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.

• I passed a theorem back when I took algebra which I think was phrased by the lecturer as "An algebraic closure is algebraically closed, aren't we lucky?". That should answer question 2 Jul 18 '17 at 8:45
• What you denote $\Bbb{QI}$ is usually denoted $\Bbb{Q}(i)$. It is the smallest field that contains all the rationals as well as the number $i$. Anyway, it is a standard fact that if a number $z$ is a zero with a polynomial $p(x)$ with coefficients in $\Bbb{Q}(i)$, then it is also a zero of a polynomial with coefficient in $\Bbb{Q}$. This is because the product $p(x)\overline{p}(x)$ has rational coefficients and obviously has $z$ as a zero. Check it out yourself! Jul 18 '17 at 8:47
• A more general result goes as follows. We say that $z$ is algebraic over a field $K$ if it is a zero of a polynomial with coefficients in $K$. The result claims that if $z$ is algebraic over a field $L$, and all the elements of $L$ are algebraic over $K$, then $z$ is algebraic over $K$. The proof is a bit more involved but not difficult at all when you are lead to think about it in terms of field extensions. Anyway, all the elements of $\Bbb{Q}(i)$ are algebraic over $\Bbb{Q}$ (basically because $i$ is), and the result follows. Jul 18 '17 at 8:51
• @5xum What do you deduce from this? Jul 18 '17 at 8:53
• Thank you all for the clarification, it is not the first time the problem lies in my inability to properly name what I am looking for. Jul 18 '17 at 8:54

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

1. $z$ as a zero,
2. its coefficients in $\Bbb{Q}(i)$, and
3. its coefficients are real.
4. 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 $K$.