I'm having a hard time understanding section 29,30,31 of Fraleigh.

In 29.16 example, what is the field $\mathbb{Q}(\pi)$? and why is it isomorphic to the field $\mathbb{Q}(x)$ of rational functions over $\mathbb{Q}$?

(According to the definition, the field $\mathbb{Q}(\pi)$ is the smallest subfield of $E$ (extension field of $\mathbb{Q}$) containing $\mathbb{Q}$ and $\pi$.)

Thank you!


Define a ring homomorphism $f\colon \mathbb{Q}[x]\to\mathbb{R}$ by $$f(p)=p(\pi)$$ so that (for example) $$f(\tfrac{1}{3})=\tfrac{1}{3},\quad f(x)=\pi,\quad f(2x^2+5)=2\pi^2+5,$$ and so on. The image of a ring homomorphism is a subring of the codomain; in the case of this particular ring homomorphism $f$, the image is given the name $\mathbb{Q}[\pi]$. It is the "smallest" subring of $\mathbb{R}$ that contains $\mathbb{Q}$ and $\pi$.

What is the kernel of this homomorphism $f$? That is, what polynomials $p\in\mathbb{Q}[x]$ have $\pi$ as a root? The answer is none (other than the obvious $p=0$). This is what it means for $\pi$ to be transcendental (in 1882, Lindemann proved that $\pi$ is transcendental). The first isomorphism theorem for rings now tells us that $$\mathbb{Q}[x]/(\ker f)\cong \mathbb{Q}[\pi]$$ but since the kernel of $f$ is trivial, this statement just says that $\mathbb{Q}[x]\cong \mathbb{Q}[\pi]$. In other words, we see that $f$ is a ring isomorphism from $\mathbb{Q}[x]$ to $\mathbb{Q}[\pi]$.

Now see if you can prove the following general fact: if two integral domains $D_1$ and $D_2$ are isomorphic, then their respective fields of fractions $\mathrm{Frac}(D_1)$ and $\mathrm{Frac}(D_2)$ are also isomorphic.

  • $\begingroup$ May I ask how to prove that the image of $f$ is $\Bbb Q[\pi]?$ $\endgroup$ – PropositionX Mar 25 '17 at 16:07

It is the smallest subfield of $\mathbb{C}$ that contains both $\pi$ and $\mathbb{Q}$. It can be shown that it is precisely


Now, why is it isomorphic to $\mathbb{Q}(T)$?

Well, consider the smallest subring of $\mathbb{C}$ containing $\mathbb{Q}$ and $\pi$. This is just $\mathbb{Q}[\pi]$ (polynomials in $\pi$). You obviously get a $\mathbb{Q}$-algebra map $\mathbb{Q}\to\mathbb{Q}[\pi]$ defined by $f(T)\mapsto f(\pi)$ (you can verify this directly, or use the general fact that polynomial rings are the free commutative $\mathbb{Q}$-algebras). Now, since $\pi$ is trancendental (it doesn't satisfy any polynomial over $\mathbb{Q}$), this map is injective, and so you can lift this to a map $\mathbb{Q}(T)\to\mathbb{Q}(\pi)=\text{Frac}(\mathbb{Q}[\pi])$ in the obvious way. By definition of $\mathbb{Q}(\pi)$ (the explicit one), this is a surjective $\mathbb{Q}$-algebra map. But, since $\mathbb{Q}(T)$ is a field, it must actually be an isomorphism.

  • $\begingroup$ It should be 'both $\pi$ and $\mathbb{Q}$' $\endgroup$ – Sigur Aug 14 '13 at 0:07
  • 1
    $\begingroup$ @DonAntonio Thanks! $\endgroup$ – Alex Youcis Aug 14 '13 at 0:07
  • $\begingroup$ what is Q[π] (polynomials in π)? is it equal to values of polynomials such that coefficients are Q and x value is π? $\endgroup$ – user88310 Aug 14 '13 at 0:11
  • $\begingroup$ @user88310 Exactly! It's just the image of the map $f(T)\mapsto f(\pi)$. $\endgroup$ – Alex Youcis Aug 14 '13 at 0:11
  • $\begingroup$ @Sigur Although more concisely it could be the smallest subfield of $\mathbb C$ containing $\pi$. $\endgroup$ – JSchlather Aug 14 '13 at 2:25

Denoting by $\,x\,$ any variable (or unknown or transcendental element over $\,\Bbb Q\,$) , define the map

$$\phi:\Bbb Q(x)\to\Bbb Q(\pi)\;,\;\;\phi(f(x)):=f(\pi)$$

Now prove the above is a bijective ring homomorphism...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.