Could you explain me about 'relation algebraic property with conjugate'?

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The discussion centers on the concept of 'relation algebraic property with conjugate' in the context of automorphisms of fields, as described in the book 'A First Course in Abstract Algebra 7th' by Fraleigh. It establishes that two elements, a and b, in an algebraic extension E of a field F, share the same algebraic property if and only if their irreducible polynomials over F are equal, denoted as irr(a, F) = irr(b, F). This equivalence implies a natural isomorphism between the field extensions F(a) and F(b), which preserves the structure of the field F.

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  • Familiarity with irreducible polynomials and their properties
  • Knowledge of isomorphisms in abstract algebra
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Students and professionals in mathematics, particularly those focusing on abstract algebra, field theory, and automorphisms. This discussion is beneficial for anyone seeking to deepen their understanding of algebraic properties and their implications in field extensions.

bw0young0math
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Hello everyone. At first, I appreciate your click this page.

I have a book named 'A first Course in Abstract Algebra 7th' by Fraleigh.

I have a question about 'relation algebraic property with conjugate' in automorhisms of fields.
in page415,
this book explains "Let E is algebraic extension of F& a,b∈E. Then a and b have the same algebraic property iff irr(a,F)=irr(b,F)."
What's mean algebraic property in that sentense? If you explain me, I will be happy:)Thanks. Have a nice day:)
 
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bw0young0math said:
Hello everyone. At first, I appreciate your click this page.

I have a book named 'A first Course in Abstract Algebra 7th' by Fraleigh.

I have a question about 'relation algebraic property with conjugate' in automorhisms of fields.
in page415,
this book explains "Let E is algebraic extension of F& a,b∈E. Then a and b have the same algebraic property iff irr(a,F)=irr(b,F)."
What's mean algebraic property in that sentense? If you explain me, I will be happy:)Thanks. Have a nice day:)
I don't have the book. But what Fraleigh is trying to say is the following:

Let $a, b\in K$ be algebraic over $F$ satisfying $irr(a, F)=irr(b, F)$. Then there is a "natural isomorphism" between $F(a)$ and $F(b)$ which is identity on $F$.

The isomorphism is given by $\phi:F(a)\to F(b)$, where $\phi(p(a))=p(b)$ for all $p(x)\in F[x]$.

The fact that $\phi$ is well defined requires $irr(a, F)=irr(b, F)$.
 
caffeinemachine said:
I don't have the book. But what Fraleigh is trying to say is the following:

Let $a, b\in K$ be algebraic over $F$ satisfying $irr(a, F)=irr(b, F)$. Then there is a "natural isomorphism" between $F(a)$ and $F(b)$ which is identity on $F$.

The isomorphism is given by $\phi:F(a)\to F(b)$, where $\phi(p(a))=p(b)$ for all $p(x)\in F[x]$.

The fact that $\phi$ is well defined requires $irr(a, F)=irr(b, F)$.
Thanks! I understand it! F(a)and F(b) are isomorphic so we can think that they have the same algebraic constructure and algebraic properties. Thank you:)
 

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