Need clarification on a theorem about field extensions/isomorphisms

[Delurker]

I'm self-studying A Book of Abstract Algebra, 2nd ed, by Pinter and I have two questions. First, the author says to consider the situation where $K$ and $K'$ are finite extensions of $F$, and furthermore that $K$ and $K'$ have a common extension $E$. Then he goes on to prove that if $K$ is a splitting field of $F$ and $h:K\rightarrow K'$ is an $F$-fixing isomorphism, then $h[K]\subseteq K$. (Pinter uses the term "root field" instead of splitting field and I've changed it to the more standard term.) Then comes the theorem:

Let $K$ and $K'$ be finite extensions of $F$. Assume $K$ is the splitting field of some polynomial over $F$. If $h:K\rightarrow K'$ is an isomorphism which fixes $F$, then $K=K'$.

Proof: From Theorem 2 [this is the primitive element theorem; we're assuming fields have characteristic zero], $K$ and $K'$ are simple extensions of $F$, say $K=F(a)$ and $K'=F(b)$. Then $E=F(a,b)$ is a common extension of $K$ and $K'$. By the comments preceding this theorem, $h$ maps every element of $K$ to an element of $K'$ [this is trivially true so the last prime must be a typo]; hence $K'\subseteq K$. Since the same argument may be carried out for $h^{-1}$, we also have $K\subseteq K'$. QED

If my understanding is correct, then we have to assume that $K$ and $K'$ are contained in a common extension for the theorem to be true. I would have just accepted the fact that Pinter didn't explicitly mention this in the theorem because he did a page ago, but the proof implies that we can always create an extension, which isn't true (it's not possible if $K$ and $K'$ are isomorphic but not equal splitting fields over F, for example). As the theorem stands, without explicitly assuming a common extension, it seems that letting $F=\mathbb{Q}, K=\mathbb{Q}(\sqrt{2})$, and $K'=\mathbb{Q}[x]/\langle x^2-2\rangle$ is a counterexample. Does that sound about right? It would clear up another confusion I had when I learned (not too well, obviously) from Fraleigh's book years ago. I didn't quite appreciate why he made sure to say that we're assuming that all work is being done in one fixed algebraic closure before developing much theory. Is it pretty much to avoid technical issues like this?

The second question is actually about the first edition of the book. He states a theorem that if $I\subseteq E\subseteq K$ with $K$ a finite extension of $E$ and $E$ a finite extension of $I$, and $K$ is a splitting field over $E$ and $E$ is a splitting field over $I$, then $K$ is a splitting field over $I$. I don't think this is true. I tried to find some counterexamples but they were too simple and didn't work. I'm not sure if I should spend more time on it considering that there is an obvious gap in his proof and the theorem doesn't appear in the second edition (and it would make later results considerably easier to prove).

 

[fresh_42]

I'm self-studying A Book of Abstract Algebra, 2nd ed, by Pinter and I have two questions. First, the author says to consider the situation where $K$ and $K'$ are finite extensions of $F$, and furthermore that $K$ and $K'$ have a common extension $E$. Then he goes on to prove that if $K$ is a splitting field of $F$ and $h:K\rightarrow K'$ is an $F$-fixing isomorphism, then $h[K]\subseteq K$. (Pinter uses the term "root field" instead of splitting field and I've changed it to the more standard term.) Then comes the theorem:

Let $K$ and $K'$ be finite extensions of $F$. Assume $K$ is the splitting field of some polynomial over $F$. If $h:K\rightarrow K'$ is an isomorphism which fixes $F$, then $K=K'$.

