 # Residue of x in polynomial ring \mathbb{Z}[x]/(f) When we say that Sullivan Pearson 2022-04-30 Answered
Residue of x in polynomial ring $\mathbb{Z}\frac{x}{f}$
When we say that α is the residue of x in $\mathbb{Z}\frac{x}{f}$, and $f={x}^{4}+{x}^{3}+{x}^{2}+x$, wouldn't $\alpha$ just be x? Because if we divide with the remainder, we would get ?
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Step 1
The elements of a quotient ring R/I (where I is a two-sided ideal of the ring R) are the cosets of I, i.e. the sets of the form $a+I\phantom{\rule{0.222em}{0ex}}=\left\{a+i:i\in I\right\}$. In particular, the elements of R/I are subsets of R!
When we talk about the residue of an element $x\in R$ in the quotient R/I, we mean the coset $x+I$.
Step 2
So, the residue of x in $\mathbb{Z}\frac{x}{f}$ is (by definition) the coset
$x+\left(f\right)=\left\{x+i:i\in \left(f\right)\right\}=\left\{x+gf:g\in \mathbb{Z}\left[x\right]\right\}.$.
Be warned: sometimes it gets annoying to keep writing "$+I$ everywhere, so people will sometimes just use "x" to refer to the residue of x in some quotient. E.g. someone might say " ${x}^{2}=0$ in R/I" to mean" ${x}^{2}+I=0+I$. It'll be up to you to keep track of the context and interpret elements as cosets when necessary!
###### Not exactly what you’re looking for? gunithd5
Solution.
Another approach, at least for an irreducible polynomial f, is to consider a hypothetical element $\alpha$ which satisfies $f\left(\alpha \right)=0$, and extend the original ring R by $\alpha$ (and whatever formal elements that makes a ring), analogously as one extends R by an element i which satisfies ${i}^{2}+1=0$ to obtain the field of complex numbers C.
Actually the quotient ring construction (as the set of cosets of the ideal I generated by f) shows that the above hypothetical construction is feasible, namely we can set $\alpha :=x+I$.