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Let's define the polynomial $p(x)$ as one of the following polynomials: $x^n-1$ or $x^{n-1}+x^{n-2}+...+1$

Does $Z[x]/(3, p(x))$ mean that we take the modulo of the elements of $Z[x]$ according to $p(x)$ and then take the modulo again according to 3? If so, does that mean $Z[x]/(3,p(x))=Z_3[x]/(p(x))$?

Also does it make any difference if the polynomial differs? I mean is it true for any polynomials?

Esra
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  • More exactly, the l.h.s. is isomorphic to the quotient of $\mathbf Z/3\mathbf Z[X]$ by the canonical image of $p$ in this polynomial ring. – Bernard Dec 06 '20 at 17:05

1 Answers1

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I wouldn't say precisely that they are the same, but they are isomorphic, which is as good as it gets for such objects. To see this for yourself, consider the ring homomorphism: $$\varphi: \mathbb{Z}[x]/(3,p(x)) \to (\mathbb{Z}/3\mathbb{Z})/(p(x))$$ given by: $$ f(x) = a_0+a_1x+...+a_nx^n \mapsto (a_0\%3) + (a_1\%3)x+...+(a_n\%3)x^n $$ Where by $a_i\%3$ I mean reduction modulo $3$. This map is clearly surjective, since any element of $(\mathbb{Z}/3\mathbb{Z})/(p(x))$ can be written as a polynomial with coefficients in $\mathbb{Z}$ by definition. Injectivity also follows rather easily by considering the kernel. This gives an isomorphism for a general $p(x)$ thus answering your second question.

Hope this helps.

Noah Solomon
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