Confused about definition of algebraic closure

In summary, the algebraic closure of a field F in another field E is a field extension of F that is algebraically closed and contains an algebraically closed subfield isomorphic to K.
  • #1
nonequilibrium
1,439
2
Hello,

After a theorem stating that the product, sum, etc of two elements of a field extension that are algebraic over the original field are also algebraic, my course states the following result (translated into english):

[itex]\textrm{Let $F \subset E$ be fields. The elements of $E$ that are algebraic across F form a subfield of $E$ (and of course a field extension of $F$).}[/itex]
[itex]\textrm{We call this subfield the algebraic closure of $F$ in $E$.}[/itex]

but later in my course it defines "the algebraic closure of F" as a field extension of F that is
(i) algebraically closed (in the sense that every polynomial has a root)
(ii) algebraic across F

These seem to be different concepts, am I right? Because the former doesn't need to be algebraically closed (despite its name...), because for example "the algebraic closure of [itex]\mathbb Q[/itex] in [itex]\mathbb R[/itex]" still has no solution for X²+1=0, yet "the algebraic closure of [itex]\mathbb Q[/itex]" (full stop) does.

So is the only difference seperating these two concepts the suffix/appendix "in E"?
 
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  • #2
mr. vodka said:
So is the only difference seperating these two concepts the suffix/appendix "in E"?

Yes, that is correct. These are two quite different concepts with the same name. So you'll need to be quite careful!
 
  • #3
If I'm not mistaken:
For every field F, we can construct a field E containing F such that E is algebraically closed. Taking the subset of E containing all elements algebraic over F will yield an algebraically closed field K containing F such that every algebraically closed field containing F will contain an algebraically closed subfield containing F isomorphic to K. In this sense such a construction is unique up to isomorphism, and informally it is this isomorphism class we refer to when we are talking about the algebraic closure of F. The important thing is that such a minimal algebraic closure exists, and any two of them are isomorphic.

The construction is not simple, but can be found in most books on abstract algebra I believe.
 
  • #4
Thank you both!
 
  • #5


Hi there,

I understand your confusion about the definition of algebraic closure. It is important to note that there are two different concepts of algebraic closure: the algebraic closure of a field and the algebraic closure of a field in a field extension.

The algebraic closure of a field, denoted by $\overline{F}$, is the smallest algebraically closed field containing $F$. This means that every polynomial with coefficients in $F$ has a root in $\overline{F}$. In your example, the algebraic closure of $\mathbb{Q}$ is not $\mathbb{R}$, but rather the set of all algebraic numbers, which includes $\mathbb{R}$ but also includes numbers like $\sqrt{2}$ and $\sqrt[3]{3}$.

On the other hand, the algebraic closure of a field in a field extension, denoted by $\overline{F}^E$, is the set of all elements in $E$ that are algebraic over $F$. This is a subfield of $E$, but it may not be algebraically closed. In your example, the algebraic closure of $\mathbb{Q}$ in $\mathbb{R}$ is indeed $\mathbb{R}$, because every element in $\mathbb{R}$ is algebraic over $\mathbb{Q}$. However, the algebraic closure of $\mathbb{Q}$ in $\mathbb{C}$ is $\overline{\mathbb{Q}}$, the set of all algebraic numbers, because not every element in $\mathbb{C}$ is algebraic over $\mathbb{Q}$.

So to answer your question, the only difference between these two concepts is the suffix "in $E$". The latter definition is specific to a particular field extension, while the former is more general and refers to the smallest algebraically closed field containing the given field. I hope this helps clarify the definition for you. Keep up the good work in your studies!
 

1. What is the definition of algebraic closure?

The algebraic closure of a field F is an extension field that contains all the roots of every polynomial in F. In other words, it is the smallest field that contains F and is algebraically closed.

2. How is the algebraic closure different from the algebraic closure of a finite field?

The algebraic closure of a finite field is the same as its algebraic closure, as every algebraic extension of a finite field is finite. However, the algebraic closure of a field F can be infinite, while the algebraic closure of a finite field is always finite.

3. Can every field be algebraically closed?

No, not every field can be algebraically closed. For a field to be algebraically closed, every polynomial in the field must have a root in the field. This is not the case for fields such as the field of rational numbers or the field of real numbers.

4. What is the importance of the algebraic closure in mathematics?

The algebraic closure plays a crucial role in many areas of mathematics, including algebra, number theory, and algebraic geometry. It allows for the study of algebraic equations and their solutions, which has numerous applications in various fields of mathematics.

5. How is the algebraic closure related to the fundamental theorem of algebra?

The fundamental theorem of algebra states that every non-constant polynomial with complex coefficients has at least one complex root. The algebraic closure of the field of real numbers is the field of complex numbers, which satisfies the fundamental theorem of algebra. Therefore, the algebraic closure is intimately related to this important theorem.

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