Let $\Bbb R$ be a field of all real numbers. Prove that $x+1$ is not a unit in $\Bbb R[x]$.


Suppose that $x+1$ is a unit in $\Bbb R[x]$. Then, there is exists $g(x) \in \Bbb R[x]$ such that $(x+1)g(x) = 1_R$.

Now, we have \begin{equation*} deg(x+1) + deg(g(x)) = deg((x+1)g(x)) = deg(1_R) = 0 \end{equation*} Thus, we have \begin{align*} deg(x+1) + deg(g(x)) &= 0 \\ 1 + deg(g(x)) &= 0 \\ deg(g(x)) &= -1 \end{align*} which is a contradiction since $deg(f(x))$ must be non-negative integers for all $f(x) \in \Bbb R[x]$. Hence, $x+1$ is not a unit in $\Bbb R[x]$.

Is the above true?


1 Answer 1


Yes, this is correct.

Another way to see this is that $\Bbb{R}[x]/(x+1)\cong \Bbb{R}$ and hence $(x+1)\ne \Bbb{R}[x]$ so that $x+1$ is not a unit.

  • $\begingroup$ Thanks Sir! Also, for that other ways. $\endgroup$
    – krewlpt
    Nov 5, 2020 at 0:02

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