# Degrees of the real and imaginary parts of an algebraic number

I am working on a theory of generalized geometric constructions, which involves generating new numbers as real roots of polynomials whose coefficients are existing numbers satisfying certain relationships. The following general questions arise, and I am wondering if anyone already knows the answers or can link to a source that would have them:

(1) If $x = a + b i$ is a root of an irreducible polynomial over the rationals of degree $n$, what is the maximum possible degree of the irreducible polynomials over the rationals for the real numbers $a$ and $b$?

(2) If $\operatorname{Alg}_n$ is the field generated by all roots of polynomials over the rationals of degree $\leq n$, and $\operatorname{RAlg}_n$ is the field generated by all REAL roots of polynomials over the rationals of degree $\leq n$, for which $n$ does $\operatorname{Alg}_n=\operatorname{RAlg}_n[i]$?

(3) The inverse function of f(x)=x^5+x is uniquely defined for all real numbers; the inverse function of f(x)=x^5-x is uniquely defined for |x|>sqrt(sqrt(0.08192)) and has two or three real values otherwise. Can either of these functions be obtained from the other (assuming one has access to all three real roots when they exist), using only operations of degree <5 and complex 5th roots (that is, real 5th roots and angle 5-sections)?

## migrated from mathoverflow.netJul 27 '13 at 4:55

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• Background of the problem, ideas, self effort...? – DonAntonio Jul 27 '13 at 9:21

For $(1)$:

If $r$ is a root of $P(x)$ and $s$ is a root of $Q(x)$ then $x+y$ is a root of the resultant of $P(x)$ and $Q(z-x)$ considering $Q(z+x)$ a polynomial in $x$, and $rs$ is a root of the resultant of $P(x)$ and $x^{deg(Q)}Q(z/x)$.

The resultant vanishes iff $P(x)$ and $Q(z-x)$ have a common root. Notice that if $z=r+s$ then $P(r)=0$ and $Q((r+s)-r)=Q(s)=0$. Likewise for $P(x)$ and $x^{deg(Q)}Q(rs/x)$. The resultant has degree equal $deg(P)deg(Q)$. We have that $a+ib$ is root of a polynomial of degree $n$, $a-ib$ is too, the same polynomial. Then $2a=(a+ib)+(a-ib)$ is a root of a polynomial of degree $2n$, and therefore $a$ is too. Therefore a bound for a is $2n$.

Likewise $ib$ is root of a polynomial of degree $2n$ and $i$ is root of a polynomial of degree $2$, $i^2+1=0$. So, from the above we get that $b$ is a root of a polynomial of degree at most $4n$.

For $(2)$:

If $m$ and $n$ are coprime and $x$ is algebraic of degree $n$ and $y$ is of degree $m$ then $x+y$ is of degree $n+m$. I think therefore, as long as we can produce two coprime numbers $\leq n$ (and $>1$) we will have $RAlg_n[i]$ larger than $ALg_n$. So for $n>2$ we shouldn't get equality.

1. For $n=1$, $Alg_n$ doesn't contain $i$.

2. For $n=2$, $Alg_n$ doesn't contain $\sqrt{2}+i\sqrt{3}$. The minimal polynomial has degree $4$.

3. For $n=3$, $Alg_n$ doesn't contain $\sqrt[3]{2}+i\sqrt[3]{3}$. The minimal polynomial has degree $18$.