Proof: From Theorem 2 [this is the primitive element theorem; we're assuming fields have characteristic zero], $K$ and $K'$ are simple extensions of $F$, say $K=F(a)$ and $K'=F(b)$. Then $E=F(a,b)$ is a common extension of $K$ and $K'$. By the comments preceding this theorem, $h$ maps every element of $K$ to an element of $K'$ [this is trivially true so the last prime must be a typo]; hence $K'\subseteq K$. Since the same argument may be carried out for $h^{-1}$, we also have $K\subseteq K'$. QED

If my understanding is correct, then we have to assume that $K$ and $K'$ are contained in a common extension for the theorem to be true. I would have just accepted the fact that Pinter didn't explicitly mention this in the theorem because he did a page ago, but the proof implies that we can always create an extension, which isn't true (it's not possible if $K$ and $K'$ are isomorphic but not equal splitting fields over F, for example). As the theorem stands, without explicitly assuming a common extension, it seems that letting $F=\mathbb{Q}, K=\mathbb{Q}(\sqrt{2})$, and $K'=\mathbb{Q}[x]/\langle x^2-2\rangle$ is a counterexample. Does that sound about right?

This has been my first idea, too. How does he change from isomorphic into to contained in? I think the common field $E$ is only needed to avoid isomorphic copies. Let's say we have $K=F(a)$ with the polynomial $f(x) \in F[x]\; , \;f(a)=0$. Now with $K'=F(b)$ we can choose $b$ such that $b=h(a)$. But $f(b)=f(h(a))=h(f(a))=0$ and as $K$ is the splitting field, we get $b \in K$ and thus $K' \subseteq K$.

It would clear up another confusion I had when I learned (not too well, obviously) from Fraleigh's book years ago. I didn't quite appreciate why he made sure to say that we're assuming that all work is being done in one fixed algebraic closure before developing much theory. Is it pretty much to avoid technical issues like this?

Hard to tell without the book. It certainly simplifies arguments.

The second question is actually about the first edition of the book. He states a theorem that if $I\subseteq E\subseteq K$ with $K$ a finite extension of $E$ and $E$ a finite extension of $I$, and $K$ is a splitting field over $E$ and $E$ is a splitting field over $I$, then $K$ is a splitting field over $I$. I don't think this is true. I tried to find some counterexamples but they were too simple and didn't work. I'm not sure if I should spend more time on it considering that there is an obvious gap in his proof and the theorem doesn't appear in the second edition (and it would make later results considerably easier to prove).

May we assume separable extensions? Then $K=E(b)=I(a)(b)=I(a,b)=I(c)$.

 

[Delurker]

This has been my first idea, too. How does he change from isomorphic into to contained in?

He uses a result from the previous page that $h$ takes elements of $K$ to elements of $K$, but this was when he assumed a common extension at the start, which he doesn't do in the proof. I think he must have meant to but messed up when rewriting the section for the second edition.

May we assume separable extensions? Then $K=E(b)=I(a)(b)=I(a,b)=I(c)$.

Pinter doesn't mention them by name and I had to look them up, but we're assuming characteristic zero so I guess we can. I understand what you wrote--I think I do at least-- but I can't see how it proves it. The proof he gives is paraphrased as follows:

Since $K$ is a finite extension of $I$, we have $K=I(a)$. Let $p(x)$ be the minimum polynomial of $a$ over $I$. Since $p(x)$ is also in $E[x]$ and $K$ contains $a$, a root of $p(x)$, then $K$ contains all roots of $p(x)$ and is the splitting field of $p(x)$ over $I$.

To justify the last assertion, he uses the theorem that if $K$ is a splitting field over $E$ and contains a root of an irreducible polynomial $p(x)$ in $E[x]$, then it contains all roots. We know that $p(x)$ is irreducible in $I[x]$, but to use the theorem we need it to be irreducible in $E[x]$. Does that follow from char 0?

If it does then I'll have another question because by removing the theorem in the second edition (he modifies it by saying that if $K$ is a splitting field over $I$ then it's also a splitting field over $E$) he creates more work for himself later on when he's trying to show that solvable polynomials have solvable Galois groups and it doesn't seem worth it. Or I'm wrong about something, which is at least as likely...

 

[fresh_42]

He uses a result from the previous page that $h$ takes elements of $K$ to elements of $K$, but this was when he assumed a common extension at the start, which he doesn't do in the proof. I think he must have meant to but messed up when rewriting the section for the second edition.

