Why is this polynomial the minimal? Let $F = \mathbb{R}(x) \cong \mathbb{R}^{(1)}$ and define $h \in \sqrt{F}$ by $h = x^{1/2} + x^{1/3}.$ Find and simplify the minimal polynomial $P(T) \in F(T)$ of $h$ over $F.$}
Here is a solution:
\textbf{Solution:}
Since $h \in \sqrt{F}$ is given by $h = x^{1/2} + x^{1/3}$ we will try to eliminate the radicals from the equation. We will do this by simplifying $(h - x^{1/2})^{3}.$\
Given $$h = x^{1/2} + x^{1/3}$$ We have
\begin{align*}
h - x^{1/2} &=  x^{1/3}\\
(h - x^{1/2})^3 &=  x &&\text{( we cubed both sides)}\\
h^3 - 3h^2 \sqrt{x} + 3hx - x \sqrt{x} &= x \\
h^3 + (3h -1)x &= \sqrt{x}(3h^2 + x)\\
[h^3 + (3h -1)x]^2 &= [\sqrt{x}(3h^2 + x)]^2&&\text{( we squared both sides)}\\
h^6 + (3h -1)^2x^2 + 2h^3(3h-1)x &= x(3h^2 + x)^2 \\
h^6 + (9h^2 - 6h + 1) x^2 + (6h^4 - 2h^3)x &= x(9h^4 + 6h^2x + x^2)\\
h^6 + (9h^2x^2 - 6hx^2 + x^2)  + (6h^4x - 2h^3x) - 9xh^4 - 6h^2x^2 - x^3&= 0\\
h^6 -3xh^4 - 2xh^3 + 3x^2 h^2 - 6x^2h + (x^2 - x^3) &= 0 &&\text{( 1)}
\end{align*}
So, we have that $$P(T) = T^6 -3xT^4 - 2xT^3 + 3x^2 T^2 - 6x^2T + (x^2 - x^3) &= 0$$
is the minimal polynomial of $h = x^{1/2} + x^{1/3}$  over $F$ as $P(h) = 0.$
My question is: do we have to prove that this polynomial is minimal to complete the solution? if so, how can I do that?
Thanks!
 A: You need some results from field theory here. First of all, the smallest field that contains $h$ and $\mathbb R(x)$, has degree (over $\mathbb R(x)$) equal to the minimal polynomial of $h$ over $\mathbb R(x)$. Therefore, if we can prove that this degree is at least $6$, we are done, since it is at most $6$ from the polynomial you produced.
You have already done some of the work. That is, when you wrote $h^3 + (3h-1)x = \sqrt x(3h^2+x)$, we get that $\sqrt x = \frac{h^3+(3h-1)x}{3h^2+x}$. So, any field that contains $h$ and $\mathbb R(x)$, also contains $\sqrt x$.
But then, $x^{\frac 13} = h - \sqrt x$, hence such a field also contains $x^{\frac 13}$.
Conclusion : The smallest field containing $h$ and $\mathbb R(x)$, also contains $x^{\frac 12}$ and $x^{\frac 13}$.

The minimal polynomial of $x^{\frac 12}$ over $\mathbb R(x)$ is the irreducible $f(T) = T^2-x$ (to show that this is the minimal polynomial is far easier than in our question). So we get $\mathbb R(x^{\frac 12})$ has degree $2$ over $\mathbb R(x)$.
Now, the minimal polynomial of $x^{\frac 13}$ over $\mathbb R(x^{\frac 12})$ is $T^3 - (x^{\frac 12})^2 = 0$, which is again easily shown to be irreducible. It follows by field theory that the extension $R(x^{\frac 12} , x^{\frac 13})$ over $\mathbb R(x)$ has degree at least $2 \times 3 = 6$, as desired.

Edit : Over $\mathbb R(x)$, the polynomial $f(T) = T^2 - x$ is irreducible. Suppose that $f(T) = g(T)h(T)$ as polynomials with coefficients in $\mathbb R(x)$. The $T$ degree of $f(T)$ is $2$, so that of $g$ and $h$ have to be $1$ so that they are non-trivial factors. Finally, $f(T) = (a(x)T + b(x))(c(x)T + d(x))$ for some polynomials $a,b,c,d$ of $x$.
Multiplying out, and comparing the $T$ coefficients, we get $a(x)c(x) = 1$, so both of $a(x)$ and $c(x)$ are constant. Then , $b(x)d(x) =-x$, so one of $b(x)$ or $d(x)$ is a constant, and the other is not. But then $ad + bc = 0$, so if three of them are constant the last one is also constant, which is a contradiction to the statement made in the previous sentence. It follows that such a non-trivial factorization can't exist.

Edit 2 : We want to show that the polynomial $T^3 - (x^{\frac 12})^2$ is irreducible over the field $\mathbb R(x^{\frac 12})$. For notational simplicity, let us take $B = x^{\frac 12}$.
Suppose $T^3 - B^2$ is reducible. Then by comparing $T$ degrees, there exist five polynomials in $B$, namely $a,b,c,d,e$ such that $$
T^3 - B^2 = (a(B)T + b(B)) (c(B)T^2 + d(B)T + e(B))
$$

Why is this? Well, if you have a cubic polynomial(in $T$) which is factorizable, then it must break into two non-trivial factors, but the sum of their degrees must equal $3$. The only combination is $1+2 = 3$, because the degrees of these polynomials have to be positive but have to sum to $3$. One of these has degree $2$, which is of the form $c(B)T^2+d(B)T+e(B)$, and the other must have degree $1$, which is of the form $a(B)T+b(B)$.

now expand and compare coefficients of degree $T$ again. First, $a(B)c(B) = 1$ so both $a,c$ are constant. Next, $b(B)e(B) = -B^2$, so one of them is a constant, and the other is of $B$-degree $2$, or else both are of degree $1$.
Imagine $e$ is constant. Then $b$ is of $B$-degree $2$. We have $ad = bc$ from comparing $T^2$ coefficients, so to have equal $B$-degrees, $d$ is of $B$-degree $2$. Then comparing $T$ coefficients, $bd=ce$ which is clearly impossible by comparing $B$-degrees.
Imagine $b$ is constant. Then $e$ is of $B$-degree $2$. From $ad=bc$, it follows that $d$ is constant, but then $bd=ce$ can't happen since the $B$-degrees don't match.
Finally, if both of them have $B$ degree $1$, then from $ad=bc$ we know that $d$ is of $B$-degree $1$, and then $bd = ce$ cannot happen because the $B$-degrees don't match.
In summary, a factorization which is non-trivial, does not exist.
