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Consider the quotient ring $R = \mathbb{R}[x]/(f(x))$, where $f(x)\in\mathbb{R}[x]$. Let $n$ be the degree of $f$.

  1. Explain why $R$ is a field if $n=1$

Attempted answer: if the degree of $f$ is 1, then $f = a_0+a_1x$ for $a_0,a_1\in\mathbb{R}$, so $f$ cannot be reduced, and hence $$ is a prime ideal, which in a PID is a maximal ideal, so $\mathbb{R}[x]/(f(x))$ is a field.

  1. Give examples showing that if $n=2$, then $R$ may or may not be a field.

Attempted answer: If $f(x)=x^2-4$, then it can be reduced to $f(x) = (x-2)(x+2)$, so it is not irreducible, and by the reasoning above, $R$ cannot be a field. However, if $f(x) = x^2+4$, then no roots exist in $\mathbb{R}$, so the polynomial is irreducible, and again $R$ is a field.

  1. Prove that if $n=3$, then $R$ is not a field.

Attempted answer: Any polynomial with degree greater than $2$ has at least one root in $\mathbb{R}$, though I'm not sure how to justify this. Either way, it means we can take a factor $(x-a)$ from the polynomial, so it is not irreducible, and hence $R$ is not a field.

Is my reasoning sound here? I'm not sure whether an appropriate answer would be more thorough.

Any input appreciated!

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    Any polynomial of odd degree has at least one root in $\mathbb{R}$. – 2017-01-23
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    I would probably have gone a different route. For instance, in the $n = 1, 2$ cases I would personally be more comfortable _explicitly_ saying that the fields in question are $\Bbb R, \Bbb C$ and give the explicit isomorphism rather than arguing through reducibility. But I don't see any major flaws in your reasoning (apart from what the comment above points out). – 2017-01-23
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    Your first two arguments look sound. You can justify 3. by the Intermediate Value Theorem. Since the degree is odd, then $\lim_{x \to -\infty} f(x) = \pm \infty$ and $\lim_{x \to \infty} f(x) = \mp \infty$, so by continuity there must be a root. – 2017-01-23

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1,2 I think are good enough. 3 is wrong; consider $x^4+1$. However, because of fundamental theorem of algebra, and also because if $f\in \mathbb{R}[x]$ and $f(z)=0$, with $z\in\mathbb{C}$, then $f(\bar{z})=0$, then if f-degree is 3, f must have a root in $\mathbb{R}$ and so on.