Let $E/F$ be a field extension. Let $A$ be the set of all elements in $E$ that are algebraic over $F$. One can show that $A$ is a subfield of $E$. The proofs I read all argue as follows:

Let $a,b \in A$. We want to show that $ab, a +b \in A$. To this end we show that $[F(a,b):F]$ is finite. Finite implies algebraic hence since $a+b, ab \in F(a,b)$, $a+b, ab$ are algebraic and hence in $A$.

I understand this proof but I was wondering if the following argument also works or whether it is flawed:

Let $a,b$ be algebraic over $F$. Then there exist $p,q \in F[x]$ with $p(a) = 0$ and $q(b) = 0$. Define $p' (x) = p(x-b)$ and $p''(x) = p(x b^{-1})$. Then $p'(a + b) = p(a) = 0$ and $p''(ab) = p(a) = 0$. Hence $ab, a+b$ are algebraic over $F$ (since $p', p'' \in F[x]$). Is this correct?

Thanks for help!

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    $\begingroup$ Be careful with using the $'$ character to indicate different polynomials, since it also might mean polynomial differentiation. $\endgroup$ Oct 2, 2012 at 15:35

2 Answers 2


Consider $a = i$ and $b = \sqrt{2}$ over $\mathbb{Q}$. Then we have $p(x) = x^2 + 1$ and $q(x) = x^2 - 2$ for our irreducible polynomials over $\mathbb{Q}$. But $p'(x) = p(x - \sqrt{2}) = x^2 - 2\sqrt{2}x + 3$ is not in $\mathbb{Q}[x]$ after all.

  • $\begingroup$ Did you mean to write "But $q'(x) = \dots$"? $\endgroup$ Oct 2, 2012 at 15:48

If $\alpha$ and $\beta$ are algebraic, the theory of the resultant gives explicit polynomials satisfied by $\alpha + \beta$ and $\alpha\beta$ in terms of the minimal polynomials for $\alpha$ and $\beta$, which I assume is what you were after:


As you can see, it's a much more difficult proof than the non-constructive one which recasts algebraicness in terms of finiteness.


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