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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 $r∈K[t]/(m)$.

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 $(a*b)$

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

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

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

Jesko Hüttenhain
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Sam
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  • You don't need to prove it. Multiplication is a associative in $\mathbb C$ so is associative in $K$ and $K[t]$ and $K[t]/m$ inherit it. – fleablood Feb 20 '18 at 22:37
  • @fleablood. In case that you are referring to multiplication (as it is often defined) of polynomials, then clearly, $(a)(b)=(ab)$. However, if you are referring to the multiplication defined above, then the claim isn't obvious to me. In such case of the latter, could you better explain your reasoning? – Sam Feb 20 '18 at 22:41
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    Polynomial $m$ should be irreducible over $K$ if I well remember. – Piquito Feb 20 '18 at 22:45
  • @Piquito. Yeah. Although, I also think I stated that. – Sam Feb 20 '18 at 22:47
  • The claim about associativity is true for any $m$, not just irreducible $m$. The explanation is essentially the same as in this MSE question about modular arithmetic. You are projecting one algebraic structure onto another and equations that hold in the first structure will hold a fortiori in the second. – Rob Arthan Feb 20 '18 at 22:52

2 Answers2

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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.

Mark Bennet
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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.

Bernard
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