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My book states that the polynomials over a field $\mathbb{R}$ mod $(x^2+1)$ form a commutative ring with addition and multiplication. Why is this? I don't understand how there is closure under multiplication.

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Do you mean the ring $R[x]/(x^2 + 1)$ as a factor ring of $R[x]$ by the ideal $(x^2 + 1)$? Why shouldn't it be closed under multiplication? –  k.stm Jan 26 '13 at 15:03
    
@K.Stm. I'm not sure I haven't seen factor rings yet. I think it's not closed under multiplication because all elements in the set are of degree $\leq$ 1. When you multiply two monomials, there is no closure. –  xcrypt Jan 26 '13 at 15:09
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@xcrypt: But *modulo $x^2+1$*, a polynomial of higher degree is *equal* to a polynomial of degree $\leq 1$. –  Zev Chonoles Jan 26 '13 at 15:13
    
Oh, that makes sense. Thanks –  xcrypt Jan 26 '13 at 15:14
    
For rings, one normally uses commutative instead of abelian. –  lhf Jan 26 '13 at 16:07

1 Answer 1

up vote 1 down vote accepted

Modulo $x^2+1$, a polynomial of higher degree is equal to a polynomial of degree $\leq 1$.

For example, $$\begin{align}x^2+12x+35=(x^2+1)+12x+34\;&\equiv\; 12x+34\bmod (x^2+1)\\ x^3+2x-3=x(x^2+1)+x-3\;&\equiv\; x-3\bmod (x^2+1)\end{align}$$

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