neq 0 [/imath] with [imath] a\cdot b=0. [/imath] Your watch has zero divisors: [imath] 4 [/imath] times [imath] 3 [/imath] sets the big hand back to [imath]12= 0. [/imath]

What does this mean for [imath] q(r) [/imath] and its degree?

Step 3: If [imath] a_0\neq 0, [/imath] then what does that mean for [imath] p(r)=0=r\cdot q(r)+a_0 \;?[/imath]

 

[logistic_guy]

Correct. But let us move forward step by step. We need every single condition that is given to us.

We now know that there is a polynomial [imath] p(x)=a_nx^n+a_{n-1}x^{n-1}+\ldots +a_2x^2+a_1x+a_0\in F[x] [/imath] such that [imath] p(r)=0. [/imath] We may also assume that this polynomial is one of the lowest degree among all polynomials [imath] \overline{p}(x) [/imath] with [imath] \overline{p}(r)=0. [/imath] One of them must have a lowest degree since the degrees of polynomials are bounded from below, by [imath] 0, [/imath] or [imath] 1 [/imath] in our case.

Step 2: If [imath] a_0=0 [/imath] then [imath] p(r)=r\cdot \underbrace{(a_nr^{n-1}+a_{n-1}r^{n-2}+\ldots +a_2r+a_1)}_{q(r)}=0 .[/imath]

Question: Does [imath] R [/imath] have zero divisors other than [imath] 0 [/imath], and if not, then why?
A zero divisor is a number [imath] a [/imath] such that there is a number [imath] b\neq 0 [/imath] with [imath] a\cdot b=0. [/imath] Your watch has zero divisors: [imath] 4 [/imath] times [imath] 3 [/imath] sets the big hand back to [imath]12= 0. [/imath]

What does this mean for [imath] q(r) [/imath] and its degree?

Step 3: If [imath] a_0\neq 0, [/imath] then what does that mean for [imath] p(r)=0=r\cdot q(r)+a_0 \;?[/imath]

you explain it nicely fresh_42 and i hope i can answer your questions

Question: Does [imath] R [/imath] have zero divisors other than [imath] 0 [/imath], and if not, then why?

if \(\displaystyle R\) is a subfield of \(\displaystyle K\) it won't have zero divisors other than \(\displaystyle 0\)
why? because a subfield by definition can't have zero divisors other than \(\displaystyle 0\)

What does this mean for [imath] q(r) [/imath] and its degree?

i don't knowbut i think the degree of \(\displaystyle q(r)\) is less than \(\displaystyle p(r)\)

Step 3: If [imath] a_0\neq 0, [/imath] then what does that mean for [imath] p(r)=0=r\cdot q(r)+a_0 \;?[/imath]

i think it mean the relation between \(\displaystyle r\) and \(\displaystyle q(r)\) won't be obvious because \(\displaystyle rq(r)\) will be a constant

 

[fresh_42]

if \(\displaystyle R\) is a subfield of \(\displaystyle K\) it won't have zero divisors other than \(\displaystyle 0\)
why? because a subfield by definition can't have zero divisors other than \(\displaystyle 0\)

Correct.

i don't knowbut i think the degree of \(\displaystyle q(r)\) is less than \(\displaystyle p(r)\)

Correct. We have [imath] p(r)=r\cdot q(r) =0 .[/imath] No zero divisors means [imath] r=0 [/imath] or [imath] q(r)=0. [/imath] The former is ruled out. Either by the fact that [imath] r=0 [/imath] doesn't need an inverse, so we only consider [imath] r\neq 0 [/imath] or by choosing [imath] r\in R\setminus F [/imath] since the elements of [imath] F [/imath] already have an inverse. Moreover, [imath] \deg p(x)=n> n-1=\deg q(x) [/imath] means that [imath] q(r)\neq 0 [/imath] since [imath] p(x) [/imath] was already of minimal possible degree among the polynomials with that property. However, if [imath] r\neq 0 \neq q(r) [/imath] then [imath] 0=p(r)= r\cdot q(r)\neq 0 [/imath] is a contradiction. This means that our assumption [imath] a_0=0 [/imath] was wrong. From here on, we know that [imath] a_0\in F\setminus \{0\} [/imath] and that [imath] a_0^{-1}\in F [/imath] exists since [imath] F [/imath] is a field.

i think it mean the relation between \(\displaystyle r\) and \(\displaystyle q(r)\) won't be obvious because \(\displaystyle rq(r)\) will be a constant

Take what I have written! We now have [imath] p(r)=0=r\cdot q(r) +a_0 [/imath] and therefore
[math] 1= a_0\cdot a_0^{-1} = (-r\cdot q(r))\cdot a_0^{-1}= r\cdot (-a_0^{-1}q(r)).[/math]
Finally, [imath] -a_0^{-1}q(r) [/imath] has coefficients in [imath] F\subseteq R [/imath] and powers of [imath] r\in R [/imath] which means that [imath] \overline{r}:=-a_0^{-1}q(r) \in R[/imath] and [imath] r\cdot \overline{r}=1. [/imath] Inverse element found!


All your arguments were correct and present. You only needed to pack them together, do a bit of algebra and write "proof" at it.

