It follows from $\sqrt{2}+\sqrt{3} \in \Q(\sqrt{2}, \sqrt{3})$ that we have $\Q(\sqrt{2}+\sqrt{3})\subset \Q(\sqrt{2}, \sqrt{3})$.

To show the reverse inclusion, consider
\[(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}\in \Q(\sqrt{2}+\sqrt{3}).\]
This yields that we have
\[\sqrt{6}\in \Q(\sqrt{2}+\sqrt{3}).\]

Now we can express $\sqrt{2}$ in terms of elements of $\Q(\sqrt{2}+\sqrt{3})$ as follows. We have
\begin{align*}
\sqrt{6}(\sqrt{2}+\sqrt{3})-2(\sqrt{2}+\sqrt{3})=\sqrt{2}\in \Q(\sqrt{2}+\sqrt{3})
\end{align*}
(Note that the numbers on the left hand side are all in the field $\Q(\sqrt{2}+\sqrt{3})$.)

Hence we also have
\[(\sqrt{2}+\sqrt{3})-\sqrt{2}=\sqrt{3}\in \Q(\sqrt{2}+\sqrt{3}).\]
Therefore the elements $\sqrt{2}, \sqrt{3}$ are in the field $\Q(\sqrt{2}+\sqrt{3})$, hence
\[\Q(\sqrt{2}, \sqrt{3}) \subset \Q(\sqrt{2}+\sqrt{3}).\]
Since we showed both inclusions, we have
\[\Q(\sqrt{2}, \sqrt{3})= \Q(\sqrt{2}+\sqrt{3}).\]

Galois Group of the Polynomial $x^2-2$
Let $\Q$ be the field of rational numbers.
(a) Is the polynomial $f(x)=x^2-2$ separable over $\Q$?
(b) Find the Galois group of $f(x)$ over $\Q$.
Solution.
(a) The polynomial $f(x)=x^2-2$ is separable over $\Q$
The roots of the polynomial $f(x)$ are $\pm […]

Degree of an Irreducible Factor of a Composition of Polynomials
Let $f(x)$ be an irreducible polynomial of degree $n$ over a field $F$. Let $g(x)$ be any polynomial in $F[x]$.
Show that the degree of each irreducible factor of the composite polynomial $f(g(x))$ is divisible by $n$.
Hint.
Use the following fact.
Let $h(x)$ is an […]

Application of Field Extension to Linear Combination
Consider the cubic polynomial $f(x)=x^3-x+1$ in $\Q[x]$.
Let $\alpha$ be any real root of $f(x)$.
Then prove that $\sqrt{2}$ can not be written as a linear combination of $1, \alpha, \alpha^2$ with coefficients in $\Q$.
Proof.
We first prove that the polynomial […]

$x^3-\sqrt{2}$ is Irreducible Over the Field $\Q(\sqrt{2})$
Show that the polynomial $x^3-\sqrt{2}$ is irreducible over the field $\Q(\sqrt{2})$.
Hint.
Consider the field extensions $\Q(\sqrt{2})$ and $\Q(\sqrt[6]{2})$.
Proof.
Let $\sqrt[6]{2}$ denote the positive real $6$-th root of of $2$.
Then since $x^6-2$ is […]

Two Quadratic Fields $\Q(\sqrt{2})$ and $\Q(\sqrt{3})$ are Not Isomorphic
Prove that the quadratic fields $\Q(\sqrt{2})$ and $\Q(\sqrt{3})$ are not isomorphic.
Hint.
Note that any homomorphism between fields over $\Q$ fixes $\Q$ pointwise.
Proof.
Assume that there is an isomorphism $\phi:\Q(\sqrt{2}) \to \Q(\sqrt{3})$.
Let […]

The Polynomial $x^p-2$ is Irreducible Over the Cyclotomic Field of $p$-th Root of Unity
Prove that the polynomial $x^p-2$ for a prime number $p$ is irreducible over the field $\Q(\zeta_p)$, where $\zeta_p$ is a primitive $p$th root of unity.
Hint.
Consider the field extension $\Q(\sqrt[p]{2}, \zeta)$, where $\zeta$ is a primitive $p$-th root of […]

Prove that $\F_3[x]/(x^2+1)$ is a Field and Find the Inverse Elements
Let $\F_3=\Zmod{3}$ be the finite field of order $3$.
Consider the ring $\F_3[x]$ of polynomial over $\F_3$ and its ideal $I=(x^2+1)$ generated by $x^2+1\in \F_3[x]$.
(a) Prove that the quotient ring $\F_3[x]/(x^2+1)$ is a field. How many elements does the field have?
(b) […]

Any Automorphism of the Field of Real Numbers Must be the Identity Map
Prove that any field automorphism of the field of real numbers $\R$ must be the identity automorphism.
Proof.
We prove the problem by proving the following sequence of claims.
Let $\phi:\R \to \R$ be an automorphism of the field of real numbers […]