Considered an irreducible polynomial with coefficient over a field $\Bbb K$ $ (f(x)\in\Bbb K[x]$) and $\deg f(x)=n$ ; in my textbook is written that: $$ {\Bbb K[x]\over \langle f(x)\rangle} = \{a_0+a_1x+...+a_{n-1}x^{n-1}+\langle f(x)\rangle \mid a_i\in\Bbb K\} $$ Is it true? and if it's how can I prove this ?(I apologize if the question may seem silly)
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1This follows immediately by transporting the quotient ring structure to the coset least-degree reps, i.e. using the remainder $,g\bmod f,$ as a normal form for the coset $,g+(f),,$ as explained here the linked dupe. – Bill Dubuque Feb 13 '24 at 08:59
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The polynomials of degree less than that of $f$ give an irredundant collection of representatives for the quotient, yes. Maybe a nod to the division algorithm for polynomials in one variable with coefficients in a field?
paul garrett
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Expanding on Paul Garrett's answer, note that $\mathbb{K}[x]$ is generated by the set $\{1,x,x^2,...\}$ as a $\mathbb{K}$-vector space. Via the division algorithm for polynomials over a field, every polynomial $g \in \mathbb{K}[x]$ can be written as $$ g = qf + r $$ with $r\equiv 0$ or $\deg r < \deg f$. Passing to the quotient, it follows that $g+(f) = r+(f)$, and $r$ can be expressed uniquely as a $\mathbb{K}$-linear combination of the set $\{1,x,...,x^{n-1}\}$.
Inbo Gottlieb-Fenves
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