What is the splitting field of x^2 + 1 over Q?

Let i be a zero of x^2+1 = 0. Now let G be the minimal field of x^2+1 over Q. That is, G contains all rational numbers and a root of x^2+1 = 0 and all the numbers that result from this; for example, 3+i would be an element of G, since 3 is rational and i satisfies the polynomial. Then x^2+1 = (x+i)(x-i) over G and so G is the splitting field of x^2+1 over Q. But note. G is NOT the complex numbers! You can't get pi from either x^2+1 = 0 or from Q. G is sometimes called Q[i]. It is the quotient field of the Gaussian integers; that is, those numbers a+bi where a and b are rational numbers. The Gaussian integers are the set of all those numbers a+bi where a and b are integers.What is the splitting field of x^2 + 1 over Q?
the field of complex numbers with rational coefficients.What is the splitting field of x^2 + 1 over Q?
Right since x^2+1 = (x+i)(x-i), the splitting field is Q(i).


You can show that Q(i) = Q[i] directly or using a theorem about algebraic numbers that says that:


If a is algebraic over a field K then Q(a) = Q[a].





Also Q[i] = Q+Qi, since 1 and i form a basis of Q(i) over Q(or you can show directly)
say what?? 0.o