# Math Help - [SOLVED] Transitivity of Algebraic extension Fields

1. ## [SOLVED] Transitivity of Algebraic extension Fields

so if we're given that L extends K, w/ L algebraic of K, and M extends L, w/ M algebraic over L, do we get that M is algebraic of K? (we may not assume finite extensions)

that is:
For all l in L, there exists p(t) in K[t] such that p(l)=0
For all m in M, there exists q(t) in L[t] such that q(m)=0

For all m in M, does there exist r(t) in K[t] such that r(m)=0?

thanks for any help

2. Originally Posted by verdi
so if we're given that L extends K, w/ L algebraic of K, and M extends L, w/ M algebraic over L, do we get that M is algebraic of K? (we may not assume finite extensions)

that is:
For all l in L, there exists p(t) in K[t] such that p(l)=0
For all m in M, there exists q(t) in L[t] such that q(m)=0

For all m in M, does there exist r(t) in K[t] such that r(m)=0?

thanks for any help

Let $m\in M\Longrightarrow \,\,\exists \,0\neq p(x)=x^n + a_{n-1}x^{n-1}+\ldots a_1x+a_0\in L[x]$ s.t. $p(m)=0$.
Now look at $F:=K(a_0,\ldots,a_{n-1})$ . This is a finite extension of $K$ ( why?) and $m$ is algebraic over $F$ , so it is algebraic over $K$ (why?) .

Tonio

3. Thanks!

I guess F:=K(a_0,\ldots,a_{n-1}) is a finite extension because, we have finitely many algebraic roots, each of which are the root of some minimal polynomial in K, so using the Tower Theorem, the degree is a product of a bunch of finite terms and thus finite. Then we simply use that finite \rightarrow algebraic (though the converse is false). So each m \in M is algebraic over K and M is algebraic over K.

Does that sound right?

4. Originally Posted by verdi
Thanks!

I guess F:=K(a_0,\ldots,a_{n-1}) is a finite extension because, we have finitely many algebraic roots, each of which are the root of some minimal polynomial in K, so using the Tower Theorem, the degree is a product of a bunch of finite terms and thus finite. Then we simply use that finite \rightarrow algebraic (though the converse is false). So each m \in M is algebraic over K and M is algebraic over K.

Does that sound right?

Sounds right and dandy. Good!

Tonio