# How to compute unit groups in quotient rings? [closed]

How do I compute the unit group of a quotient ring $\mathbb{R}[x]/(f(x)\mathbb{R}[x])$, for example $f(x)=x^2+2x+1$?

## closed as off-topic by user26857, zz20s, Shailesh, user91500, John BApr 18 '16 at 8:50

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• How do you compute units in $\mathbb{Z}/n\mathbb{Z}$ ? – Captain Lama Apr 17 '16 at 21:15
• Hint number two: $\Bbb R[x]$ is a Euclidean domain, so what can we do in such things? – David Wheeler Apr 17 '16 at 21:20
• Forgive me for nit-picking, but I see you wrote $\mathbb R[x]/(f(x)\mathbb R[x])$. Please keep in mind that we usually say that $f$ is an element of $\mathbb R[X]$. When we write $f(x)$, we mean the output of $f$ under the evaluation homomorphism $\phi_x$. Pedantic but important, because it basically makes the expression $\mathbb R[x]/(f(x)\mathbb R[x])$ meaningless. This is why a lot of authors write $\mathbb R[X]$, with a capital. – gebruiker Apr 17 '16 at 21:23
• @gebruiker I kind of disagree with your comment. Writing $f(x)$ is very common even amongst algebraists, and if we really want to say $f(x)$ is an evaluation, then it's the evaluation of $f$ at the generic point, which is very rigourously equal to $f$. – Captain Lama Apr 17 '16 at 21:26
• So, because $\mathbb{R}[x]$ is an Euclidean domain I can use the $\gcd(p,q)$ for polynomials $p$ and $q$. But I do struggle at the moment with the quotient ring $\mathbb{R}[x]/ \left((x^2+2x+1)\mathbb{R}[x]\right)$. Which elements does it contain? – monoid Apr 17 '16 at 21:32

The elements of $R=\mathbb R[X]/((X+1)^2)$ are of the form $a+bx$, that is, residue classes modulo the ideal $((X+1)^2)$. Now note that $R$ has only one maximal ideal: the ideal generated by $x+1$. This means that $R$ is local, and then its unit group is $R\setminus (x+1)$, that is, the elements of the form $a+bx$ with $a\ne b$.
Another approach: a residue class $a+bx$ is invertible iff there is another residue class $c+dx$ such that $(a+bx)(c+dx)=1$, that is, $(a+bX)(c+dX)-1\in ((X+1)^2)$. This can be restated as $\gcd(a+bX,X+1)=1$ or $X+1\nmid a+bX$ which is equivalent to $a\ne b$.