# Math Help - Field Extension

1. ## 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?

2. 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

3. Originally Posted by tonio

Define ; it's easy to check this is a well-defined homomorphism of rings (why?) , and since it isn't

trivial and 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 of rational functions over in the indeterminate .", 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 , since this isn't the zero function, the kernel is {0} so it is 1-1).

Using this method with the isomorphism $\phi(f(x)) = f(\pi)$.