Residue of x in polynomial ring \(\displaystyle{\mathbb{{{Z}}}}\frac{{{x}}}{{{f}}}\) When

Oxinailelpels3t14 2022-04-02 Answered
Residue of x in polynomial ring Zxf
When we say that α is the residue of x in Zxf, and f=x4+x3+x2+x, wouldn't α just be x? Because if we divide with the remainder, we would get x=0 f+x?
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Answers (2)

Alexzander Evans
Answered 2022-04-03 Author has 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={a+i:iI}. In particular, the elements of R/I are subsets of R!
When we talk about the residue of an element xR in the quotient R/I, we mean the coset x+I.
Step 2
So, the residue of x in Zxf is (by definition) the coset
x+(f)={x+i:i(f)}={x+gf:gZ[x]}..
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 " x2=0 in R/I" to mean" x2+I=0+I. It'll be up to you to keep track of the context and interpret elements as cosets when necessary!
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Aidyn Wall
Answered 2022-04-04 Author has 10 answers
Solution.
Another approach, at least for an irreducible polynomial f, is to consider a hypothetical element α which satisfies f(α)=0, and extend the original ring R by α (and whatever formal elements that makes a ring), analogously as one extends R by an element i which satisfies i2+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 α:=x+I.
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