Direct Sum algebraic number I'm doing a course in algebraic number theory and I don't understand the direct sum notation. How can the direct sum of fields be equal to a field? I thought the direct sum is a Cartesian product, where if you take the direct sum of $n$ fields, then each element of the direct sum is an $n$-tuple? 
For example take a field $K$ and an element a which is algebraic over $K$ of degree $n$. Then $K(a)=\bigoplus_{i=1}^{n-1} K(a^i)$. 
I'm very confused. If somebody can assist me here that would be great! 
 A: The direct sum (or rather the product) of fields is not even an integral domain. 
In your context, $K(\alpha)$ is the direct sum of the $K\alpha^i$s as a $K$-vector space.
A: In your first paragraph it is a bit unclear what direct sums you are talking about. You are right that a non-trivial finite product of fields is not a field.
But the direct sum decomposition that you give as an example is one of $K$-vector spaces, not rings (resp., fields). So, more explicitly, the isomorphism preserves addition and multiplication with scalars (elements of $K$), but does not preserve multiplication.
For instance, $$\mathbb Q(\sqrt{2}) \cong \mathbb Q \oplus \mathbb Q\sqrt 2.$$ (as $\mathbb Q$-vector spaces)
by sending $1$ to $(1,0)$ and $\sqrt{2}$ to $(0,\sqrt{2})$. However, on the left the multiplication gives, for instance:
$$
(1+\sqrt{2})(1+\sqrt{2}) = 1 + 2 \sqrt{2} + 2 = 3+2\sqrt{2}.
$$
The right hand side, $\mathbb Q \oplus \mathbb Q\sqrt 2$, has a natural ring structure, but it is different. E.g.,
$$
(1,\sqrt{2})\cdot (1,\sqrt{2}) = (1\cdot1, \sqrt{2}\cdot \sqrt{2}) = (1,2).
$$
So the isomorphism above is not an isomorphism of rings, if you consider the natural multiplication on the left hand side and the componentwise multiplication on the right hand side.
