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Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13, page 282

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am reading Nicholson: Introduction to Abstract Algebra, Section 6.2 Algebraic Extensions.

Example 13 on page 282 (see attachment) reads as follows:

"If [TEX] u = \sqrt[3]{2} [/TEX] show that [TEX] \mathbb{Q}(u) = \mathbb{Q}(u^2) [/TEX]"

In the third line of the explanation - see page 282 of attachment - we read:

"But [TEX] [\mathbb{Q}(u^2) \ : \ \mathbb{Q}] \ne 1 [/TEX] because [TEX] u^2 \notin \mathbb{Q} [/TEX] ... ... "

Can someone explain why it follows that [TEX] u^2 \notin \mathbb{Q} \Longrightarrow [\mathbb{Q}(u^2) \ : \ \mathbb{Q}] \ne 1 [/TEX]

Peter

[This has also been posted on MHF]
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
I am reading Nicholson: Introduction to Abstract Algebra, Section 6.2 Algebraic Extensions.

Example 13 on page 282 (see attachment) reads as follows:

"If [TEX] u = \sqrt[3]{2} [/TEX] show that [TEX] \mathbb{Q}(u) = \mathbb{Q}(u^2) [/TEX]"

In the third line of the explanation - see page 282 of attachment - we read:

"But [TEX] [\mathbb{Q}(u^2) \ : \ \mathbb{Q}] \ne 1 [/TEX] because [TEX] u^2 \notin \mathbb{Q} [/TEX] ... ... "

Can someone explain why it follows that [TEX] u^2 \notin \mathbb{Q} \Longrightarrow [\mathbb{Q}(u^2) \ : \ \mathbb{Q}] \ne 1 [/TEX]

Peter

[This has also been posted on MHF]
We prove the following general fact.

Let $E$ be an extension of a field $F$.
Let $a\in E$ be such that $[F(a):F]=1$.
Then $a\in F$.

What is the minimal polynomial of $a$ over $F$? It's degree is $1$. So say $x-b\in F[x]$ is the minimal polynomial of $a$ over $F$.
Now we show that $f(a)\in F$ for any polynomial $f(x)\in F[x]$. Note that $f(x)=(x-a)g(x)+r(x)$ for some $r(x)\in F[x]$ with $\deg r(x)=1$. Thus $f(a)=r(a)$. Since $\deg r(x)=1$, $r(a)\in F$ and hence $f(a)\in F$.
We also knew that $F(a)=\{f(a):f(x)\in F[x]\}$ since $a$ is algebraic over $F$.
So we conclude $F(a)=F$ and hence $a\in F$.

Using the above discussion you can easily solve your original problem.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
We prove the following general fact.

Let $E$ be an extension of a field $F$.
Let $a\in E$ be such that $[F(a):F]=1$.
Then $a\in F$.

What is the minimal polynomial of $a$ over $F$? It's degree is $1$. So say $x-b\in F[x]$ is the minimal polynomial of $a$ over $F$.
Now we show that $f(a)\in F$ for any polynomial $f(x)\in F[x]$. Note that $f(x)=(x-a)g(x)+r(x)$ for some $r(x)\in F[x]$ with $\deg r(x)=1$. Thus $f(a)=r(a)$. Since $\deg r(x)=1$, $r(a)\in F$ and hence $f(a)\in F$.
We also knew that $F(a)=\{f(a):f(x)\in F[x]\}$ since $a$ is algebraic over $F$.
So we conclude $F(a)=F$ and hence $a\in F$.

Using the above discussion you can easily solve your original problem.


Thanks caffeinemachine, most helpful of you. Your post enables me to progress on with Field Theory.

Peter
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
We prove the following general fact.

Let $E$ be an extension of a field $F$.
Let $a\in E$ be such that $[F(a):F]=1$.
Then $a\in F$.

