MHB Show that Q[π] is not a subset of Q(π)

[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. $K=F$
   

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)

 

[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. $K=F$
   

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)

Let $F=\mathbf Q$. Let $f:F[x]\to F[\pi]$ be the map which is identity on $F$ and sends $x$ to $\pi$. Using the transcendence of $\pi$ show that $f$ is a ring isomorphism. Then note that $F[x]\subsetneq F(x)$. This is because $x$ is not invertible in $F[x]$.

 

[mathmari]

Let $F=\mathbf Q$. Let $f:F[x]\to F[\pi]$ be the map which is identity on $F$ and sends $x$ to $\pi$. Using the transcendence of $\pi$ show that $f$ is a ring isomorphism. Then note that $F[x]\subsetneq F(x)$. This is because $x$ is not invertible in $F[x]$.

$$f(a(x)+b(x))=f((a+b)(x))=(a+b)(\pi)=a(\pi)+b(\pi)=f(a(x))+f(b(x)) \\ f(a(x)\cdot b(x))=f((a\cdot b)(x))=(a\cdot b)(\pi)=a(\pi)\cdot b(\pi)=f(a(x))\cdot f(b(x))$$
So, $f$ is a ring homomorphism.

For all $c(\pi)=\sum_{i=0}^nq_i\pi^i \in F[\pi]$ we have that $c(\pi)=\sum_{i=0}^nq_i\pi^i =\sum_{i=0}^nf(q_i)f(x)^i =f(\sum_{i=0}^nq_ix^i)=:f(b(x))$.
So, $f$ is surjective.

If $f(a(x))=f(b(x))$ then we have that $a(\pi)=b(\pi)\Rightarrow (a-b)(\pi)=0$. SInce $\pi$ is transcedent we conclude that $a-b\equiv 0\Rightarrow a(x)=b(x)$.
So, $f$ is injective.

Therefore, $f$ is a ring isomorphism.

Is this correct? (Wondering)

How do we show that $x$ is not invertible in $F[x]$ ? (Wondering)

 

[caffeinemachine]

$$f(a(x)+b(x))=f((a+b)(x))=(a+b)(\pi)=a(\pi)+b(\pi)=f(a(x))+f(b(x)) \\ f(a(x)\cdot b(x))=f((a\cdot b)(x))=(a\cdot b)(\pi)=a(\pi)\cdot b(\pi)=f(a(x))\cdot f(b(x))$$
So, $f$ is a ring homomorphism.

For all $c(\pi)=\sum_{i=0}^nq_i\pi^i \in F[\pi]$ we have that $c(\pi)=\sum_{i=0}^nq_i\pi^i =\sum_{i=0}^nf(q_i)f(x)^i =f(\sum_{i=0}^nq_ix^i)=:f(b(x))$.
So, $f$ is surjective.

If $f(a(x))=f(b(x))$ then we have that $a(\pi)=b(\pi)\Rightarrow (a-b)(\pi)=0$. SInce $\pi$ is transcedent we conclude that $a-b\equiv 0\Rightarrow a(x)=b(x)$.
So, $f$ is injective.

Therefore, $f$ is a ring isomorphism.

Is this correct? (Wondering)

How do we show that $x$ is not invertible in $F[x]$ ? (Wondering)

That's correct. To see $x$ is not invertible in $F[x]$, assume on the contrary that it is. Then we can find a polynomial $p(x)\in F[x]$ such that $xp(x)=1$. Now compare the degrees on both the sides.

 

[mathmari]

To see $x$ is not invertible in $F[x]$, assume on the contrary that it is. Then we can find a polynomial $p(x)\in F[x]$ such that $xp(x)=1$. Now compare the degrees on both the sides.

We assume that $x$ is invertible in $F[x]$, so there is a polynomial $p(x)\in F[x]$ such that $xp(x)=1$.
We have that $\deg (xp(x))=\deg (1) \Rightarrow \deg (x)+\deg (p(x))=\deg (1) \Rightarrow 1+\deg (p(x))=0$, a contradiction, since $\deg (p(x))\geq 0$.
Is this correct? (Wondering)

Why did we need to show that $f$ is a ring isomorphism? (Wondering)

 

[caffeinemachine]

We assume that $x$ is invertible in $F[x]$, so there is a polynomial $p(x)\in F[x]$ such that $xp(x)=1$.
We have that $\deg (xp(x))=\deg (1) \Rightarrow \deg (x)+\deg (p(x))=\deg (1) \Rightarrow 1+\deg (p(x))=0$, a contradiction, since $\deg (p(x))\geq 0$.
Is this correct? (Wondering)

Why did we need to show that $f$ is a ring isomorphism? (Wondering)

If $f:R\to S$ is a ring isomorphism between two rings, then an element $r\in R$ is invertible in $R$ iff $f(r)$ is invertible in $S$. Since $x$ is not invertible in $F[x]$, as you have just shown, we have $\pi$ is not invertible in $\mathbf Q[\pi]$. Is it clear now?

 

[mathmari]

If $f:R\to S$ is a ring isomorphism between two rings, then an element $r\in R$ is invertible in $R$ iff $f(r)$ is invertible in $S$.

When an element $r\in R$ is invertible in $R$ then $\exists r^{-1} : rr^{-1}=1$.
Then $f(rr^{-1})=f(1) \Leftrightarrow f(r)f(r^{-1})=f(1) \Leftrightarrow f(r)(f(r))^{-1}=f(1)=$, since $f$ is an homomorphism.
Therefore, $f(r)$ is invertible in $S$.

Is this correct? (Wondering)

 

[caffeinemachine]

When an element $r\in R$ is invertible in $R$ then $\exists r^{-1} : rr^{-1}=1$.
Then $f(rr^{-1})=f(1) \Leftrightarrow f(r)f(r^{-1})=f(1) \Leftrightarrow f(r)(f(r))^{-1}=f(1)=$, since $f$ is an homomorphism.
Therefore, $f(r)$ is invertible in $S$.

Is this correct? (Wondering)

Yes this is fine.

 

[mathmari]

Yes this is fine.

Thank you very much! (Sun)