# Uniqueness of representation of elements of quotient by minimal polynomial in a polynomial ring

Let $$F$$ be a subfield of $$K$$. It is easy to see that if $$\alpha$$ is algebraic over $$F$$, then its minimal polynomial $$p(x)$$ is (monic) unique and irreducible.

Taking the quotient, $$F[x]/p(x)F[x]$$ is a field (as $$p(x)$$ is irreducible), and any element $$g\in F$$ can be represented uniquely as $$p(x)F(x)$$+ a polynomial of $$degree, where $$n=deg \ p(x).$$
I can immediately see that if instead of taking the minimal polynomial $$p(x)$$, we take any polynomial $$f$$ and quotient by its ideal $$(f)$$, we get a ring (not necessarly a field).

In this case, the representation does not need to be unique, right?

• yes, thank you. Jul 6 '17 at 17:44
• The representation of what is not unique ? $\ \mathbb{Q}[x]/(x^2+1) \cong \mathbb{Q}(i) = \mathbb{Q}(di+e) \cong \mathbb{Q}[x]/(ax^2+bx+c)$ whenever $\frac{b^2-4ac}{4a^2} = -d^2$ Jul 6 '17 at 17:47
• any element $g\ in F[x]$ can be repesented ,by quotinent, $F[x]/p(x)F[x]$ as ${a_0+a_1*x+...+a_{n-1}*x^{n-1}}$. Is this representations unique if $f$ is a random poly? Jul 6 '17 at 21:52
• Yes of course it is unique : the elements of the quotient are of the form $\{ f(x) + (p(x)), deg(f) < deg(p) \}$ Jul 6 '17 at 23:10

No, the representation is unique regardless of what (nonzero) polynomial $f$ is. For any polynomials $f$ and $g$ with $f$ nonzero, there exist unique polynomials $q$ and $r$ with $\deg r<\deg f$ such that $$g=qf+r.$$ This is just the statement of polynomial division with remainder, which has nothing to do with whether the polynomials are irreducible.