Residue of x in polynomial ring ZxfWhen 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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Residue of x in polynomial ring ZxfWhen we say that α is the residue of x in Zxf, and f=x4+x3+x2+x, wouldn&#039;t α just be x? Because if we divide with the remainder, we would get x=0 f+x?

Alexzander Evans · Accepted Answer

Step 1The 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:i∈I}. In particular, the elements of R/I are subsets of R!When we talk about the residue of an element x∈R in the quotient R/I, we mean the coset x+I.Step 2So, the residue of x in Zxf is (by definition) the cosetx+(f)={x+i:i∈(f)}={x+gf:g∈Z[x]}..Be warned: sometimes it gets annoying to keep writing &quot;+I everywhere, so people will sometimes just use &quot;x&quot; to refer to the residue of x in some quotient. E.g. someone might say &quot; x2=0 in R/I&quot; to mean&quot; x2+I=0+I. It&#039;ll be up to you to keep track of the context and interpret elements as cosets when necessary!

Aidyn Wall · Accepted Answer

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.

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

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