# Simplify $(x+2+I)(x^2+3+I)$ where $I=<4x,2>$ an Ideal of $Z[x]$

Simplify (x+2+I)(x^2+3+I) where $I=<4x,2>$ Ideal of $Z[x]$ where $$(x+2+I)(x^2+3+I)=((x+2)(x^2+3)+I)=(x^3+2x^2+3x+6+I)$$

Now I have handwriting that it is a subring. Not 100% that $if r\in Z[x] \wedge i\in I \Rightarrow ir \in <4x,2>$. Letting $r=r_n x^m+\dots+r_0$ $r*i=(r_n x^m+\dots+r_0)*(4x*q_1(x)+2q_2(x))$ where ofcourse $q_1(x),q_2(x) \in Z[x]$. So, $=r_nx^{n+1}q_(x)+\dots +r_0 q_2(x)$ it is messy, I am just going to take it for granted that it is an Ideal, Hopefully.

My real concern is that I am unsure what are the elements of $Z[x]/I$. I know that $Z[x]/(4x+2)$ are polynomials of degree less than of $4x+2$ but since $$<4x,2>=\{ q_1(x)*4x+q_2(x)*(2) : q_1(x),q_2(x) \in Z[x] \}$$ So there are alot, the elements of $z_[x]/<4x+2>$ are those not divisble by the combination. So, degrees bigger than 1,right?

Also, let $i \in I$ $$Z[x]/(4x+5),Z[x]/(5),\dots ,Z[x]/i, \dots,Z[x]/(4x^5+x^3)$$ Is there a way to combine them to express $Z[x]/I$? Guessing union.

All of these questions might help the main question if the polynomial can be simpliefied further.

Appreciate Constructive input, Thanks

• I do mean (4x,2) which i did not realize it is (2). Able to hand write it and prove it to myself. Thought somehow $Z[x]/(4x,2)=Z[x]/(2)=Z_2[x]$ would translate to $Z/(4x+2)$ but it is not the case, It makes sense now. Since deg of $Z_2[x]$ has polynomials degree larger than 1 where $Z[x]/(4x+2)$ Does not. Commented Dec 5, 2015 at 21:10

Since the ideal is generated by $4x$ and $2$. So when taking quotients you will consider $4x \equiv 0$ and $2\equiv 0$. In fact if you combine these congruences you get that $2x \equiv 0$ as well.
So in the quotient ring, any integer $n$ will either be $n\equiv 1 \text{ or } 0$. Likewise we have $(2k)x^j \equiv 0$ as well. The only non-trivial case is when we have an odd coefficient of a power of $x$.
Consider $(2k+1) x^j \equiv 2k x^j + x^j \equiv x^j$.
So in such a ring something like $5x^4+4x^3+7x^2-11x+17 \equiv x^4+x^2+x+1$.
Note: if you carefully observe then $\langle 4x, 2 \rangle = \langle 2 \rangle$. Thus this becomes a polynomial ring over $\mathbb{Z}_2$.