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Let $R=\mathbb{Z}/4\mathbb{Z}$. How do I prove that $1+2r\in R[X]^*$ for all $r\in R[X]$?

I don't fully understand the question, so I would really appreciate it if you could also explain what you are doing?

Also is every $u\in R[X]^*$ of the form $u=1+2r$ and why (not)?

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    $\begingroup$ $(1+2r)^2=1$... $\endgroup$ – user26857 Sep 5 '15 at 16:12
  • $\begingroup$ @user26857 How does this help me? $\endgroup$ – JimmyP Sep 5 '15 at 17:19
  • $\begingroup$ @JimmyP It says that 1+2r is its own multiplicative inverse. $\endgroup$ – Cihan Sep 5 '15 at 17:22
  • $\begingroup$ @Cihan and how does that help me prove the above? $\endgroup$ – JimmyP Sep 5 '15 at 17:24
  • $\begingroup$ @JimmyP Your first question is proving 1+2r is invertible. Showing that it squares to 1 does that. $\endgroup$ – Cihan Sep 5 '15 at 17:27
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For your first question, as user26857 has mentioned, $(1 + 2r)^2 = 1$.

For the second if $u \in R[x]^{*}$ then it is necessarily also a unit in $R[x]/2R[x] \cong \mathbb{Z}/2[x]$. But $\mathbb{Z}/2$ is a field, and so you know that the only unit of $\mathbb{Z}/2[x]$ is just the class of $1$. So $u \in 1 + 2R[x]$.

Your question is tied to the fact that if $I$ is a nilpotent ideal in a commutative ring $A$, then $u \in A$ is a unit iff it is a unit in $A/I$.

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Set $R = \mathbb{Z}/4 \mathbb{Z}$.

  1. We want to prove: if $r \in R$, then $q_r := 1 + 2r \in R[X]^*$ i.e. there exists a $p \in R[X]$ such that $p \cdot q_r = 1$.

(comment by user26857) For a fixed $r \in R$, observe that $(1+2r)^2 = 1 + 2r + 2r + 4 r^2 = 1 + 4 (r+r^2)$. This reduces to $(1+2r)^2 = 1$, since $ 4 = 0$ in $R$, and we find that $p = q_r$.

  1. If $u \in R[X]^*$, does there exists an $r \in R[X]$ such that $u = 1 + 2r$?

The answer is yes, which we will prove by induction over the degree of $u$. If $\deg(u) = 0$, then $u \in \{1,3\}$ and we can write $3 = 1 + 2 \cdot 1$.

Assume that if $\deg(u) \leq n-1$, then $u = 1 + 2 r$ for some $r \in R$. Let $n > 0$ and $u = \sum_0^n a_i x^i$, $v = \sum_0^m b_j x^j$, $u \cdot v = 1$ and $a_n \neq 0 \neq b_m$. Since $u \cdot v = 1$, we must have $a_n \cdot b_m = 0$, which implies $a_n = b_m = 2$. If we can prove that $ u - 2x^n$ is a unit, we are done. Indeed: By the induction hypothesis $u - 2x^n = 1 + 2r$, thus $u = 1 + 2(r+x^n)$.

To see that $U := u - 2x^n$ is a unit, observe that $\alpha := (u - 2x^n) \cdot (v - 2x^m) = uv - 2ux^m - 2vx^n + 4 x^{n+m} = 1 - 2(ux^m - vx^n) = 1 + 2 r$, where we used $2 = -2$ and set $r := (ux^m - vx^n)$. Thus, the product $\alpha$ is a unit by (1.) i.e. $\alpha^2 = (u - 2x^n)^2 \cdot (v - 2x^m)^2 = 1$. Defining $V = (u - 2x^n) \cdot (v - 2x^m)^2$, we get $U \cdot V = 1$.

Hope this helps.

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