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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 $degree<n$, 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?

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  • $\begingroup$ yes, thank you. $\endgroup$ – Dan Leonte Jul 6 '17 at 17:44
  • $\begingroup$ 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 $\frac{b^2-4ac}{4a^2} = -d^2$ $\endgroup$ – reuns Jul 6 '17 at 17:47
  • $\begingroup$ 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? $\endgroup$ – Dan Leonte Jul 6 '17 at 21:52
  • $\begingroup$ Yes of course it is unique : the elements of the quotient are of the form $\{ f(x) + (p(x)), deg(f) < deg(p) \}$ $\endgroup$ – reuns Jul 6 '17 at 23:10
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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.

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