Let, $E$ be an extension over $F$, $a,bin E$ is algebric over $E$ such that $[F(a):F]=n,[F(b):F]=m$. Prove $[F(a,b):F]le nm$.

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I have solved the problem in the following manner-

As per Tower law of field extension, we know that $[F(a,b):F]=[F(a,b):F(a)][F(a):F]=[F(a,b):F]n$

Now, $[F(a,b):F(a)]=[(F(a))(b):F(a)]le m$

Because, $[F(b):F]=mimplies$the minimal polynomial $p(x)$(say)(obviously irreducible over $F$) of $b$ over $F$ has degree $m$. So, the minimal polynomial of $b$ over $F(a)$ has degree not exceeding $m$(since $p(x)in Fsubseteq F(a)$ with $p(a)=0$.

So, $[F(a,b):F]le nm$

Is this method correct?

My next and actual question is- Can I get an example where $[F(a,b):F]$ is strictly less than $nm$ i.e. Can I get an example where $[F(a,b):F(a)]=[(F(a))(b):F(a)]$ is strictly less than $m$.






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  • $b=a$ or $b=1+a$ for a trivial example. $a=root3of2$, $b=e^2pi i/3root3of2$ for a slightly less trivial example. Did you search the site? I'm sure we have had examples just like this.
    – Jyrki Lahtonen
    Sep 1 at 5:19















up vote
0
down vote

favorite












I have solved the problem in the following manner-

As per Tower law of field extension, we know that $[F(a,b):F]=[F(a,b):F(a)][F(a):F]=[F(a,b):F]n$

Now, $[F(a,b):F(a)]=[(F(a))(b):F(a)]le m$

Because, $[F(b):F]=mimplies$the minimal polynomial $p(x)$(say)(obviously irreducible over $F$) of $b$ over $F$ has degree $m$. So, the minimal polynomial of $b$ over $F(a)$ has degree not exceeding $m$(since $p(x)in Fsubseteq F(a)$ with $p(a)=0$.

So, $[F(a,b):F]le nm$

Is this method correct?

My next and actual question is- Can I get an example where $[F(a,b):F]$ is strictly less than $nm$ i.e. Can I get an example where $[F(a,b):F(a)]=[(F(a))(b):F(a)]$ is strictly less than $m$.






abstract-algebra field-theory extension-field







share|cite|improve this question





















  • $b=a$ or $b=1+a$ for a trivial example. $a=root3of2$, $b=e^2pi i/3root3of2$ for a slightly less trivial example. Did you search the site? I'm sure we have had examples just like this.
    – Jyrki Lahtonen
    Sep 1 at 5:19













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I have solved the problem in the following manner-

As per Tower law of field extension, we know that $[F(a,b):F]=[F(a,b):F(a)][F(a):F]=[F(a,b):F]n$

Now, $[F(a,b):F(a)]=[(F(a))(b):F(a)]le m$

Because, $[F(b):F]=mimplies$the minimal polynomial $p(x)$(say)(obviously irreducible over $F$) of $b$ over $F$ has degree $m$. So, the minimal polynomial of $b$ over $F(a)$ has degree not exceeding $m$(since $p(x)in Fsubseteq F(a)$ with $p(a)=0$.

So, $[F(a,b):F]le nm$

Is this method correct?

My next and actual question is- Can I get an example where $[F(a,b):F]$ is strictly less than $nm$ i.e. Can I get an example where $[F(a,b):F(a)]=[(F(a))(b):F(a)]$ is strictly less than $m$.






abstract-algebra field-theory extension-field







share|cite|improve this question













I have solved the problem in the following manner-

As per Tower law of field extension, we know that $[F(a,b):F]=[F(a,b):F(a)][F(a):F]=[F(a,b):F]n$

Now, $[F(a,b):F(a)]=[(F(a))(b):F(a)]le m$

Because, $[F(b):F]=mimplies$the minimal polynomial $p(x)$(say)(obviously irreducible over $F$) of $b$ over $F$ has degree $m$. So, the minimal polynomial of $b$ over $F(a)$ has degree not exceeding $m$(since $p(x)in Fsubseteq F(a)$ with $p(a)=0$.

So, $[F(a,b):F]le nm$

Is this method correct?

My next and actual question is- Can I get an example where $[F(a,b):F]$ is strictly less than $nm$ i.e. Can I get an example where $[F(a,b):F(a)]=[(F(a))(b):F(a)]$ is strictly less than $m$.






abstract-algebra field-theory extension-field




abstract-algebra field-theory extension-field






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asked Sep 1 at 5:06









Biswarup Saha

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  • $b=a$ or $b=1+a$ for a trivial example. $a=root3of2$, $b=e^2pi i/3root3of2$ for a slightly less trivial example. Did you search the site? I'm sure we have had examples just like this.
    – Jyrki Lahtonen
    Sep 1 at 5:19

















  • $b=a$ or $b=1+a$ for a trivial example. $a=root3of2$, $b=e^2pi i/3root3of2$ for a slightly less trivial example. Did you search the site? I'm sure we have had examples just like this.
    – Jyrki Lahtonen
    Sep 1 at 5:19
















$b=a$ or $b=1+a$ for a trivial example. $a=root3of2$, $b=e^2pi i/3root3of2$ for a slightly less trivial example. Did you search the site? I'm sure we have had examples just like this.
– Jyrki Lahtonen
Sep 1 at 5:19





$b=a$ or $b=1+a$ for a trivial example. $a=root3of2$, $b=e^2pi i/3root3of2$ for a slightly less trivial example. Did you search the site? I'm sure we have had examples just like this.
– Jyrki Lahtonen
Sep 1 at 5:19
















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