What is the minimal polynomial of $a$ over $F$? It's degree is $1$. So say $x-b\in F[x]$ is the minimal polynomial of $a$ over $F$.
Now we show that $f(a)\in F$ for any polynomial $f(x)\in F[x]$. Note that $f(x)=(x-a)g(x)+r(x)$ for some $r(x)\in F[x]$ with $\deg r(x)=1$. Thus $f(a)=r(a)$. Since $\deg r(x)=1$, $r(a)\in F$ and hence $f(a)\in F$.
We also knew that $F(a)=\{f(a):f(x)\in F[x]\}$ since $a$ is algebraic over $F$.
So we conclude $F(a)=F$ and hence $a\in F$.

Using the above discussion you can easily solve your original problem.
Hi caffeinemachine,

I was just revising and reflecting on your post and have a couple of points needing clarification.

You write:

"So say \(\displaystyle x-b\in F[x] \) is the minimal polynomial of a over F."

Do you mean:

"So say \(\displaystyle x-a \in F[x] \) is the minimal polynomial of a over F."?



Further, later in the post you write:

"Note that \(\displaystyle f(x)=(x-a)g(x)+r(x) \) for some \(\displaystyle r(x)\in F[x] \) with deg r(x)=1."

Given that the divisor (x - a) is of degree 1 does not this mean that the remainder is of degree less than 1, that is deg r(x) = 0?



Can you please clarify these two points?

Peter
 

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Hi caffeinemachine,

I was just revising and reflecting on your post and have a couple of points needing clarification.

You write:

"So say \(\displaystyle x-b\in F[x] \) is the minimal polynomial of a over F."

Do you mean:

"So say \(\displaystyle x-a \in F[x] \) is the minimal polynomial of a over F."?



Further, later in the post you write:

"Note that \(\displaystyle f(x)=(x-a)g(x)+r(x) \) for some \(\displaystyle r(x)\in F[x] \) with deg r(x)=1."

Given that the divisor (x - a) is of degree 1 does not this mean that the remainder is of degree less than 1, that is deg r(x) = 0?



Can you please clarify these two points?

Peter
Thank you Peter. I see that my solution is messed up. I can't believe I made those mistakes.

Here.

We know that the minimal polynomial of $a$ over $F$ has degree $1$. So say $x-b\in F[x]$ is the minimal polynomial of $a$ over $F$.

(Note: Here I don't mean that $x-a\in F[x]$ is the minimal polynomial of $a$ over $F$. We cannot say this because we don't know yet that $a\in F$. Saying $x-a\in F[x]$ would automatically imply that $a\in F$, but that is what we have to prove!)

Now, since $x-b\in F[x]$ is the minimal polynomial of $a$ over $F$, we know that $a$ satisfies $x-b$. Thus $a-b=0$. thus giving $a=b$. Since $b\in F$ (Why? Simply because $x-b\in F[x]$), we have $a\in F$.

Forget about my previous solution. It is an embarrassment.

Tell me if you have any further doubts.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
Thank you Peter. I see that my solution is messed up. I can't believe I made those mistakes.

Here.

We know that the minimal polynomial of $a$ over $F$ has degree $1$. So say $x-b\in F[x]$ is the minimal polynomial of $a$ over $F$.

(Note: Here I don't mean that $x-a\in F[x]$ is the minimal polynomial of $a$ over $F$. We cannot say this because we don't know yet that $a\in F$. Saying $x-a\in F[x]$ would automatically imply that $a\in F$, but that is what we have to prove!)

Now, since $x-b\in F[x]$ is the minimal polynomial of $a$ over $F$, we know that $a$ satisfies $x-b$. Thus $a-b=0$. thus giving $a=b$. Since $b\in F$ (Why? Simply because $x-b\in F[x]$), we have $a\in F$.

Forget about my previous solution. It is an embarrassment.

Tell me if you have any further doubts.
Thanks caffeinemachine! as always, I appreciate your help

Peter