Prove that the following set is a field

Question

Let $K$ be a subfield of $ℂ$, and $K[t]$ be the set of all the polynomials over $K$ of the variable $t$.

Now, for a polynomial $m$ over $K$, let $K[t]/(m)$ be the set of the reduced forms of all polynomials over $K$, in other words, if $f∈K[t]$, and $$f=pm+r$$(by the division algorithm), then $r∈K[t]/(m)$.

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For $a, b∈K[t]/(m)$ define $$(a)+(b)=(a+b)$$ and $$(a)*(b)=R(a*b),$$ where $R(a*b)$ is the reduced form of $(a*b)$

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The author then continues to prove that $K[t]/(m)$ is a field (under the two operations defined above, call them addition and multiplication respectively) iff $m$ is irreducible over $K$. However, he forgets to prove that multiplication is associative in $K[t]/(m)$, in other words, that $$[(a)*(b)]*(c)=(a)*[(b)*(c)]$$

or

$$R[R(a*b)*(c)]=R[(a)*R(b*c)]$$

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I tried to prove it myself, but didn't achieve anything. I would really appreciate any help.

Answer 1

Note that $$((am+r)(bm+s))(cm+t)\equiv rst \equiv (am+r)((bm+s)(cm+t))\bmod m$$

Now $rst$ is not necessarily a reduced form, but since the reduced form of $rst$ is unique, both outside products are equivalent to the same reduced form.

This is the kind of thing you should only need to understand once - then it becomes obviously applicable to homomorphisms across a wide variety of algebraic contexts - homomorphisms respect multiplication, so they preserve associativity of multiplication.

Answer 2

Actually, as multiplication in $K[t]/(m)$ is deduced from multiplication in $ K[t]$ by quotienting, and multiplication in $K[t]$ has all the desired properties, all you have to prove is that this multiplication is well-defined, i.e. if $a\equiv a'\mod m$ and $b\equiv b'\mod m$, then $a'b'\equiv ab \mod m$.

Now this is easy: $$a \equiv a'\mod m\iff m \mid a-a',\qquad b \equiv b'\mod m\iff m \mid b-b', $$ so let's write $a-a'=qm$, $b-b'=rm$ for some $q, r\in K[t]$. We deduce: $$ab-a'b'=(a-a')b+a'b-a'b'=(a-a')b+a'(b-b')=qmb-arm=(qb-ar)m, $$ which proves the assertion.