I have learned in my class that quotient ring of $\mathbb{R}[x] / (x^2 + 1) \cong \mathbb{C}$. Just from curiosity, I was interested in knowing if $$ \mathbb{R}[x] / (x^2 + ax + b) \cong \mathbb{C} $$ holds for any $a,b \in \mathbb{R}$? Thanks!
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3What happens when both roots of the polynomial are real? – RougeSegwayUser Aug 11 '14 at 23:33
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1The quotient is $\Bbb R^2$, $\Bbb R[\epsilon]/(\epsilon^2)$ or $\Bbb C$ depending on if $b^2-4ac$ is positive, zero or negative. – anon Aug 12 '14 at 00:27
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Half of the quadratics have real roots, so the quotient ring would be $\mathbf{R}^2$.
The other half have complex roots, and $\mathbf{R}(\alpha) = \mathbf{C}$ for any nonreal complex number $\alpha$
A handful of quadratics have a double root, and the quotient ring would be $\mathbf{R}[\epsilon]/\epsilon^2$.
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Are $\mathbb{R}^2$ and $\mathbb{R}[\epsilon]/\epsilon^2$ both isomorphic to $\mathbb{C}$? – Tom Mosher Aug 11 '14 at 23:46
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@Tom: As rings? No: the former has a pair of nonzero numbers that multiply to zero. The latter has a nonzero number whose square is zero. As $\mathbf{R}$-vector spaces? Yes. As additive abelian groups? Yes. – Aug 11 '14 at 23:47
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