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I understand that $\mathbb{R}[x]/\langle x^2 + 1\rangle \cong \mathbb{C}$, so it is a field, but I'm struggling to understand what are the units of this factor ring of cosets. Every element in $\mathbb{R}[x]/\langle x^2 + 1\rangle$ has the form $ax + b$, and the unity of this field is $1 + \langle x^2 + 1\rangle$. However, for any polynomial in a field, $$\deg f(x)g(x) = \deg f(x) + \deg g(x)$$ So how can any polynomial with degree 1 in this factor ring, say $2x + 1$ have a multiplicative inverse?

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  • $\begingroup$ $x$ essentially acts like the complex number $i$. So to figure out your inverse for $2x+1$, just invert $2i+1$, then replace all the occurrences of $i$ in your answer with $x$. $\endgroup$ Dec 7, 2017 at 6:18
  • $\begingroup$ Ah right, I completely forgot that I need to divide my result by $x^2 + 1$ afterwards.. $\endgroup$
    – q.Then
    Dec 7, 2017 at 6:20

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Think of how $1+2i$ can have an inverse in $\mathbb{C}$: it's inverse is $\frac{1}{5}-\frac{2}{5}i$.

What happens when you multiply $1+2x+\langle x^2+1\rangle$ and $\frac{1}{5}-\frac{2}{5}x+\langle x^2+1\rangle$ in the ring $\mathbb{R}[x]/\langle x^2+1\rangle$?

You get $$\begin{align*} (1+2x)(\tfrac{1}{5}-\tfrac{2}{5}x)+ \langle x^2+1\rangle&=(\tfrac{1}{5}-\tfrac{4}{5}x^2)+\langle x^2+1\rangle \\ &=1+(-\tfrac{4}{5})(x^2+1)+\langle x^2+1\rangle\\ &=1+\langle x^2+1\rangle \end{align*}$$

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Your proof will be exactly similar to the proof that $\mathbb{C}$ is a field.

For an arbitrary element in your field, a simple calculation shows that: $$(a+bx + \langle x^2+1\rangle) \cdot (a-bx + \langle x^2+1 \rangle)=a^2-b^2x^2+\langle x^2 +1 \rangle $$ $$= a^2+b^2-b^2(x^2+1)+\langle x^2+ 1\rangle =a^2+b^2 + \langle x^2+1 \rangle$$

Where the last step is true because $-b^2(x^2+1) \in \langle x^2+1 \rangle$.

Therefore, if you take: $$(a+bx + \langle x^2+1\rangle)^{-1}=\frac{a-bx}{a^2+b^2}+\langle x^2+1 \rangle$$

You will be done because your ring is commutative and it definitely works both from the left and from the right as shown in our calculation.

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