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]