Past exam of the options II course of the University of Oxford / 2006 / B (Introduction to Fields)

An irreducible polynomial has positive degree, so its ideal is proper: otherwise $1=fh$ would contradict degrees. Every nonzero class $[g]$ has a representative not divisible by $f$. Irreducibility gives $\gcd (g,f)=1$, and the Euclidean algorithm supplies polynomials $u,v$ with $ug+vf=1$. Passing to the quotient gives $[u][g]=1$. Thus the quotient is a nonzero commutative ring in which every nonzero element has an inverse, hence a field. Constants embed injectively because no nonzero constant lies in the positive-degree ideal $(f)$.

Polynomial division gives a unique remainder of degree less than $n$ in every coset. Thus the classesspan the quotient as an $F$-vector space. A linear combination of them that vanishes would be a polynomial of degree less than $n$ divisible by $f$, so it is the zero polynomial. They are therefore linearly independent, proving

Take $a=[x]$. Every quotient class is a polynomial expression in $a$ with coefficients from $F$, so $K=F[a]$. Substitution gives $f(a)=[f(x)]=0$. ThusThis constructs a field containing a root even when no root existed in the original field.

First address the general splitting-field request. A splitting field is an extension in which the polynomial factors into linear factors and which is generated over the base field by its roots. For a degree-$m$ polynomial, adjoin a root of an irreducible factor. The quotient construction in the preceding parts gives an extension of degree at most $m$. Remove the resulting linear factor and repeat, now with a polynomial of degree $m-1$ over that extension. Induction and the tower law giveThe resulting field is generated by the roots and is a splitting field. Repeated roots or inseparability can only reduce the required degrees; separability was not used.

For the specified cubic, put $a=\sqrt[3]{2}$ and $\omega =e^{2\pi i/3}$. Its splitting field is $\mathbb{Q}(a,\omega )$. Eisenstein criterion at two gives $[\mathbb{Q}(a):\mathbb{Q}]=3$. This cubic field is real and therefore does not contain the nonreal $\omega$, whose quadratic polynomial is $x^2+x+1$. Consequently

Let $\zeta =e^{\pi i/4}$. Its odd powers are exactly the four roots, so $\mathbb{Q}(\zeta )$ is the splitting field. The polynomial $x^4+1$ is irreducible: after the invertible substitution $x\mapsto x+1$, it becomes $x^4+4x^3+6x^2+4x+2$, which satisfies Eisenstein criterion at two. HenceEquivalently the field is $\mathbb{Q}(\sqrt{2},i)$, since $\zeta =(1+i)/\sqrt{2}$, ${\zeta }^2=i$ and $\zeta +{\zeta }^{-1}=\sqrt{2}$.

FactorThe roots are $\pm \sqrt{2},\pm i$, so the splitting field is $\mathbb{Q}(\sqrt{2},i)$. The real quadratic field $\mathbb{Q}(\sqrt{2})$ does not contain $i$, which adds a further degree two. Therefore