In which of the cases $F = \mathbb{Z}_2$ or $F = Q$ is the quotient ring $F[x]/(x^2+3)$ a field?
I have no idea where to even begin with this
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2$R/I$ is a field precisely when $I$ is a maximal ideal. For polynomials, $(f(x))$ is maximal precisely if $f(x)$ is irreducible. Is $x^2+3$ irreducible over $\mathbb{Q}$? What about over $\mathbb{Z}/2\mathbb{Z}$? – Mark Schultz-Wu Dec 16 '17 at 06:57
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Since $x^2+3$ is irreducible over $\mathbb{Q}[x]$, the ideal $(x^2+3)$ is maximal, hence $\mathbb{Q}[x]/(x^2+3)$ is a field (Used the following result), whereas, $x^2+3 $ is reducible over $\mathbb{Z}_2$, as $1$ is a root . So $\mathbb{Z}[x]/(x^2+3)$ is not a field.
Sachchidanand Prasad
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Note that $x^2+3$ is reducible over $\mathbb{Z}/2\mathbb{Z}$ because $x^2+3\equiv x^2+1\equiv x^2+2x+1\equiv (x+1)^2\mod 2$. – Mark Schultz-Wu Dec 16 '17 at 07:01
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@Mark That is what I have written in my answer that $(x^2+3)$ is reducible, as $1$ is a root. – Sachchidanand Prasad Dec 16 '17 at 07:03
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I agree, I just thought it might be interesting for OP to see the explicit factorization as well, but didn't think that it was enough to be a full answer. – Mark Schultz-Wu Dec 16 '17 at 07:05
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Hint: Let $F$ be a field and $p(x)\in F[x]$, then
$$F[x]/(p(x))\; \text{is a field} \iff p(x)\; \text{is irreducible in F}.$$
I hope you can carry on from here.
Naive
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