# Fixed roots and quotient rings

I'm reading a paper by Smart and Vercauteren on homomorphic encryption (http://eprint.iacr.org/2011/133). I don't understand a specific statement around quotient rings. The authors state that for a polynomial $F(X)$ in indeterminate $X$, elements of quotient rings such as $\mathbb F_2 (X)/F(X)$ and $\mathbb Q(X)/F(X)$ can be represented as polynomials in some fixed root of $F(X)$ in the algebraic closure of the base field. I don't understand this statement, could someone please explain?. Would some of the more comprehensive texts on algebra such as Dummit and Foote be helpful in explaining such constructs? If not, any other recommendations for texts?

• Pick $\alpha \in \overline{\mathbb Q}$ such that $F(\alpha) = 0$ ($\alpha$ is the fixed root) and map $\mathbb Q[x]/F(x) \to \overline{\mathbb Q}$ by sending $g(x)$ to $g(\alpha)$. And yes, Dummit and Foote's sections on field and Galois theory would be helpful to you. – Jim Nov 26 '13 at 20:17

The statement is only true if $F$ is irreducible over the base field, which is to say that it cannot be factored into two polynomials of lower degree over that field.
Let $k$ be a field and $f(x)\in k[x]$ an irreducible polynomial over $k$. Let $\alpha$ be a root of $f$ in $\bar k$, the algebraic closure of $k$.
Consider the homomorphism $$\phi_\alpha: k[x]\to \bar k$$ given by evaluation at $\alpha$, i.e. $\phi_\alpha(g) = g(\alpha)$. Since $f$ is irreducible, we have that $g(\alpha)=0$ if and only if $g(x) = q(x)f(x)$ for some $q\in k[x]$. In other words, the kernel of $\phi_\alpha$ is precisely $(f)$, the principal ideal generated by $f$. Note also that the image of $\phi_\alpha$ is $k[\alpha]$, the subring of $\bar k$ generated by $k$ and $\alpha$. Then by the first isomorphism theorem, $\phi_\alpha$ descends to an isomorphism
$$k[x]/(f(x))\stackrel{\sim}{\to} k[\alpha].$$
This is the correspondence between elements of the quotient ring and polynomials in the fixed root $\alpha$.