MHB Show That F[u]=F(u): Proving with Theorem

[mathmari]

Hey!

We have the following theorem:
Let $F$ be a field and $p(x)\in F[x]$ irreducible. Then there is a field $K$, of which $F$ is a subfield with the following properties:

1. $\exists u\in K$ with $p(u)=0$, i.e., $p$ has a root in $K$.
2. $K=F$
   

Referring to this theorem, I want to show that $F=F(u)$.

The elements of $F(u)$ are of the form $\frac{f(u)}{g(u)}$, where $f(u), g(u)\in F$ and $g(u)\neq 0$.
The elements of $F$ are the polynomials of $u$ with coefficients in $F$.
So, we have that $F\subseteq F(u)$ for $g(u)=1$, right?

At the theorem we have that $K=F$ is a field, that means that each non-zero element has an inverse. Therefore, $\forall f(u)\in F \ \ \exists g(u)\in F : \frac{f(u)}{g(u)}\in F \Rightarrow F(u)\subseteq F$.

So, we conclude that $F=F(u)$.

Is everything correct? Could I improve something? (Wondering)

 

[caffeinemachine]

Hey!

We have the following theorem:
Let $F$ be a field and $p(x)\in F[x]$ irreducible. Then there is a field $K$, of which $F$ is a subfield with the following properties:

1. $\exists u\in K$ with $p(u)=0$, i.e., $p$ has a root in $K$.
2. $K=F$
   

Referring to this theorem, I want to show that $F=F(u)$.

The elements of $F(u)$ are of the form $\frac{f(u)}{g(u)}$, where $f(u), g(u)\in F$ and $g(u)\neq 0$.
The elements of $F$ are the polynomials of $u$ with coefficients in $F$.

The elements of $F$ are the polynomials in $F[x]$ evaluated at $u$.

So, we have that $F\subseteq F(u)$ for $g(u)=1$, right?

At the theorem we have that $K=F$ is a field, that means that each non-zero element has an inverse. Therefore, $\forall f(u)\in F \ \ \exists g(u)\in F : \frac{f(u)}{g(u)}\in F \Rightarrow F(u)\subseteq F$.

Saying that each nonzero element of $F$ is invertible in $F$ is not the same as what you have written. It means that
$$\forall f(u)\in F, \exists g(u)\in F \text{ such that } f(u)g(u)=1$$
This helps because it tells us that whenever $g(u)\in F$ is nonzero, we have $1/g(u)\in F$. Therefore $f(u)/g(u)\in F$ for all $f(u)\in F$ and all nonzero $g(u)\in F$, showing that $F(u)\subseteq F$.

So, we conclude that $F=F(u)$.

Is everything correct? Could I improve something? (Wondering)

I have added my comments in red in the above.

 

[mathmari]

I have added my comments in red in the above.

Let $F$ be a field and $p(x)\in F[x]$ irreducible. Then there is a field $K$, of which $F$ is a subfield with the following properties:

1. $\exists u\in K$ with $p(u)=0$, i.e., $p$ has a root in $K$.
2. $K=F$
   

The elements of $F(u)$ are of the form $\frac{f(u)}{g(u)}$, where $f(u), g(u)\in F$ and $g(u)\neq 0$.
The elements of $F$ are the polynomials in $F[x]$ evaluated at $u$.

So, we have that $F\subseteq F(u)$ for $g(u)=1$.

At the theorem we have that $K=F$ is a field, that means that each nonzero element of $F$ is invertible in $F$, so
$$\forall f(u)\in F, \exists g(u)\in F \text{ such that } f(u)g(u)=1$$
So, whenever $g(u)\in F$ is nonzero, we have $1/g(u)\in F$. Therefore $f(u)/g(u)\in F$ for all $f(u)\in F$ and all nonzero $g(u)\in F$, showing that $F(u)\subseteq F$.

Why do we have that whenever $g(u)\in F$ is nonzero, we have $1/g(u)\in F$ and not whenever $f(u)\in F$ is nonzero, we have $1/f(u)\in F$ ?
We know that for each $f(x)$ there is a $g(x)$, but we don't know that for each $g(x)$ there is a $f(x)$, do we? (Wondering)