Suppose \(K/F\) is a field extension. Prove that the elements of \(K\) that are algebraic over \(F\) form a subfield of \(K\).
Find the minimal polynomial of \(2\sqrt{2} - \sqrt{3}\) over \(\mathbb{Q}\).
Solution
1. The elements algebraic over \(F\) form a subfield of \(K\).
Every element of \(F\) is algebraic over \(F\), being a root of a linear polynomial. Suppose \(\alpha,\beta\in K\) are algebraic over \(F\), with monic annihilating polynomials of degrees \(m,n\), respectively. Reduction by these equations shows that the subring \(A=F[\alpha,\beta]\subseteq K\) is spanned over \(F\) by \[ \{\alpha^i\beta^j:0\leq i<m,\ 0\leq j<n\}. \] Thus \(A\) is a finite-dimensional \(F\)-vector space. For \(0\ne a\in A\), multiplication by \(a\) is an injective \(F\)-linear endomorphism of \(A\), since \(A\) is a subring of a field. An injective endomorphism of a finite-dimensional space is surjective, so \(ab=1\) for some \(b\in A\). Therefore \(A\) is itself a field.
Each \(z\in A\) is algebraic over \(F\): if \(d=\dim_F A\), the \(d+1\) powers \(1,z,\ldots,z^d\) are linearly dependent, yielding a nonzero annihilating polynomial in \(F[T]\). In particular \(\alpha-\beta\), \(\alpha\beta\), and \(\alpha^{-1}\) when \(\alpha\ne0\) are algebraic. Together with containment of \(F\), these closure properties prove the subfield assertion.
2. The field \(E=\mathbb Q(\sqrt2,\sqrt3)\) has degree \(4\)over \(\mathbb Q\).
A square of a nonzero rational number has an even exponent at every prime in its factorization. Hence \(2\) is not a rational square, and \(1,\sqrt2\) are a basis of \(\mathbb Q(\sqrt2)\) over \(\mathbb Q\). If \((a+b\sqrt2)^2=3\) for \(a,b\in\mathbb Q\), comparison of these two basis coefficients gives \(a^2+2b^2=3\) and \(2ab=0\). For \(b=0\) this gives \(a^2=3\), impossible because the exponent of \(3\) is odd. For \(a=0\) it gives \(b^2=3/2\), also impossible because the exponent of \(3\) is odd. Thus \(T^2-3\) has no root in \(\mathbb Q(\sqrt2)\) and is irreducible over that field. Adjoining \(\sqrt3\) has degree \(2\), and the tower law gives \([E:\mathbb Q]=2\cdot2=4\).
3. The minimal polynomial in part (b) is\[ \boxed{T^4-22T^2+25}. \]
Put \(\gamma=2\sqrt2-\sqrt3\). One has \(\gamma^2=11-4\sqrt6\), so \((\gamma^2-11)^2=96\). Expanding gives \(\gamma^4-22\gamma^2+25=0\).
The element \(\gamma\) has degree \(4\) over \(\mathbb Q\). Indeed, \[ (2\sqrt2-\sqrt3)(2\sqrt2+\sqrt3)=8-3=5. \] Thus \(\gamma\ne0\) and \(5/\gamma=2\sqrt2+\sqrt3\). Consequently \[ \sqrt2=\frac{\gamma+5/\gamma}{4},\qquad \sqrt3=\frac{5/\gamma-\gamma}{2}. \] Both radicals belong to \(\mathbb Q(\gamma)\), so this field equals \(E\). Step 2 shows that \(\gamma\) has degree \(4\). Its minimal polynomial is therefore the monic degree-four annihilating polynomial displayed above.