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I wanted to prove that if $F$ is a field and we consider the fraction field $F(x)$, then $\sqrt{x} \notin F(x)$.

I said that if $\sqrt{x} \in F(x)$ then there are $f,g \in F[x]$ such that $\bigl(\frac{f(x)}{g(x)}\bigr)^2=x$ so $f(x)^2=xg(x)^2$ so $2(\deg(f)-\deg(g))=1$ which is not possible. Is this true? Is there a quicker way to see this?

egreg
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roi_saumon
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  • You might want to pay special attention to the cases $f(x) = 0$ or $g(x) = 0$ since then, depending on custom, $\deg 0$ is either undefined, set to $-\infty$, or set to 0 but with the proviso that then $\deg(fg) = \deg(f) + \deg(g)$ is only valid when $f, g \ne 0$. Otherwise, it looks good. – Daniel Schepler Jan 16 '19 at 22:58
  • Another very similar argument would be instead of using the degree valuation, use the valuation of order wrt $x$, or in other words the multiplicity of 0 as a root. – Daniel Schepler Jan 16 '19 at 23:00
  • @DanielSchepler If $x=(f(x)/g(x))^2$, then $f(x)\ne0$ and $g(x)\ne0$. – egreg Jan 16 '19 at 23:28
  • @egreg Exactly, I was just commenting that since the zero polynomial is an exceptional case for degree in some way, no matter what convention you use, it would be a good idea to mention that fact explicitly in the proof. – Daniel Schepler Jan 17 '19 at 00:00
  • A variant of the argument using valuation at 0: assume $f,g$ are relatively prime. Then $f(x)^2 = x g(x)^2$ implies $x \mid f(x)$ implies $g(x) = x (f(x) / x)^2$ implies $x \mid g(x)$, contradiction. – Daniel Schepler Jan 17 '19 at 00:03

1 Answers1

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That's correct and I can't see a quicker way. You should observe that $f(x)\ne0$ and $g(x)\ne0$, but otherwise it's fine.

Probably I'd write $$ 2\deg f(x)=1+2\deg g(x) $$ and observe that the left-hand side is even, while the right-hand side is odd, but it's just personal preference.

By the way, if one defines a “degree” in $\mathbb{Z}$ by declaring that $\deg n$ is, for $n\ne0$, the number of prime factors in $n$, so $\deg 1=0$, $\deg 2=1$ and $\deg12=3$; this function is well defined because of uniqueness of factorization and $\deg(mn)=\deg m+\deg n$. Using this degree, irrationality of $\sqrt{2}$ is proved in exactly the same way; more generally, this proves the irrationality of $\sqrt{p}$ for every prime $p$. Using the “$p$-degree”, defined in an obvious way, one can prove the irrationality of $\sqrt{d}$ for every square free positive integer $d$.

egreg
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