We don't need $E$. What is used is, that $K$ already contains all roots of the minimal polynomial, so $h$ has to be an automorphism.

Pinter doesn't mention them by name and I had to look them up, but we're assuming characteristic zero so I guess we can. I understand what you wrote--I think I do at least-- but I can't see how it proves it. The proof he gives is paraphrased as follows:

I only asked because this way I didn't have to think about whether I may use primitive elements $a,b,c$ or not, i.e. whether we can use single minimal polynomials. I tend to confuse the various properties: separable, normal, complete, primitive. Characteristic zero is fine.

Since $K$ is a finite extension of $I$, we have $K=I(a)$. Let $p(x)$ be the minimum polynomial of $a$ over $I$. Since $p(x)$ is also in $E[x]$ and $K$ contains $a$, a root of $p(x)$, then $K$ contains all roots of $p(x)$ and is the splitting field of $p(x)$ over $I$.

We had $I\subseteq E \subseteq K$. Now if we already have $K=I(a)$ with $p(x)$, then only the splitting is left to show, which he did. I thought the splitting fields $I \subseteq E$ and $E \subseteq K$ were given and the extension $K=I(a)$ were sought. In any case, we can glue the two since we have the full bases and all coefficients of polynomials in $E[x]$ can be rewritten as polynomials over $I$ and powers of a primitive element.

To justify the last assertion, he uses the theorem that if $K$ is a splitting field over $E$ and contains a root of an irreducible polynomial $p(x)$ in $E[x]$, then it contains all roots. We know that $p(x)$ is irreducible in $I[x]$, but to use the theorem we need it to be irreducible in $E[x]$. Does that follow from char 0?

Yes. The case $\operatorname{char}F =0$ implies that all roots are simple, that is separable extensions. If all irreducible polynomials have only simple roots, then it is a complete field, and a extension is normal if it is algebraic and one root in implies already all roots are in. Fields of characteristic zero are complete. And algebraic, separable extensions are primitive, i.e. generated by adjunction of only one (primitive) element.

If it does then I'll have another question because by removing the theorem in the second edition (he modifies it by saying that if $K$ is a splitting field over $I$ then it's also a splitting field over $E$) he creates more work for himself later on when he's trying to show that solvable polynomials have solvable Galois groups and it doesn't seem worth it. Or I'm wrong about something, which is at least as likely...

I cannot answer this, because I don't know what exactly is going on in this particular presentation of Galois theory. However the connection between solvability by radicals and solvability of the Galois group requires some work to do. This is basically due to the fact that solvable polynomials is a bit of a vague expression which first has to be translated into algebraic terms and not so much because of the composition series, neither the groups' nor the fields'. You work with vector space bases here, many elements at prior and the entire symmetric group, all of which add up to long summations.

 

[Delurker]

We don't need $E$. What is used is, that $K$ already contains all roots of the minimal polynomial, so $h$ has to be an automorphism.

But what if $K$ and $K'$ are "different," i.e. isomorphic but not equal, spitting fields? Isn't a map $h:\mathbb{Q}(\sqrt{2})\rightarrow \mathbb{Q}[x]/\langle x^2-2\rangle$ fixing $\mathbb{Q}$ and matching roots of $x^2-2$ a counterexample?

Or, taking a different approach, if we follow the given proof and create a common extension $E$, what would it look like if $K=\mathbb{Q}(\sqrt{2})$ and $K'=\mathbb{Q}[x]/\langle x^2-2\rangle$? Wouldn't $x^2-2$ have four roots?

Thanks for the comments on my second question. I'll take some time to look over them in detail.

 

[Erland]

The second question is actually about the first edition of the book. He states a theorem that if $I\subseteq E\subseteq K$ with $K$ a finite extension of $E$ and $E$ a finite extension of $I$, and $K$ is a splitting field over $E$ and $E$ is a splitting field over $I$, then $K$ is a splitting field over $I$. I don't think this is true. I tried to find some counterexamples but they were too simple and didn't work. I'm not sure if I should spend more time on it considering that there is an obvious gap in his proof and the theorem doesn't appear in the second edition (and it would make later results considerably easier to prove).

