How to simplify an ideal (f,g) division with remainder suppose we have an ideal $(f,g)\subseteq K[X]$ for an euclidean domain $K$.
Where $f,g\in K[X]$. 
Can we always simplify the ideal by using division with remainder?
Example:
$(1+x^2, 5+x^3)\subseteq\mathbb{Z}[X]$
It is $(1+x^2, 5+x^3)=(x-5, 26)$
How do we attempt the division with remainder?
$(x^3+5)\div (x^2+1)=x~~ R~~5-x$ ($R~~ 5-x$ means that the remainder is $5-x$)
Since $x^3+5=x(x^2+1)+(5-x)$
Now we use a 2nd division with remainder:
$(x^2+1)\div (-x+5)= -x+5~~ R~~ 26$
Since $x^2+1=(-x+5)(-x-5)+26$
And we conclude $(x^3+5, x^2+1)=(x-5, 26)$
Can this always be done? We need an euclidean domain for this, right? (Else we can not use division with remainder)
The "algorithm" is always(?) like above?
We first divide $f\div g$  (with $\deg f\geq \deg g$) and get a remainder $R_1$, then we divide
$g$ by $R_1$ and get a remainder $R_2$ and our ideal reduces too:
$(f,g)=(R_1, R_2)$
Can you approve this?
Thanks in advance.
 A: 
suppose we have an ideal $(f,g)\subseteq K[X]$ for an euclidean domain $K$.
  Where $f,g\in K[X]$. 
  Can we always simplify the ideal by using division with remainder?

Yes, in a Euclidean domain, you can always divide one nonzero element by another, and simplify every set of generators of an ideal to a single element that generates the entire ideal.

Example:
  $(1+x^2, 5+x^3)\subseteq\mathbb{Z}[X]$ ...

You do realize that $\mathbb Z[x]$ is not a Euclidean domain, right?

Can this always be done? We need an euclidean domain for this, right? (Else we can not use division with remainder)
  The "algorithm" is always(?) like above?

You can always divide by a monic polynomial in any polynomial ring $R[x]$. You don't have to have a Euclidean domain to do that.  The division you did does in fact produce an equivalent set of generators with lesser degrees in cases like you describe.
A: I guess not. Remember that $\mathbf{Z}[x]$ is not an euclidean domain. In fact, $I = (2, x)$ is not principal. So, it's not possible do this algorithm for every pairs of polynomials. If $F$ is a field, then $F[x]$ is euclidean. At this case, if $I=(f,g)$ with $\deg f \geq \deg g$, there is $q, r \in F[x]$ such that $f = q g + r$ and $\deg r < \deg g$ or $r = 0$. So $(f, g) = (r, g)$. We can continue this process and it'll end in a finite number of steps. Everything is ok and you can prove that I is principal. It's not a surprise: we can use Bezout's Theorem to obtain $I = (\gcd(f, g))$.
Note that the only problem with your algorithm is the existence of a division process.
