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

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The discussion centers on the concept of algebraic properties in the context of automorphisms of fields, specifically as described in Fraleigh's "A First Course in Abstract Algebra." It explains that two elements, a and b, in an algebraic extension E of a field F, share the same algebraic properties if their irreducible polynomials over F are equal. This leads to a natural isomorphism between the field extensions F(a) and F(b), which preserves the structure of F. The isomorphism is defined such that it maps polynomials evaluated at a to those evaluated at b, confirming their equivalence in terms of algebraic properties. Understanding this relationship clarifies how algebraic extensions relate to each other through their irreducible polynomials.
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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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