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Let $A$ be the set of all algebraic numbers. Show that $A$ is closed under addition and multiplication.

Attempt

Let $a, b \in A$, $\deg_{\mathbb{Q}}(a) = m$, and $deg_{\mathbb{Q}}(b) = n$.

Then $\mathbb{Q}(a,b) = \{g(a)hb : g, h \in \mathbb{Q}[x], \deg(g) \leq m - 1, deg(n) \leq n - 1.\}$

Then $\mathbb{Q}(a,b)$ is a finite dimensional $\mathbb{Q}$-subalgebra of $\mathbb{C}$.

Okay from here if I can find a spanning set with $\leq mn$ elements I think I am good, however this is where I am a little unsure.

Mr Mathster
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Daniel Smith
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  • Does your text or notes write "$\deg_{\Bbb Q}(a)$"? I've not seen that notation before. – anon Oct 29 '13 at 04:14
  • Ya it does, however I believe it would be no different then having it without the rational number symbol to be honest. – Daniel Smith Oct 29 '13 at 04:17
  • Anyway. Your definition of $\Bbb Q(a,b)$ is a bit weird. Mind fixing it? And then... how do you conclude $\Bbb Q(a,b)$ is finite-dimensional? That conclusion is actually the whole meat of the proof. Rather than concluding $\Bbb Q(a,b)$ is finite-dimensional and then seeking a spanning set, what you need to do is exhibit a spanning set in order to show that $\Bbb Q(a,b)$ is finite-dimensional! Then you're home free: since $\Bbb Q(ab)$ and $\Bbb Q(a+b)$ are subalgebras of $\Bbb Q(a,b)$, they are subspaces and so finite-dimensional over $\Bbb Q$, hence $ab$ and $a+b$ are algebraic. – anon Oct 29 '13 at 04:21
  • Apply http://math.stackexchange.com/questions/482747/showing-ku-is-a-field-when-u-is-algebraic-over-k twice. – Prahlad Vaidyanathan Oct 29 '13 at 04:22
  • Now, take a stab at the most obvious spanning set of $\Bbb Q(a,b)$. As a hint, what do the elements of a polynomial ring in two variables, $\Bbb Q[x,y]$, look like? Which of those elements can be simplified and are therefore redundant if $a$ and $b$ are algebraic of specified degrees? – anon Oct 29 '13 at 04:23
  • I thought about the spanning set, and would my spanning set me ${a^{i}b^{j}}$ where $0 < i < m$ and $0 < j < n$? – Daniel Smith Oct 29 '13 at 21:10
  • You can use the fact that if $V=[\gamma_1,\gamma_2,\ldots, \gamma_t]$ is a $\mathbb{Q}$-module and that $\alpha\in\mathbb{C}$ is such that $\alpha\gamma\in V$ for all $\gamma\in V$, then $\alpha$ is an algebraic number, and then consider the $\mathbb{Q}$-module from forming all linear combinations of elements $\alpha_1^i\alpha_2^j$. Closure is not far off. – tc1729 Feb 16 '14 at 04:01

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