Field

[vagrant apex]

How can $k[x] / (f(x))$ be a field with infinite elements if there are only finitley many residue classes modulo $f(x)$?

[last terrace]

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[sullen bison]

The quotient is in bijection with K^(deg(f))

Like for concreteness, R[x]/(x^2+1) is {[ax+b]|a,b in R, x^2=-1}, so each equivalence class maps to a point in the plane

So... there's not finitely many equivalence classes (unless K itself is finite)

[viscid lake]

Yeah I think simplest example would be K[x]/(x), which is just K again

You take polynomials over K, you remove all multiples of x (so elements of the ideal of x), what you're left over with is just the constant term in K, so if K is infinite K[x]/(x) will be infinite too.

[viscid lake]

Am I saying weird or incorrect stuff @sullen bison? 😰

[sullen bison]

Better to have let them respond to the already given response.

[viscid lake]

Just trying to clarify but alright 🤷‍♂️

Thought K[x]/(x) was simplest way to see what happens. Although the construction of C is a neat example indeed, one which you might actually see in the wild.

[vagrant apex]

Could you explain in detail how all these cosets are reduced to linear polynomials? From my understanding $(x^2 + 1)$ is the set of all multiples of $x^2 + 1$?

[last terrace]

Addict

[sullen bison]

The quotient contains remainders when you divide by the ideal generator

Dividing by x^2+1 gives remainders of degree at most 1

It's clear that, for any polynomial f, g is equivalent mod (f) to a polynomial of degree at most deg(f)-1.
g=fq+r implies g-r is in (f) and deg(r)<deg(f)

So the equivalence classes mod (f) can be represented by remainders when dividing by f

[vagrant apex]

Oh, ok thanks

That's what I thought originally, but for some reason I thought there would be only one equivalence class of linear polynomials but there is not

[wanton portal]

@vagrant apex

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