Franklin's Notes

Decomposable form

A form $F(x_1,\cdots,x_n)$ with coefficients in $\mathbb Q$ is called decomposable if it factors into linear factors in some extension $\Omega/\mathbb Q$. It is a theorem that if a form is decomposable, i.e. splits into linear factors in some extension, then it does so in some algebraic number field :

Proposition 1. Any rational decomposable form splits into linear factors in some algebraic number field.

Proof. Let us suppose that the form $F$ is decomposable, so that there exists an extension $\Omega/\mathbb Q$ such that for some constants $\alpha_{ij}\in\Omega$. Let us factor out the coefficients of $x_1$ from each linear term to obtain a constant $A=\alpha_{11}\alpha_{21}\cdots\alpha_{m1}$, such that Now, if we choose a specific $x_k$ with $k\ne 1$ and set $x_k=1$ and $x_j=0$ for all $j\ne k$, then we obtain Notice that since $F$ is a polynomial with rational coefficients, the above must be a polynomial with rational coefficients, so that $-\alpha_{1k},\cdots,-\alpha_{nk}$, and therefore $\alpha_{1k},\cdots,\alpha_{nk}$ are the roots of some degree $n$ polynomial with rational coefficients, making them algebraic numbers. We may repeat a similar argument to prove that this is also true for $k=1$. Hence, if we consider the finite-degree extension field of $\mathbb Q$ generated by all of the $\alpha_{ij}$, we will have that $F$ splits into linear factors over this extension field, and the theorem is proven. $\blacksquare$

It can easily be shown that every form in $n=2$ variables is decomposable:

Proposition 2. Every rational form $F(x,y)$ with $n=2$ variables is decomposable.

Proof. Because all terms of a form $F$ have the same degree $d$, we may write it in the form $F(x,y)=y^d G(x/y)$ where $G$ is a polynomial of one variable with degree $d$. This polynomial must split in some extension of $\mathbb Q$, so that for some $\alpha_i\in\Omega/\mathbb Q$ with $\Omega$ an algebraic number field. Then we have that so that $F$ splits into linear factors over $\Omega$ as desired. $\blacksquare$

However, in three variables, we have the following indecomposable form: