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I am being asked to take a grad math course with no formal training in proofs and especially in Number Theory and Galois Theory(my bachelor's was not in mathematics at all). However I would like to work in the direction pointed out. All I want to know is how do I do the following assignment question. A hint would go a long way in helping me.

a) Let $f(x) \in \mathbb{C}[x]$. Show that $f(x)$ is a square in $\mathbb{C}(x)$ iff $f(x)$ is a square in $\mathbb{C}[x]$. Conclude $y^2-f(x)$ is irreducible in $\mathbb{C}(x)[y]$ iff $f(x)$ is not a square in $\mathbb{C}[x]$

b) Show that $f(x,y)=y^2-f(x)$ is an irreducible element of $\mathbb{C}[x,y]$ (Hint: Notice $\mathbb{C}[x,y] = \mathbb{C}[x][y]$. Formulate and use an analogue of Gauss's lemma.

I haven't started thinking about b) yet but any hint in that direction would be greatly appreciated.

For a) I know (<=) is immediate as $\mathbb{C}[x] \subset \mathbb{C}(x)$(its field of fractions). For (=>) let $f(x)=(G(x)/H(x))^2$ where $G(x), H(x)$ have coefficients in $\mathbb{C}$ and $H(x)$ is assumed to be monic(From Wikipedia). Now how do I go about proving that $f(x)$ is a square of some polynomial in $\mathbb{C}[x]$. I am so lost as in I don't even understand how to proceed from here.

Thank you.

Papa Delta
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2 Answers2

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Here's one way to think about it. $F = \mathbb{C}(X)$ is the set of fractions $\frac{f}{g}$, where $f, g \in R = \mathbb{C}[X]$. Now, every nonzero element of $R$ can be written uniquely as $c(X-c_1)^{n_1} \cdots (X - c_m)^{n_m}$, where $c,c_1, ... , c_m \in \mathbb{C}$, $n_1, ... , n_m$ are nonnegative integers, and $c \neq 0$.

It follows that every nonzero element of $F$ can be uniquely written as $c(X-c_1)^{n_1} \cdots (X - c_m)^{n_m}$, except now the $n_i$ can be integers, not just natural numbers. Such an element lies in $R$ if and only if all the integers $n_i$ are nonnegative. If $f$ is such an element of $F$, then

$$f^2 = c^2 (X-c_1)^{2n_1} \cdots (X - c_m)^{2n_m}$$

and this expression for $f^2$ is unique. It follows that $f^2$ cannot be in $R$ unless $f$ already is.

D_S
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  • Wow, you make it seem so easy. How did you get the idea to write an element of $R$ in such a way? Please do tell. – Papa Delta Nov 10 '16 at 18:54
  • Using the fundamental theorem of algebra :) – b00n heT Nov 10 '16 at 18:59
  • In $\mathbb{C}$, every polynomial factors into linear factors. In other words, $\mathbb{C}$ is an algebraically closed field. This is a result of the fundamental theorem of algebra. – Fimpellizzeri Nov 10 '16 at 19:01
  • Well, $\mathbb{C}[X]$ is a unique factorization domain, just like $\mathbb{Z}$. So if you understand how $\mathbb{Z}$ works, this is the same kind of reasoning. The essential property here is that in any UFD $R$, with quotient field $F$, you can choose a complete set $S$ of nonassociate prime elements of $R$, so that every nonzero element $x$ of $F$ can be written uniquely as $$\prod\limits_{p \in S} p^{n_p}$$ where $n_p$ are integers, all but finitely many of them are $0$, and you have $x \in R$ if and only all the $n_p$ are $\geq 0$. – D_S Nov 10 '16 at 19:04
  • I gave a similar answer here http://math.stackexchange.com/questions/2000092/how-to-prove-sqrt18-is-irrational-without-using-proof-by-contradiction/2000106#2000106 to show that an integer is a square in $\mathbb{Q}$ if and only if it is a square in $\mathbb{Z}$ (hence numbers like $\sqrt{2}, \sqrt{18}$ are not rational). – D_S Nov 10 '16 at 19:05
  • As noted implicitly by D_S the same result is true if $\mathbb C$ is replaced by a completely arbitrary field $k$. Algebraic closedness is irrelevant: if $k$ is not algebraically closed the $X-c_i$'s will be replaced by the family of irreducible monic polynomials $p_i(x)\in k[x]$ and the proof goes through word for word. – Georges Elencwajg Nov 10 '16 at 19:35
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Hint $\ \Bbb C[x]\, $ enjoys Euclidean division-with-remainder, so any two elements have a gcd, so the high-school proof of RRT = Rational Root Test remains valid. Thus if $\ \color{#c00}1\cdot y^2 - f\ $ has a root $\, y = g/h\in \Bbb C(x)$ then wlog $\,\gcd(g,h)=1\,$ so RRT $\,\Rightarrow\,h\mid\color{#c00} 1\,$ in $\,\Bbb C[x],\,$ so $\, y\in \Bbb C[x].$

Remark $\ $ This is simply a polynomial analog of this proof of irrationality of $\sqrt 2\,$ using RRT, where we have replaced the gcd domain $\,\Bbb Z\,$ by the GCD domain $\,\Bbb C[x].\, $ The linked simple proof of RRT works over any gcd domain, i.e. an integral domain where any two elements have a gcd.

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Bill Dubuque
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