Extension of algebraic numbers I did not find a question like this, so I hope I am not copying another question. If I am, I am sorry.
Before I ask my question, first let $\Bbb Q(i)$ denote numbers, which are all numbers of the form $a+bi$, where $a,b \in \Bbb Q$. And I will note algebraic numbers as $\Bbb A$.


*

*My first  question is, whether numbers, that are defined as roots of polynomials with coefficients from $\Bbb{Q}(i)$ are proper superset of algebraic numbers or not?

*My second question is,  whether numbers, that are defined as roots of polynomials with coefficients from $\Bbb{A}$ are proper superset of algebraic numbers or not?
I think that both of them are equal to algebraic numbers, because such a polynomial can always be extended to a polynomial with at least real coefficients using a method described here https://en.wikipedia.org/wiki/Complex_conjugate_root_theorem .
Because when there is a factor $(x+t+si)$, I simply start multiplying the polynomial with $(x+t-si)$, until this conjugate root has the same power in the polynomial as the root of that factor has. That would result in a polynomial with real coefficients for sure. But whether they are rational, or the polynomials can further be extended to have rational coefficients, that I do not know (how to prove or disprove), but I think they are, or can be extended.
 A: A summary of textbook material on algebraic extensions of fields and algebraic elements (off the top of my head, just to get this question off the unanswered list).
Definition. If $K\subseteq L$ are fields and $z\in L$, we say that $z$ is algebraic over $K$, if there is a non-zero polynomial $p(x)\in K[x]$ such that $p(z)=0$. We say that $L$ is algebraic over $K$, if all the elements of $L$ are algebraic over $K$.
Basic observation. In the above setting $z$ is algebraic over $K$ if and only if there exists a vector space $V\subseteq L,V\neq\{0\}$, finite dimensional over $K$, such that $V$ is closed under multiplication by $z$.
Proof (sketch). If $z$ is a zero of a polynomial $p(x)\in K[x]$ of degree $n$, then $$V=K\cdot 1+K\cdot z+\cdots+K\cdot z^{n-1}$$ is closed under multiplication by $z$ because $z^n$ can be written as a $K$-linear combination of lower powers of $z$. For the other direction we can consider the characteristic polynomial $m(x)\in K[x]$ of the $K$-linear transformation $\rho:V\to V, \rho(x)=zx$. By basic linear algebra $m(z)=0$.
Corollary 1. If $[L:K]<\infty$ then $L$ is algebraic over $K$.
Proof. In this case we can use $V=L$ for all $z\in L$.
In particular because $\Bbb{Q}(i)$ is a 2-dimensional vector space over $\Bbb{Q}$, all its elements are algebraic over $\Bbb{Q}$.
Corollary. If $K\subseteq L\subseteq M$ are fields, $z\in M$ is algebraic over $L$, and $L$ is algebraic over $K$, then $z$ is algebraic over $K$.
Proof (Sketch). Let $p_L(x)=a_0+a_1x+\cdots a_nx^n\in L[x]$ be a non-zero polynomial such that $p_L(z)=0$. Because all the coefficients $a_i\in L$ there exist finite dimensional $(/K)$ subspaces $V_i,i=0,\ldots,n$, such that $V_i$
stable under multiplication by $a_i$. We then easily see that the space
$$
V=V_0\cdot V_1\cdots V_n\cdot W,
$$
where $W=K+K\cdot z+\cdots K\cdot z^n$, is stable under multiplication by $z$. Here I define the product $U_1\cdot U_2$ of two subspaces, $U_1$ and $U_2$, to be the space of finite $K$-linear combinations
$$
U_1U_2=\{\sum_{j=1}^m a_ju_{1j}u_{2j}\mid a_j\in K, u_{1j}\in U_1, u_{2j}\in U_2\}.
$$
We easily see that $\dim_K U_1\cdot U_2\le \dim_K U_1\cdot \dim_K U_2$. Consequently
$$
\dim_KV\le n\cdot\prod_{i=0}^n\dim_KV_i
$$
is finite. The claim follows.
Your claim is answered in the affirmative as a consequence of Corollary 2. Any number that is algebraic over $\Bbb{Q}(i)$ is necessarily also algebraic over $\Bbb{Q}$.

In your case an easier route to the destination is to observe that if 
$$p(x)=a_0+a_1x+\cdots+a_nx^n$$
has coefficients in $\Bbb{Q}(i)$ (and $p(z)=0$ for some $z$), then, denoting
polynomial gotten by conjugating all the coefficients
$$
\overline{p}(x)=\overline{a_0}+\overline{a_1}x+\cdots+\overline{a_n}x^n,
$$
we see that the product $p(x)\overline{p}(x)$ has


*

*$z$ as a zero,

*its coefficients in $\Bbb{Q}(i)$, and

*its coefficients are real.

*Therefore its coefficients are actually rational.


A similar argument is available more generally when $L/K$ is a Galois extension, when we can apply all the field automorphisms to the coefficients of $p(x)\in L[x]$ and as the product get a polynomial with coefficients in the smaller field $K$.
