Field Theory - Nicholson - Algebraic Extensions - Section 6.2 - Example 13

[Math Amateur]

xample 13 from Nicholson: Introduction to Abstract Algebra, Section 6.2, page 282 reads as follows: (see attachment)

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Example 13: If u = \sqrt[3]{2} show that \mathbb{Q}(u) = \mathbb{Q}(u)^2

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The solution comes down to the following:

Given \mathbb{Q}(u) \supseteq \mathbb{Q}(u)^2 \supseteq \mathbb{Q}

so [ \mathbb{Q}(u) \ : \ \mathbb{Q}] = [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] \ [ \mathbb{Q}(u)^2 \ : \ \mathbb{Q} ]

Now Nicholson shows that [ \mathbb{Q}(u) \ : \ \mathbb{Q}] = 3 and [ \mathbb{Q}(u)^2 \ : \ \mathbb{Q} ] = 3

so [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] = 1

Then Nicholson (I think) concludes that \mathbb{Q}(u) = \mathbb{Q}(u)^2

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My problem is as follows:

How (exactly) does it follow that:

[ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] = 1 \Longrightarrow \mathbb{Q}(u) = \mathbb{Q}(u)^2

Can someone help?

Peter

 

[caffeinemachine]

xample 13 from Nicholson: Introduction to Abstract Algebra, Section 6.2, page 282 reads as follows: (see attachment)

------------------------------------------------------------------------------------------------------------------

Example 13: If u = \sqrt[3]{2} show that \mathbb{Q}(u) = \mathbb{Q}(u)^2

------------------------------------------------------------------------------------------------------------------

The solution comes down to the following:

Given \mathbb{Q}(u) \supseteq \mathbb{Q}(u)^2 \supseteq \mathbb{Q}

so [ \mathbb{Q}(u) \ : \ \mathbb{Q}] = [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] \ [ \mathbb{Q}(u)^2 \ : \ \mathbb{Q} ]

Now Nicholson shows that [ \mathbb{Q}(u) \ : \ \mathbb{Q}] = 3 and [ \mathbb{Q}(u)^2 \ : \ \mathbb{Q} ] = 3

so [ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] = 1

Then Nicholson (I think) concludes that \mathbb{Q}(u) = \mathbb{Q}(u)^2

----------------------------------------------------------------------------------------------------------

My problem is as follows:

How (exactly) does it follow that:

[ \mathbb{Q}(u) \ : \ \mathbb{Q}(u)^2] = 1 \Longrightarrow \mathbb{Q}(u) = \mathbb{Q}(u)^2

Can someone help?

Peter

I'd like to point put that the notation you are using is incorrect (or at least not standard). You write $\mathbb Q(u)^2$. I think you mean $\mathbb Q(u^2)$.

Now. We prove the following general fact.

Let $E$ be an extension of a field $F$ with $[E:F]=1$. Then $E=F$.

Proof: Suppose not. Then there is an element $a \in E\setminus F$. Clearly $a$ is algebraic over $F$. But since $[E:F(a)][F(a):F]=[E:F]=1$, we must have $[F(a):F]=1$. This forces $a\in F$ (as discussed in one of your previous threads). Thus we arrive at a contradiction. Thus $E=F$.

From here it is easy to see that $[\mathbb Q(u):\mathbb Q(u^2)]=1$ gives $\mathbb Q(u)=\mathbb Q(u^2)$ by putting $E=\mathbb Q(u)$ and $F=\mathbb Q(u^2)$ in the above claim.

 

[Math Amateur]

I'd like to point put that the notation you are using is incorrect (or at least not standard). You write $\mathbb Q(u)^2$. I think you mean $\mathbb Q(u^2)$.

Now. We prove the following general fact.

Let $E$ be an extension of a field $F$ with $[E:F]=1$. Then $E=F$.

Proof: Suppose not. Then there is an element $a \in E\setminus F$. Clearly $a$ is algebraic over $F$. But since $[E:F(a)][F(a):F]=[E:F]=1$, we must have $[F(a):F]=1$. This forces $a\in F$ (as discussed in one of your previous threads). Thus we arrive at a contradiction. Thus $E=F$.

From here it is easy to see that $[\mathbb Q(u):\mathbb Q(u^2)]=1$ gives $\mathbb Q(u)=\mathbb Q(u^2)$ by putting $E=\mathbb Q(u)$ and $F=\mathbb Q(u^2)$ in the above claim.

Thanks caffeinemachine.

Yes, notation was wrong ... Was a typo, but you are correct to point it out!

Peter

 

[Deveno]

Another approach:

Suppose $[E:F] = 1$ with $E$ algebraic over $F$, and assume $a \in E\setminus F$.

Then $a$ satisfies a polynomial of degree 1 in $F[x]$, say:

$f(x) = c_0 + c_1x$, where $c_1 \neq 0$.

Thus $c_0 + c_1a = 0$ so that: $a = -\dfrac{c_0}{c_1} \in F$, contradiction.

Thus $E = F$.

Yet another way to see this:

If $[E:F] = 1$, then $E$ as a vector space over $F$ has dimension 1. Since $1 \neq 0$ in $F$, we see that $\{1\}$ is a basis for $E$ over $F$, and hence any element $a \in E$ is of the form:

$a = c_0(1) = c_0$ for some $c_0 \in F$, that is: $E \subseteq F$, and since $F \subseteq E$ is trivial, we must have $E = F$.

 

[Math Amateur]

Another approach:

Suppose $[E:F] = 1$ with $E$ algebraic over $F$, and assume $a \in E\setminus F$.

Then $a$ satisfies a polynomial of degree 1 in $F[x]$, say:

$f(x) = c_0 + c_1x$, where $c_1 \neq 0$.

Thus $c_0 + c_1a = 0$ so that: $a = -\dfrac{c_0}{c_1} \in F$, contradiction.

Thus $E = F$.

Yet another way to see this:

If $[E:F] = 1$, then $E$ as a vector space over $F$ has dimension 1. Since $1 \neq 0$ in $F$, we see that $\{1\}$ is a basis for $E$ over $F$, and hence any element $a \in E$ is of the form:

$a = c_0(1) = c_0$ for some $c_0 \in F$, that is: $E \subseteq F$, and since $F \subseteq E$ is trivial, we must have $E = F$.

Thanks Deveno. Appreciate your help.

Peter