 

[logistic_guy]

Correct.
All your arguments were correct and present. You only needed to pack them together, do a bit of algebra and write "proof" at it.

thank fresh_42 very much

do you think i'll master ring theory in no time by this rate of understanding?

 

[fresh_42]

do you think i'll master ring theory in no time by this rate of understanding?

This is hard to impossible to tell from a distance. Your answers here were way better than in some other threads.

I gave you some advice somewhere hidden in a seeming joke: Nike: Just do it! I meant that seriously. It is not enough to read solutions, you must do them. I didn't know the answer in this thread either and actually found it a good question. I first thought a bit about a possible counterexample, and as there wasn't an immediate one, I took a closer look at the given conditions. The word "algebraic" was an eyecatcher! That gave me [imath] p(x)\in F[x] [/imath] with [imath] p(r)=0. [/imath] But every multiple [imath] f(x)\cdot p(x) [/imath] also has [imath] f(r)\cdot p(r)=0. [/imath] It is a standard assumption to assume lowest degree in order to make [imath] p(x) [/imath] unique - up to scaling factors from [imath] F. [/imath] Then, I wrote down [imath] p(x)=a_nx^n+\ldots+a_0 [/imath] and "saw" that it is [imath] p(x)=x\cdot q(x)+ a_0. [/imath] That was easy since I was interested in finding something like [imath] r\cdot \text{ something }=1. [/imath] Keep your goal in mind! This led me to consider [imath] a_0 [/imath] and to find a way that excluded the possibility [imath] a_0=0. [/imath] Once I knew that [imath] a_0\neq 0 [/imath] it was easy to multiply my equation by [imath] a_0^{-1} [/imath] and get [imath] 0=a_0^{-1}\cdot r\cdot q(r) + 1. [/imath] That was the clue.

So now you know "how" I found the solution. I didn't look it up in a book or know it right away. I took some elementary steps. And that's what you should do. Play with what you know and find a way through the jungle. Be confident and try different paths. You won't succeed automatically on your first attempt. No problem, take another attempt. If you only knew how many scribblings in my life went directly into the bin! You have to try. And I think this is more of a difficulty that you need to overcome. Try some things and don't be afraid to fail.

I don't know whether this will be helpful, but I wrote an article about problem-solving: How to Write a Math Proof and Their Structure
It's probably not my best article. Nevertheless, it might contain some helpful tips. Practice and training is the key.

 

[logistic_guy]

This is hard to impossible to tell from a distance. Your answers here were way better than in some other threads.

I gave you some advice somewhere hidden in a seeming joke: Nike: Just do it! I meant that seriously. It is not enough to read solutions, you must do them. I didn't know the answer in this thread either and actually found it a good question. I first thought a bit about a possible counterexample, and as there wasn't an immediate one, I took a closer look at the given conditions. That gave me [imath] p(x)\in F[x] [/imath] with [imath] p(r)=0. [/imath] But every multiple [imath] f(x)\cdot p(x) [/imath] also has [imath] f(r)\cdot p(r)=0. [/imath] It is a standard assumption to assume lowest degree in order to make [imath] p(x) [/imath] unique - up to scaling factors from [imath] F. [/imath] Then, I wrote down [imath] p(x)=a_nx^n+\ldots+a_0 [/imath] and "saw" that it is [imath] p(x)=x\cdot q(x)+ a_0. [/imath] This led me to consider [imath] a_0 [/imath] and to find a way that excluded the possibility [imath] a_0=0. [/imath] Once I knew that [imath] a_0\neq 0 [/imath] it was easy to multiply my equation by [imath] a_0^{-1} [/imath] and get [imath] 0=a_0^{-1}\cdot r\cdot q(r) + 1. [/imath] That was the clue.

So now you know "how" I found the solution. I didn't look it up in a book or know it right away. I took some elementary steps. And that's what you should do. Play with what you know and find a way through the jungle. Be confident and try different paths. You won't succeed automatically on your first attempt. No problem, take another attempt. If you only knew how many scribblings in my life went directly into the bin! You have to try. And I think this is more of a difficulty that you need to overcome. Try some things and don't be afraid to fail.

I don't know whether this will be helpful, but I wrote an article about problem-solving: How to Write a Math Proof and Their Structure
It's probably not my best article. Nevertheless, it might contain some helpful tips. Practice and training is the key.

i'll not be cocky to compare myself with your great knowledge and insight in the rings
but i think i master \(\displaystyle 99\%\) of ring theory
remain this tiny fragments which can be handle easily by a small look up

i'm not mathematcian to fully understand proofs
but i find it more entertaining to proof than to solve

i'm advance engineering but neither engineering nor mathematics is my interest
i'm an artist

i'll try to read every website you give me in my spare time
beside numbers i'm also try to improve my english
i feel i'm master \(\displaystyle 99\%\)

soon i'll master \(\displaystyle 41\) subjects and \(\displaystyle 6\) language
each subject have at least \(\displaystyle 1000\) questions
so i've to answer \(\displaystyle 41000\) questions to gain all the skills in the material
i give myself to finish at the end of \(\displaystyle 2025\)