Oxinailelpels3t14

2022-04-02

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 $x=0\text{}f+x$ ?

When we say that α is the residue of x in

Alexzander Evans

Beginner2022-04-03Added 9 answers

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$a+I{\textstyle \phantom{\rule{0.222em}{0ex}}}=\{a+i:i\in I\}$ . 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

$x+\left(f\right)=\{x+i:i\in \left(f\right)\}=\{x+gf:g\in \mathbb{Z}\left[x\right]\}.$ .

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!

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

When we talk about the residue of an element

Step 2

So, the residue of x in

Be warned: sometimes it gets annoying to keep writing "

Aidyn Wall

Beginner2022-04-04Added 10 answers

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$\alpha :=x+I$ .

Another approach, at least for an irreducible polynomial f, is to consider a hypothetical element

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

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