Help with Dummit & Foote Exercise 1, Section 13.2 - Algebraic Extensions

In summary, the conversation is about Exercise 1 of Dummit and Foote, Section 13.2 in which the person is having trouble starting the problem and is seeking help. They discuss the definition of "characteristic" of a field and its link to the prime subfield. They also mention that the finite field has a finite dimension over the prime subfield, and they provide a possible solution to the problem. They ask for feedback on their solution and if it is rigorous.
  • #1
Math Amateur
Gold Member
MHB
3,998
48
I need help with Exercise 1 of Dummit and Foote, Section 13.2 : Algebraic Extensions ..

I have been unable to make a meaningful start on the problem ... ... Exercise 1 of Dummit and Foote, Section 13.2 reads as follows:
View attachment 6608
I have been unable to make a meaningful start on this problem ...BUT ... further I hope to understand why D&F put this exercise at the end of a section on algebraic extensions ... the exercise seems a bit remote from the subject matter ... ...
Hope someone can help ...Peter
 
Physics news on Phys.org
  • #2
I would start with the definition of "characteristic" of a field. What is that definition?
 
  • #3
HallsofIvy said:
I would start with the definition of "characteristic" of a field. What is that definition?
Thanks HallsofIvy ...

Thought about definition and its link to the prime subfield ... and had considerable help from people on the Physics Forums ...Have the following solution ... please critique it ...============================================================================

\(\displaystyle F\) finite field of characteristic \(\displaystyle p\)\(\displaystyle \Longrightarrow\) prime subfield of \(\displaystyle F\) is isomorphic to \(\displaystyle \mathbb{F}_p\) and \(\displaystyle p\) must be prime ... (Lidl and Niederreiter, Introduction to Finite Fields ... ... , Theorem 1.78 ... ...also \(\displaystyle F\) finite means that \(\displaystyle F\) has finite dimension over \(\displaystyle \mathbb{F}_p\), say dimension of \(\displaystyle F\) over \(\displaystyle \mathbb{F}_p = n\) \(\displaystyle \Longrightarrow\) Basis for \(\displaystyle F\) over \(\displaystyle \mathbb{F}_p\) has \(\displaystyle n\) elements\(\displaystyle \Longrightarrow\) all elements of \(\displaystyle F\) are uniquely expressible as \(\displaystyle c_1 v_1 + c_2 v_2 + \ ... \ ... \ \ c_n v_n\) ... ... ... (1)where \(\displaystyle c_1, c_2, \ ... \ ... \ \ , c_n \in \mathbb{F}_p\)

and

\(\displaystyle v_1, v_2, \ ... \ ... \ \ , v_n \in F\)
\(\displaystyle \Longrightarrow\) number of elements in \(\displaystyle F, |F| = p^n\)

since each \(\displaystyle c_i\) in expression (1) has \(\displaystyle p\) possibilities ... ... ( and the \(\displaystyle v_i\) are fixed)=========================================================================Is the above correct?Could someone please critique solution/proof ... is it rigorous ...?Peter
 
Last edited:

Related to Help with Dummit & Foote Exercise 1, Section 13.2 - Algebraic Extensions

What is an algebraic extension?

An algebraic extension is a field extension where every element is the root of a polynomial with coefficients in the base field. This means that every element in the extension field can be expressed as a solution to a polynomial equation with coefficients in the base field.

How do I determine if an extension is algebraic?

To determine if an extension is algebraic, you can check if every element in the extension field is the root of a polynomial with coefficients in the base field. If this is the case, then the extension is algebraic.

What is the degree of an algebraic extension?

The degree of an algebraic extension is the degree of the minimal polynomial of any element in the extension field over the base field. This is the smallest degree of a polynomial that has the element as a root.

How do I prove that an extension is algebraic?

To prove that an extension is algebraic, you can show that every element in the extension field is the root of a polynomial with coefficients in the base field. This can be done by explicitly constructing the minimal polynomial for each element, or by using the Tower Law to show that the extension is generated by algebraic elements.

Can I have a transcendental element in an algebraic extension?

No, an algebraic extension only contains algebraic elements. This means that all elements in the extension field can be expressed as solutions to polynomial equations with coefficients in the base field. Transcendental elements, on the other hand, cannot be expressed as solutions to any polynomial equation with coefficients in the base field.

Similar threads

Replies
5
Views
2K
Replies
4
Views
1K
Replies
1
Views
933
Replies
1
Views
1K
  • Linear and Abstract Algebra
Replies
3
Views
2K
  • Linear and Abstract Algebra
Replies
3
Views
1K
Replies
1
Views
2K
Replies
2
Views
1K
  • Linear and Abstract Algebra
Replies
2
Views
1K
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
Back
Top