How to calculate $ \mathbb{Z}[x] /\langle2x-1\rangle $?

Question

I don't understand very well the quotient, so, I do not know if calculate is the correct word, but I need to simplify the expression $$ \mathbb{Z}[x] /\langle2x-1\rangle $$

Answer 1

I believe by "simplification" you mean "to find an algebraic structure isomorphic to $\mathbb{Z}[x]/(2x-1)$ but of a simpler form".

First we want to know what the elements in $\mathbb{Z}[x]/(2x-1)$ look like. For any $f(x)\in\mathbb{Z}[x]$, we divide it by $2x-1$ to obtain $$f(x)=q(x)(2x-1)+r(x)$$ with $\deg r(x)<\deg(2x-1)=1$, then $r(x)=c$ is a constant, and $f(x)\mod (2x-1)$ is $$\overline{f(x)}=\overline{q(x)(2x-1)+c}=\overline{q(x)(2x-1)}+\bar c=\bar c$$ because $\overline{2x-1}=\bar0$. This means any element in $\mathbb{Z}[x]/(2x-1)$ can be represented by a number in $\mathbb{Z}$. Conversely, it is easy to see two different numbers in $\mathbb{Z}$ do not represent the same equivalence class in $\mathbb{Z}[x]/(2x-1)$. So one should make a guess that $$\mathbb{Z}[x]/(2x-1)\cong\mathbb{Z}$$ and this is indeed the answer. It remains to verify this isomorphism, i.e., find such an isomorphism. Can you take it from here?

$\textbf{Hint}:$ My argument above gives a well-defined mapping $$\mathbb{Z}[x]\to\mathbb{Z}$$ $$f(x)\mapsto c$$ Verify this is a homomorphism, and then an epimorphism, and find its kernel.

Answer 2

Hint:

Consider the ring homomorphism from the ring of polynomials to the dyadic fractions: \begin{align} f:\mathbf Z[x]&\longrightarrow \mathbf Z_2\\ x&\longmapsto \tfrac12 \end{align} Show that it is surjective and that its kernel is generated by $2x-1$.