- #1
mathmari
Gold Member
MHB
- 5,049
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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:
I want to show that $\mathbb{Q}[\pi]\subsetneqq \mathbb{Q}(\pi)$.
The elements of $\mathbb{Q}(\pi)$ are of the form $\frac{f(\pi)}{g(\pi)}$, where $f(\pi), g(\pi)\in \mathbb{Q}[\pi]$ and $g(\pi)\neq 0$.
The elements of $\mathbb{Q}[\pi]$ are the polynomials of $\pi$ with coefficients in $\mathbb{Q}$.
So, we have that $\mathbb{Q}[\pi]\subseteq \mathbb{Q}(\pi)$ for $g(\pi)=1$, right?
We have that $\pi$ is not a root of a polynomial with rational coefficients, right?
How could we prove this? (Wondering)
That means that the first property of the above theorem is not satisfied. Do we conclude from that that $\mathbb{Q}[\pi]$ is not a field, and so $\mathbb{Q}[\pi]\subsetneqq \mathbb{Q}(\pi)$ ? (Wondering)
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:
- $\exists u\in K$ with $p(u)=0$, i.e., $p$ has a root in $K$.
- $K=F$
I want to show that $\mathbb{Q}[\pi]\subsetneqq \mathbb{Q}(\pi)$.
The elements of $\mathbb{Q}(\pi)$ are of the form $\frac{f(\pi)}{g(\pi)}$, where $f(\pi), g(\pi)\in \mathbb{Q}[\pi]$ and $g(\pi)\neq 0$.
The elements of $\mathbb{Q}[\pi]$ are the polynomials of $\pi$ with coefficients in $\mathbb{Q}$.
So, we have that $\mathbb{Q}[\pi]\subseteq \mathbb{Q}(\pi)$ for $g(\pi)=1$, right?
We have that $\pi$ is not a root of a polynomial with rational coefficients, right?
How could we prove this? (Wondering)
That means that the first property of the above theorem is not satisfied. Do we conclude from that that $\mathbb{Q}[\pi]$ is not a field, and so $\mathbb{Q}[\pi]\subsetneqq \mathbb{Q}(\pi)$ ? (Wondering)