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I want to prove the theorem below:

Theorem. Let $F \subseteq E$ be a totally transcendental extension and let $f\in F[X]$ be irreducible. Show that $f$ is irreducible in $E[X]$

I could not go further than this:

Suppose that $L$ is the splitting field of $f$ over $E$, so that, $f(X)=(X-\alpha_1)\dots(X-\alpha_n)$. We can say that none of $\alpha_i$'s belongs to $F$ since otherwise $f$ would be reducible in $F[X]$. Now we should use totally transcendentalility somehow but I do not know a reasonable conclusion.

How can I complete the proof?

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Suppose that $f=gh$ is a non-trivial decomposition of $f$ over $E$. Without loss of generality we may assume $f$, $g$ and $h$ monic.

Correspondingly, the set $A$ of the roots of $f$ can be written as $A_g\cup A_h$, where $A_g$ (resp. $A_h$) is the set of roots of $g$ (resp. $h$) and neither subset is empty.

Then, the coefficients of, say, $g$ are the elementary symmetric functions in $\deg g$ variables evaluated at the elements in $A_g$ and as such are algebraic over $F$. Same for $h$. But the only elements in $E$ algebraic over $F$ are those in $F$ already.

So, the decomposition holds in $F[X]$ and this is a contradiction.

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  • $\begingroup$ By non-trivial decomposition, do you mean that both $g,h$ has degree at least $1$? Also, what is the non trivial decomposition of roots, is that mean they do not share a common root? $\endgroup$ – Ninja May 8 '17 at 12:01
  • $\begingroup$ @Ninja : Yes, $g$ and $h$ are not units in $E[X]$. I slightly changed the second sentence to avoid the ambiguity that you mention (nowhere is written that is a disjoint union) $\endgroup$ – Andrea Mori May 8 '17 at 13:44
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Consider any factor of $f$ over some algebraic closure $\bar{F}$. Then the coefficients of the factor are algebraic over $F$. So if $g$ is a monic factor of $f$ in $ E[X]$ all the coefficients lie in $F$ . Thus $f=g$

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Assuming the theorem is true, then the intuition going forward is that, if $f$ did reduce in $E[X]$, we should arrive at a contradiction with regards to $E$ being a totally transcendental extension. In other words, a reduction should imply that $E$ would have to contain algebraic elements. Let's fill in the details to try and elicit this contradiction:

Suppose $f$ did reduce as $f(x) = g(x)h(x)$ for some $g, h \in E[X]$. The fact that we could not reduce it as such in $F[X]$ implies that some coefficients of $g$ or of $h$ were not elements of $F$, but are elements of $E$.

What do we know about the coefficients of a polynomial? Assuming $g$ and $h$ are monic, their coefficients are elementary symmetric polynomials evaluated at their roots** (or an algebraic multiple of such evaluations if not monic). The roots of $g$ and $h$ are also roots of $f \in F[X]$ (thus are algebraic over $F$), and it is a fact that the set of algebraic elements is closed under addition and multiplication, so what can we say about these coefficients?

And given that every element in $E \setminus F$ was supposed to be transcendental...




**As an intuition-builder, imagine we have factored $f$ completely in its splitting field: $$f(x) = (x-\alpha_1)(x-\alpha_2) \cdots (x-\alpha_n)$$ Now imagine multiplying out back into the form $f(x) = x^n + c_{n-1}x^{n-1} + \cdots + c_1x + c_0$. Notice that these coefficients $c_k$ will be in terms of the roots of $f$. For example, $c_0 = (-1)^n \alpha_1 \alpha_2 \cdots \alpha_n$.

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  • $\begingroup$ The theorem is exactly from Martin Isaac's Algebra. It says that the extension is totally transcendental, but not purely. $\endgroup$ – Ninja May 8 '17 at 12:02
  • $\begingroup$ I believe the theorem is true; I was just trying to give a stream of consciousness so that it doesn't seem like I'm pulling magic out of a hat in constructing a proof. And oops, I didn't even realize "purely" and "totally" transcendental extensions were different things. I was assuming what you meant is that every element in $E$ but not in $F$ is transcendental over $F$. $\endgroup$ – Kaj Hansen May 8 '17 at 12:05
  • $\begingroup$ Yes, I meant it. What will change now? $\endgroup$ – Ninja May 8 '17 at 12:07
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    $\begingroup$ See my previous comment; I added some explanation. I think we're good :) $\endgroup$ – Kaj Hansen May 8 '17 at 12:16
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    $\begingroup$ Hope that helps! I know there's a lot of ink on that page I linked; try not to get too bogged down trying to parse all of it. Instead, try some small examples, like expanding a quadratic $(x-a)(x-b)$ or a cubic $(x-a)(x-b)(x-c)$ and look at how all the coefficients end up expressed in terms of the roots $a, b, c$. For example, you can see pretty easily that the constant term for both is going to be the (signed) product of all the roots. I'm going to get some sleep now, but I'll be up again in a few hours. Good luck! $\endgroup$ – Kaj Hansen May 8 '17 at 12:39

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