You are right. This is not true. Not even if the extension is assumed to be separable.
As a counterexample, let $I=\mathbb Q$, $E=\mathbb Q(\sqrt 2)$, and $K=\mathbb Q(\sqrt[4] 2)$, all being considered as subfields of both $\mathbb R$ and $\mathbb C$. Then, $E$ is a splitting field of the polynomial $x^2-2$ over $I=\mathbb Q$, and $K$ is a splitting field of the polynomial $x^2-\sqrt 2$ over $E$.
The extension $K/I$ is separable, since $\mathbb Q$ has characterstic zero.
Assume now that $K$ is a splitting field of some polynomial over $I$. Then, the extension $K/I$ is normal. Now, the polynomial $x^4-2$ is irreducible over $I=\mathbb Q$, by Eisenstein's criterion. This polynomial has the zeros $\pm\sqrt[4] 2\in K$. Since the extension $K/I$ is normal, this polynomial splits in $K$. But this not the case: the zeros $\pm i \sqrt[4] 2\in\mathbb C$ do not lie in $\mathbb R$, and hence not in $K$.
This is a contradiction, which shows that $K$ is not a splitting field of any polynomial over $I$.
So, the above claim is false, even if we assume that the extension $K/I$ is separable.

 

[Erland]

If my understanding is correct, then we have to assume that $K$ and $K'$ are contained in a common extension for the theorem to be true. I would have just accepted the fact that Pinter didn't explicitly mention this in the theorem because he did a page ago, but the proof implies that we can always create an extension, which isn't true (it's not possible if $K$ and $K'$ are isomorphic but not equal splitting fields over F, for example). As the theorem stands, without explicitly assuming a common extension, it seems that letting $F=\mathbb{Q}, K=\mathbb{Q}(\sqrt{2})$, and $K'=\mathbb{Q}[x]/\langle x^2-2\rangle$ is a counterexample. Does that sound about right? It would clear up another confusion I had when I learned (not too well, obviously) from Fraleigh's book years ago. I didn't quite appreciate why he made sure to say that we're assuming that all work is being done in one fixed algebraic closure before developing much theory. Is it pretty much to avoid technical issues like this?

Yes, I agree, the claim is obviously false if we don't assume that $K$ and $K'$ are contained in a common field.

I also read Fraleigh decades ago, and you may very well be right about his fondness for algebraic closures. :)

 

[Delurker]

You are right. This is not true. Not even if the extension is assumed to be separable.
As a counterexample, let $I=\mathbb Q$, $E=\mathbb Q(\sqrt 2)$, and $K=\mathbb Q(\sqrt[4] 2)$, all being considered as subfields of both $\mathbb R$ and $\mathbb C$. Then, $E$ is a splitting field of the polynomial $x^2-2$ over $I=\mathbb Q$, and $K$ is a splitting field of the polynomial $x^2-\sqrt 2$ over $E$.
The extension $K/I$ is separable, since $\mathbb Q$ has characterstic zero.
Assume now that $K$ is a splitting field of some polynomial over $I$. Then, the extension $K/I$ is normal. Now, the polynomial $x^4-2$ is irreducible over $I=\mathbb Q$, by Eisenstein's criterion. This polynomial has the zeros $\pm\sqrt[4] 2\in K$. Since the extension $K/I$ is normal, this polynomial splits in $K$. But this not the case: the zeros $\pm i \sqrt[4] 2\in\mathbb C$ do not lie in $\mathbb R$, and hence not in $K$.
This is a contradiction, which shows that $K$ is not a splitting field of any polynomial over $I$.
So, the above claim is false, even if we assume that the extension $K/I$ is separable.

