Totally transcendental extension and irreducibility.

Question

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?

Answer 1

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...

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**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, $c_0 = (-1)^n \alpha_1 \alpha_2 \cdots \alpha_n$.

Answer 2

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.

Answer 3

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 $f=g$