If F < E are fields, how is it possible for a representation X, to be irreducible as an F-representation, but reducible as an E-representation? Studding Character theory, and been bouncing back and forth between reading Dummit and Foote, and Character theory of finite groups by Martin Issacs.
In section 18.1 of Dummit and Foote, we are given a Bijection:
$$\{V: V  \text{ is an } FG-\text{module}\} \leftrightarrow \{ (\phi, V): \text{is an } F-\text{ represention from } G \rightarrow GL(V)\}$$
at the start of chapter 2 of Character theory of finite groups, It states that it is possible to have fields $F < E$, and an irreducible $F$-representation $X$, such that $X$ is reducible as an $E$-representation.
I don’t understand how $X$ can be irreducible as an $F$-representation and reducible as an $E$-representation. Since if $X$ was a representation with corresponding module $V$, and if $\phi$ and $\psi$ were $E$-representations with corresponding $E$-modules $W$ and $U$, such that  $X = \phi + \psi$. Then shouldn’t $W$ and $U$ also be $F$-modules and $V = W + U$, so there would have the be $F$-representations that sum up to $X$?
Somone gave me the example: "$X : C_3 \rightarrow GL_2(\mathbb{R})$,  where  $C_3 = \langle g \rangle$  is cyclic of order $3$ and  $X(g)$  is the matrix whose first row is  $[ 0 \ \ 1 ]$  and
whose second row is $[ -1 \ \  -1 ]$.  The eigenvalues of this matrix are not real (they are the two primitive cube roots of $1$). Therefore,  $X$  is irreducible as an $\mathbb{R}$-representation of  $C_3$. Since  $X(g)$  is diagonalizable over the complex numbers  $\mathbb{C}$, it follows that as a representation,  $X : C_3 \rightarrow GL_2(\mathbb{C})$ reduces.  So, over $\mathbb{C}$,  the representation is equivalent to a block diagonal sum of two linear representations of  $C_3$
I understand the example, but I'm still not seeing where and why my logic fails?

Let's apply your logic to the example, since clearly something must go wrong.
Here $F = \mathbb{R}$, $E = \mathbb{C}$, $X: C_3 \rightarrow GL_2(\mathbb{R})$ given by $g \mapsto X(g) = \begin{pmatrix} 0 & 1 \\ -1 & -1 \end{pmatrix}$.
Then as a $\mathbb{C}-$representation $(X,V)$ decomposes into two one dimensional representations, and the generators of these one dimensional representations are the eigenvectors of $X(g)$, you can work these out to be $\begin{pmatrix} \omega \\ 1 \end{pmatrix}$ and $\begin{pmatrix} \omega^2 \\ 1 \end{pmatrix}$ where $\omega$ is a cube root of unity.
Thus as a $\mathbb{C}-$ representation $(X,V)$ decomposes into $\big \langle \begin{pmatrix} \omega \\ 1 \end{pmatrix} \big \rangle \,\oplus \, \big \langle \begin{pmatrix} \omega^2 \\ 1 \end{pmatrix} \big \rangle $  (this is your $U$ and $W$, if you like) .
However as an $\mathbb{R}-$representation you cannot take the span of these eigenvectors since $\omega \notin \mathbb{R}$, so you no longer have an eigenvector decomposition and so the representation doesn't decompose.