Great, I thought there must be a simple counterexample. To come full circle to the supposed proof, since the minimum polynomial of $\sqrt[4]{2}$ in $I[x]$ is $x^4-2$ and this is reducible in $E[x]$ as $(x^2+\sqrt{2})(x^2-\sqrt{2})$, this is where the proof fails.

And the other proof breaks down because it's not always possible to find a common extension field for $K$ and $K'$ that satisfy the conditions in the theorem, but at least this theorem is easily fixed by sticking to a fixed algebraic closure. Thanks to both of you!

 

[Erland]

Great, I thought there must be a simple counterexample. To come full circle to the supposed proof, since the minimum polynomial of $\sqrt[4]{2}$ in $I[x]$ is $x^4-2$ and this is reducible in $E[x]$ as $(x^2+\sqrt{2})(x^2-\sqrt{2})$, this is where the proof fails.

Yes, this is where this proof fails for my counterexample, but there is also another error in the proof, namely in the sentence

Since $K$ is a finite extension of $I$, we have $K=I(a)$.

There is no guarantee that $K$ is a simple extension of $I$. More than one element from $K\setminus I$ may be needed to generate $K$, although one element was sufficient in my counterexample.

 

[Delurker]

There is no guarantee that $K$ is a simple extension of $I$. More than one element from $K\setminus I$ may be needed to generate $K$, although one element was sufficient in my counterexample.

I forgot to mention in my first post that we're assuming fields of characteristic zero here, so this part should be ok.

 

[Erland]

Is it really so that if $E$ and $F$ are fields of characteritic zero, and $F/E$ is a finite extension, then this is a simple extension, that is, there exists an $\alpha\in F$ such that $F=E(\alpha)$?
Then, for example, there would exist an $\alpha\in \mathbb Q(\sqrt 2, i)$ such that $\mathbb Q(\alpha)=\mathbb Q(\sqrt 2, i)$ (and this is a splitting field of the polynomial $(x^2-2)(x^2+1)$ over $\mathbb Q$). I don't think it is so, but I can't prove it right now.

I think there is a mix-up in earlier posts between a simple root and a simple extension. These are not the same thing.

 

[fresh_42]

I think there is a mix-up in earlier posts between a simple root and a simple extension. These are not the same thing.

Not really. Separability refers to single roots, which I have meant. Then we have
https://en.wikipedia.org/wiki/Primitive_element_theorem#Classical_Primitive_Element_Theorem

 

[Erland]

Not really. Separability refers to single roots, which I have meant. Then we have
https://en.wikipedia.org/wiki/Primitive_element_theorem#Classical_Primitive_Element_Theorem

I am so sorry! This theorem was unknown to me. I thought you meant something else. Sorry, once again...

 

[Infrared]

Is it really so that if $E$ and $F$ are fields of characteritic zero, and $F/E$ is a finite extension, then this is a simple extension, that is, there exists an $\alpha\in F$ such that $F=E(\alpha)$?
Then, for example, there would exist an $\alpha\in \mathbb Q(\sqrt 2, i)$ such that $\mathbb Q(\alpha)=\mathbb Q(\sqrt 2, i)$ (and this is a splitting field of the polynomial $(x^2-2)(x^2+1)$ over $\mathbb Q$). I don't think it is so, but I can't prove it right now.

Let $\alpha=\sqrt{2}+i$. By considering $1/\alpha$, you can verify that $\sqrt{2}-i$, and hence $\sqrt{2}$, is an element of $\mathbb{Q}(\alpha)$. From that, it's easy to conclude $\mathbb{Q}(\alpha)=\mathbb{Q}(\sqrt{2},i)$.

 

[Erland]

Let $\alpha=\sqrt{2}+i$. By considering $1/\alpha$, you can verify that $\sqrt{2}-i$, and hence $\sqrt{2}$, is an element of $\mathbb{Q}(\alpha)$. From that, it's easy to conclude $\mathbb{Q}(\alpha)=\mathbb{Q}(\sqrt{2},i)$.

Yes, yes, indeed. I just made an *ss of myself...