TITLE: Is every element of $\mathbb{C}$ is algebraic over $\mathbb{R}$?
QUESTION [5 upvotes]: Is every element of $\mathbb{C}$ is algebraic over $\mathbb{R}$? I believe that this is true since elements of $\mathbb{C}$ are of the form $a + b\mathfrak{i}$ for $a$ and $b$ elements of $\mathbb{R}$. I am just unsure how to approach a justification for my answer.

REPLY [11 votes]: True.
For
$a + bi \in \Bbb C, \; a, b \in \Bbb R, \tag 1$
we have
$(x - (a + bi))(x - (a - bi)) = x^2 - 2ax + (a^2 + b^2) \in \Bbb R[x], \tag 2$
so the roots of this quadratic are $a \pm bi$, which are thus algebraic over $\Bbb R$.