Reducing the Coefficients of a Polynomial Modulo an Ideal

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The explanation can be found in the documentation for QuotientRing:

ASSUMPTION:

I has a method I.reduce(x) returning the normal form of elements $x \in R$. In other words, it is required that I.reduce(x)==I.reduce(y) $\iff x-y \in I$, and x-I.reduce(x) in I, for all $x,y \in R$.

That is, elements of a QuotientRing are represented by normal forms (usually obtained by polynomial division).

We can see that $(2) \subset \mathbf{Z}[x]$ in Sage does not possess such a method:

R.<x> = ZZ[]
I = R.ideal(2)
I.reduce??

This shows the source code of I.reduce which is the default implementation lambda f: return f, which doesn't satisfy the property that QuotientRing assumes: e.g. I.reduce(2) != I.reduce(0) but $2 \in I$.

The implementation of I.reduce is the same for I = R.ideal(2,x) which explains your last result.

Note that $\mathbf{Z}[x]$ is not a Euclidean ring because not every ideal is principal (e.g. the ideal $(2,x)$ is a non-principal), so it is impossible to have a normal form for elements in a quotient via naive polynomial division in $\mathbf{Z}[x]$, in general. There is a theory of Groebner bases for univariate polynomial rings over PIDs which would apply to $\mathbf{Z}[x]$, but it doesn't seem to be implemented in Sage.

Of course not all is lost, because the ideals you consider are nice enough, e.g. $\mathbf{Z}[x]/(2) \cong \mathbf{F}_2[x]$ and $\mathbf{Z}[x]/(2,x) \cong \mathbf{F}_2[x]/(x) \cong \mathbf{F}_2$ and the objects on the right-hand sides can be represented easily in Sage:

sage: (2*t1 + a*t2).change_ring(GF(2)[a])
a*t2
sage: (2*t1 + a*t2).change_ring(GF(2)[a].quotient(a))
0

Or, if you prefer the name abar for the image of a in the quotient:

sage: S.<abar> = GF(2)[]
sage: (2*t1 + a*t2).change_ring(S)
abar*t2
sage: (2*t1 + a*t2).change_ring(S.quotient(abar))
0