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Due to the Kronecker's theorem, $x^2+1\in\mathbb Q[x]$ splits over $\mathbb Q[x]/\langle x^2+1\rangle.$ But how to show that $\mathbb Q[x]/\langle x^2+1\rangle$ is a splitting field of $x^2+1$ over $\mathbb Q.$

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    $\begingroup$ Doesn’t it have all the roots of your polynomial? And isn’t it the case that no smaller field has the roots? That’s my idea of splitting field. $\endgroup$ – Lubin Apr 3 '14 at 13:15
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I'm not sure what is meant by Kronecker's theorem; perhaps the statement that $x^2+1$ admits a root in $K:=\mathbf{Q}[x]/(x^2+1)$, namely the image of $x$ (call it $i$). Since $-i$ is also a root of $x^2+1$ in $K$ (distinct from $i$ since $K$ has characteristic zero), $x^2+1$ indeed splits over $K$.

Now, what is a splitting field of $x^2+1$ over $\mathbf{Q}$? It is an extension $L$ of $\mathbf{Q}$ with two properties: first, $x^2+1$ must split over $L$, and second, if the roots are $\alpha,\beta$, then we require that $L=\mathbf{Q}(\alpha,\beta)$, i.e., that the roots of $x^2+1$ in $L$ generate it over $\mathbf{Q}$. We've already observed that $x^2+1$ splits over $K$, and by construction, $K=\mathbf{Q}(i)=\mathbf{Q}(i,-i)$, so $K$ satisfies the two conditions necessary to be called a splitting field of $x^2+1$ over $\mathbf{Q}$.

Note that the condition that $L$ be generated over $\mathbf{Q}$ by the roots of $x^2+1$ is equivalent to requring that $x^2+1$ does not split over any proper subfield of $L$.

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This is a general fact for degree two polynomials.

If $F$ is a field and $f(x)$ is an irreducible degree two monic polynomial in $F[x]$, we know that $F[x]/(f(x))$ is an extension field of $F$ where $f$ has a root. If we denote by $\alpha$ this root, namely $x+(f(x))$, we have that $F[x]/(f(x))=F[\alpha]$ (under the usual identification of the elements of $F$ with the cosets of the constant polynomials).

Now $f(x)$, as a polynomial in $F[\alpha][x]$ has a root in $F[\alpha]$, so it factors as $f(x)=(x-\alpha)(x-\beta)$ with $\beta\in F[\alpha]$. Hence $F[\alpha]$ is a splitting field for $f$ because $f$ factors in it and $F[\alpha]=F(\alpha,\beta)$.

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In $\mathbb Q[x]/\langle x^2+1\rangle$ both of the roots of $x^2+1$ viz. $x+\langle x^2+1\rangle,-x+\langle x^2+1\rangle$ exist. Thus $\mathbb Q(x+\langle x^2+1\rangle,-x+\langle x^2+1\rangle)$ is a splitting field of $x^2+1$ over $\mathbb Q.$

Note that $\mathbb Q(x+\langle x^2+1\rangle,-x+\langle x^2+1\rangle)\\=\mathbb Q(x+\langle x^2+1\rangle)\\=\{(a+\langle x^2+1\rangle)+(b+\langle x^2+1\rangle)(x+\langle x^2+1\rangle):a,b\in\mathbb Q\}\\=\{(a+bx)+\langle x^2+1\rangle:a,b\in\mathbb Q\}\\=\mathbb Q[x]/\langle x^2+1\rangle\text{ (due to the division algorithm)}$

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