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Let $\alpha\in\mathbb R$. Prove that $\mathbb Q(\alpha)\cong\mathbb Q(x)$ if and only if $\alpha$ is transcendental

So is this saying that we need everything (all expressions involving x) to split a transcendental? This chapter is defeating me. I don't see what is being said here.

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You can easily define an isomorphism $\phi: \mathbb{Q}[x]\to \mathbb{Q}[\alpha]$ namely $\phi(p(x))=p(\alpha)$ (it's clearly surjective and the kernel is $0$ since $\alpha$ is trascendental). From this, it follows that $\mathbb{Q}(x)$ is isomorphic to $\mathbb{Q}(\alpha)$ (remember that $\mathbb{Q}(x)$ is the field of fractions of the ring $\mathbb{Q}[x]$)

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  • $\begingroup$ Thanks, this is a good lead. $\endgroup$ – OLP Apr 4 '15 at 20:54
  • $\begingroup$ Why is it surjective? Aren't there other elements in the extension? $\endgroup$ – OLP Apr 12 '15 at 4:27
  • $\begingroup$ It's surjective since every element is $\mathbb{Q}[\alpha]$ is a polynomial in $\mathbb{Q}[x]$ evaluated on $\alpha$ (it is a useful characterizaton of $\mathbb{Q}[\alpha]$, it's even a definition in some books) $\endgroup$ – Daniel Apr 13 '15 at 22:54
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Think about what happens if $\alpha$ is something like $\sqrt2$. Then in $\mathbb Q(\alpha)$ you have things happening like $\alpha^2-2=0$. But in $\mathbb Q(x)$, $x$ is a variable and so $x^2-2\not=0$, in fact no polynomial is zero in $\mathbb Q(x)$ except 0 itself. So if $\alpha$ is transcendental it does not satisfy any polynomial so $P(\alpha)$ is not zero for any polynomial, just like in $\mathbb Q(x)$.

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  • $\begingroup$ Thanks. So Q(x) is Q[x]. This helps, but I'm still thrown off by the notation Q(x), exactly because x is a variable and not an element. $\endgroup$ – OLP Apr 4 '15 at 20:53
  • $\begingroup$ $\mathbb{Q}(x)$ IS NOT $\mathbb{Q}[x]$. The first one is a field, but the second one is not. Are you sure you know the definition of each one? $\endgroup$ – Daniel Apr 4 '15 at 21:07
  • $\begingroup$ $ℚ[X] ≠ ℚ(X)$ if $ℚ[X]$ denotes the ring of polynomials @OLP. By the way, I recommend using capital letters if you are refering to indeterminates for polynomials/rational functions. $\endgroup$ – k.stm Apr 4 '15 at 21:07
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    $\begingroup$ $\mathbf{Q}(x)\neq\mathbf{Q}[x]$ ($x$ an indeterminate). $\mathbf{Q}(x)$ is the smallest field extension of $\mathbf{Q}$ containing $x$, or equivalently the fraction field of $\mathbf{Q}[x]$. Explicitly, $\mathbf{Q}[x]$ is polynomials in $x$ with $\mathbf{Q}$ coefficients, and $\mathbf{Q}(x)$ is rational functions in $x$ with $\mathbf{Q}$ coefficients. If $\alpha$ is algebraic over $\mathbf{Q}$, then it is true that $\mathbf{Q}(\alpha) = \mathbf{Q}[\alpha]$, but the algebraic assumption on $\alpha$ is essential: $1/x$ is not a polynomial in $x$ (no polynomial has poles, but $1/x$ does). $\endgroup$ – Stahl Apr 4 '15 at 21:07
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This is essentially the same as Daniel’s answer, but a bit fleshed out with more background.


Look up localization and especially its universal property if you don’t know about this yet.

There’s a ring homomorphism $φ\colon ℚ[X] → ℚ(α),~X ↦ α$ with $ℚ[α] ⊂ \operatorname{img} φ$. This map has trivial kernel if and only if $α$ isn’t root of any polynomials with rational coefficients, i.e. if and only if $α$ is transcendental. But in this case, the situation of the universal property of localization for $S = ℚ[X]\setminus \{0\}$ is given because $ℚ(α)$ is a field and now $φ(S) ⊂ ℚ(α) \setminus \{0\}$.

The localization of $ℚ[X]$ by $S$ is $S^{-1}ℚ[X] = ℚ(X)$. You only need to show that the resulting map $ℚ(X) → ℚ(α)$ stays injective and becomes surjective.

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