# Field Extension

• Jun 10th 2010, 10:02 AM
Horizon
Field Extension
For a field extension $E$ of the field $F$, with a transcendental element $\alpha\in E$ over $F$, my textbook says the following:
" $E$ contains the field of fractions of $F[\alpha]$, which is the smallest subfield containing $F$ and $\alpha$, we denote this field by $F(\alpha)$."

It also says that $F[\alpha]$ is defined to be the image of the evaluation homomorphism $\phi_\alpha : F[x]\to E$. I think this corresponds to the usual notation of meaning polynomials in $\alpha$, with coefficients form $F$, right?

Anyway, my first question is how do we know that $F(\alpha)$ is the smallest field containing $F$ and $\alpha$?

Next, in an example it says that $\pi$ is transendental in $\mathbb{Q}$, the field $\mathbb{Q}(\pi)$ is isomorphic to the field $\mathbb{Q}(x)$ of rational functions over $\mathbb{Q}$ in the indeterminate $x$.

This is very confusing. How does it conclude this? $\mathbb{Q}(\pi)$ by the definition should equal the fraction field of $\mathbb{Q}[\pi]$. If $\mathbb{Q}[\pi]$ is a field then we can conclude that its fraction field is itself, and by the evaluation homomorphism $\phi_\alpha$above (with codomain restricted to its image, making it an isomorphism), we do see $\mathbb{Q}(\pi)=\mathbb{Q}[\pi]\cong\mathbb{Q}[x]$, which is almost the result, I think $\mathbb{Q}[x]$ does actually equal $\mathbb{Q}(x)$, but I'm not sure why, can anyone explain? Also, I don't see how $\mathbb{Q}[\pi]$ is a field, in fact I think it's only a integral domain.

So can someone explain how this example works?

Also, in the exercise section it asks, for a indeterminate $x$ whether the statement: $\mathbb{Q}[\pi]\cong\mathbb{Q}[x]$ is true or false. It should be similar to the above example, can someone explain this too?
• Jun 10th 2010, 10:34 AM
tonio
Quote:

Originally Posted by Horizon
For a field extension $E$ of the field $F$, with a transcendental element $\alpha\in E$ over $F$, my textbook says the following:
" $E$ contains the field of fractions of $F[\alpha]$, which is the smallest subfield containing $F$ and $\alpha$, we denote this field by $F(\alpha)$."

It also says that $F[\alpha]$ is defined to be the image of the evaluation homomorphism $\phi_\alpha : F[x]\to E$. I think this corresponds to the usual notation of meaning polynomials in $\alpha$, with coefficients form $F$, right?

Right

Anyway, my first question is how do we know that $F(\alpha)$ is the smallest field containing $F$ and $\alpha$?

Let $K$ be any field containing $F,\,\alpha\Longrightarrow f(\alpha)\in K\,\,\forall f(x)\in F[x]\Longrightarrow \frac{1}{f(\alpha)}\in K$ (why? Transcendence of $\alpha$ over $F$ is important here) $\forall 0\neq f(x)\in K\Longrightarrow F(\alpha)\subset K$

Next, in an example it says that $\pi$ is transendental in $\mathbb{Q}$, the field $\mathbb{Q}(\pi)$ is isomorphic to the field $\mathbb{Q}(x)$ of rational functions over $\mathbb{Q}$ in the indeterminate $x$.

This is very confusing. How does it conclude this? $\mathbb{Q}(\pi)$ by the definition should equal the fraction field of $\mathbb{Q}[\pi]$. If $\mathbb{Q}[\pi]$ is a field

It can't possibly be a field: we know that if $K$ is any field and $w$ is an element that belongs to some extension field of $K$ , then $K[w]$ is a field iff $w$ is algebraic over $K$ , and if this is the case then we have that $K[w]=K(w)$

then we can conclude that its fraction field is itself, and by the evaluation homomorphism $\phi_\alpha$above (with codomain restricted to its image, making it an isomorphism), we do see $\mathbb{Q}(\pi)=\mathbb{Q}[\pi]\cong\mathbb{Q}[x]$, which is almost the result, I think $\mathbb{Q}[x]$ does actually equal $\mathbb{Q}(x)$

, but I'm not sure why, can anyone explain? Also, I don't see how $\mathbb{Q}[\pi]$ is a field, in fact I think it's only a integral domain.

So can someone explain how this example works?

Define $\phi:\mathbb{Q}(x)\rightarrow \mathbb{Q}(\pi)\,\,\,by\,\,\,\phi\left(\frac{f(x)} {g(x)}\right):=\frac{f(\pi)}{g(\pi)}$ ; it's easy to check this is a well-defined homomorphism of rings (why?) , and since it isn't

trivial and $\mathbb{Q}(x),\,\,\mathbb{Q}(\pi)$ are fields it automatically is 1-1 and onto

Also, in the exercise section it asks, for a indeterminate $x$ whether the statement: $\mathbb{Q}[\pi]\cong\mathbb{Q}[x]$ is true or false. It should be similar to the above example, can someone explain this too?

If you understood the above this last must be easier now to attack.

Tonio
• Jun 10th 2010, 07:14 PM
Horizon
Quote:

Originally Posted by tonio

Define http://www.mathhelpforum.com/math-he...36163430-1.gif ; it's easy to check this is a well-defined homomorphism of rings (why?) , and since it isn't

trivial and http://www.mathhelpforum.com/math-he...a8841fbf-1.gif are fields it automatically is 1-1 and onto

Hmm, so should $\mathbb{Q}(x)$ means the the field of fractions of $\mathbb{Q}[x]$, i.e. treating the indeterminate x as a transcendental, or should x be algebraic, so that $\mathbb{Q}[x]= \mathbb{Q}(x)$, since in the example it says: "the field http://www.mathhelpforum.com/math-he...362d580a-1.gif of rational functions over http://www.mathhelpforum.com/math-he...a4249b74-1.gif in the indeterminate http://www.mathhelpforum.com/math-he...155c67a6-1.gif.", or is this a typo?

Also, why is a homomorphism between two fields onto? (I can see it must be 1-1 since $\ker\phi$ is an ideal and the only ideal is {0} or http://www.mathhelpforum.com/math-he...362d580a-1.gif, since this isn't the zero function, the kernel is {0} so it is 1-1).

Using this method http://www.mathhelpforum.com/math-he...834411a9-1.gif with the isomorphism $\phi(f(x)) = f(\pi)$.