Show That $C$ Is Algebraic Closure for $F$

  • MHB
  • Thread starter mathmari
  • Start date
  • Tags
    closure
In summary, $C$ is the algebraic closure of $E$ and it also contains all the algebraic elements of $F$, making it the algebraic closure of $F$ as well. Good job!
  • #1
mathmari
Gold Member
MHB
5,049
7
Hey! :eek:

Let $E/F$ be an algebraic extension and $C$ the algebraic closure of $E$. I want to show that the field $C$ is the algebraic closuree also for $F$.

We have that $C=\{c\in E\mid c \text{ algebraic over } E\}$, i.e., every polynomial $f(x)\in E[x]$ splits completely in $C$.

Since $F\leq E$ we have that so every polynomial of $F[x]$ is also a polynomial of $E[x]$. Therefore, these polynomials have also their roots in $C$, and so $C$ is also the algebraic closure for $F$.

Is this correct? (Wondering)
 
Physics news on Phys.org
  • #2


Hello! That is correct, good job! Your reasoning is spot on. Since $C$ contains all the algebraic elements of $E$, it also contains all the algebraic elements of $F$, since $F$ is a subfield of $E$. Therefore, $C$ is the algebraic closure of $F$. Keep up the good work!
 

FAQ: Show That $C$ Is Algebraic Closure for $F$

What is the definition of algebraic closure for a field?

The algebraic closure of a field $F$ is an extension field $C$ that contains all the roots of every polynomial in $F[x]$, where $x$ is an indeterminate. In other words, every element in $C$ is algebraic over $F$, meaning it is a root of a polynomial with coefficients in $F$.

Why is algebraic closure important in mathematics?

Algebraic closure plays a crucial role in many areas of mathematics, including algebra, number theory, and geometry. It allows us to solve polynomial equations and understand the properties of fields and their extensions. Additionally, it provides a deeper understanding of the structure and behavior of numbers and other mathematical objects.

How do you show that $C$ is algebraic closure for $F$?

To show that $C$ is algebraic closure for $F$, we need to prove two things: (1) every element in $C$ is algebraic over $F$, and (2) $C$ is a field. To prove the first condition, we can use the fact that $C$ contains all the roots of every polynomial in $F[x]$. To prove the second condition, we can use the fact that $C$ is a field extension of $F$, meaning it inherits all the properties of $F$ and is closed under addition, subtraction, multiplication, and division.

Can a field have more than one algebraic closure?

Yes, a field can have more than one algebraic closure. This is because algebraic closure is not unique and depends on the choice of the field $F$. For example, the field of rational numbers $\mathbb{Q}$ has infinitely many algebraic closures, such as $\mathbb{C}$ and $\overline{\mathbb{Q}}$, the algebraic closure of $\mathbb{Q}$.

What are some examples of algebraically closed fields?

The most well-known example of an algebraically closed field is the field of complex numbers $\mathbb{C}$, which contains all the roots of every polynomial with complex coefficients. Other examples include the field of algebraic numbers $\overline{\mathbb{Q}}$ and the field of Puiseux series $\mathbb{C}((t))$, both of which are algebraic closures of the field of rational numbers $\mathbb{Q}$.

Similar threads

MHB The extension KE/F is finite and separable
Replies
1
Views
1K
MHB Show that the mapping is surjective
Replies
2
Views
2K
MHB The irreducible polynomial is not separable
Replies
1
Views
2K
MHB It is an algebraically dependent set over F
Replies
4
Views
2K
MHB Algebraic element - Minimal polynomial
Replies
1
Views
1K
MHB The extension is Galois iff E is a splitting field of a separable polynomial of F[x]
Replies
1
Views
2K
I Splitting Fields: Anderson and Feil, Theorem 45.5 ....
Replies
9
Views
2K
MHB Splitting Fields: Anderson and Feil, Theorem 45.5
Replies
1
Views
1K
I Splitting ring of polynomials - why is this result unfindable?
Replies
9
Views
2K
MHB Is $\text{tr.deg}(E_1E_2/F) \leq \text{tr.deg}(E_1/F) + \text{tr.deg}(E_2/F)$?
Replies
1
Views
1K

Hot Threads

Recent Insights

Back
Top