Showing that $7+\sqrt[3]{2}$ is an algebraic number

How do I go about showing that $7+\sqrt[3]{2}$ is an algebraic number?

I need to show that it is the root of an integer valued formal polynomial? How do I solve these problems in general? I haven't a clue where to start on this one, so it is hard to give context.

The problem is perhaps solved using methods of ring theory.

• This particular one is easy to find a polynomial for. One could also prove that any number obtained by arithmetic operations from algebraics is algebraic. May 27 '15 at 5:10
• When I see the revision history, I think that it is worth pointing out that (algebraic-numbers) tag is now discussed on meta. May 31 '15 at 9:45

Well if you manage to find a polynomial with that as its root, then it is an algebraic number. The most obvious choice would be its minimal polynomial.

$x=7+\sqrt[3]{2}$

$(x-7)^3=2$

Expand that and move all terms to one side, and there's your polynomial.

I would also like to add that not all problems are this trivial. For example, showing that $\sqrt{2}+\sqrt{3}$ requires a little more work. Even worse is something like $\cos \left ( k \pi \right )$ where $k\in\mathbb{Q}$

• The $x=\cos(\frac pq\pi)$ isn't so hard. Note that $\cos(qA)$ is always a polynomial of $\cos(A)$, because you can use the addition formulae a bunch of times. That means that $\cos(p\pi)$ (an integer) is a polynomial of $\cos(\frac pq\pi)=x$… May 27 '15 at 20:02

So, first you write down the silliest thing: your number is a solution of $$x = 7 + \sqrt[3]{2}$$ (Always do the silliest thing first.) Well, you say, that's not a polynomial with integer coefficients. The constant term has that $\sqrt[3]{2}$ in it, not an integer at all. So you want to get rid of that $\sqrt[3]{2}$. I know, you think, I'll cube everything! That'll get rid of that nasty $\sqrt[3]{2}$! $$x^3 = (7 + \sqrt[3]{2})^3 = 7^3 + 3\cdot49\sqrt[3]{2} + 3\cdot7\cdot\sqrt[3]{4} + 2$$ Alas, now there are even more $\sqrt[3]{\cdot}$ than when we started. What went wrong? The 7 interfered, created all those cross terms. We should cube the $\sqrt[3]{2}$ without interference. So, move everything to the other side, where it won't get in the way. $$x - 7 = \sqrt[3]{2}$$ Then cube. $$(x-7)^3 = 2$$ All integers now, hurrah.

• Ive always wanted to really put the effort in showing the intuition-behind-the-intuition. I am very glad you took the time to go that second level step of why we ought to even shift things over in the first place. Often its the motivation behind that motivation, such as here, that many don't know how to connect May 27 '15 at 5:25

The other responses are great (+1), but what if you had been given a more complicated number like $\sqrt{3} + \sqrt[4]{5} + \sqrt[3]{17} + 1/\sqrt{2}$? Indeed, it'd be nigh-impossible to find a polynomial with this number as a root!

Well, the set of all algebraic numbers is a field, and as such, with a more complicated number as above, it would suffice to show that $\sqrt{3}$, $\sqrt[4]{5}$, $\sqrt[3]{17}$, and $\sqrt{2}$ are all algebraic. Then you can apply the field axioms to conclude that the number in question is algebraic.

You could apply this approach to your problem, since it is fairly obvious that $7$ and $\sqrt[3]{2}$ are algebraic. But of course, as the other responses have shown, it is also fairly easy to prove it directly by constructing a polynomial for which $7+\sqrt[3]{2}$ is a root.

Footnotes:

You can prove that the algebraic numbers form a field in a few different ways.

• One way is to follow Ragib's approach in his answer here, where he is applying the fact that every element of an extension of a field $F$ of finite degree is algebraic over $F$. This approach is perhaps most accessible to a beginner.

• In greater generality, Zorn's lemma can be used to show that every field $F$ has what is called an algebraic closure, which is an algebraically closed extension field of $F$ whose elements are all algebraic over $F$. See here for a proof.

• Proving, in general, that the algebraic numbers form a field might be more difficult than the specific approaches in the other answers.
– robjohn
May 27 '15 at 5:17
• No doubt about that @robjohn. The other responses are objectively better, especially if that result isn't yet known. But I think that this perspective has at least a little pedagogical value, and the OP will be able to appreciate it sometime soon, if not already. May 27 '15 at 5:21
• I do appreciate it and I will now go about trying to prove it. May 27 '15 at 5:23
• @OceansBleed, I've added some footnotes to my answer that you might find helpful. Don't click on them yet if you want to take a stab at a proof on your own. May 27 '15 at 5:40

I'd like to strike a middle ground between these excellent answers. I shall use two results without proof. You might try to see if you can prove these, yourself.

1. If $R$ is an extension ring of a field $F$, then $R$ is a vector space over $F$.

2. If $u$ is algebraic over a field $F$, then $F[u]$ is a finite-dimensional vector space over $F$.

(Hint: suppose $p(x)\neq 0 \in F[x]$ is such that $p(u) = 0$. Show that if $m = \text{deg }p$, that $\{1,u,u^2,\dots,u^m\}$ is linearly dependent over $F$. Why is this enough?)

Now clearly $\Bbb Q$ is a field, and $\sqrt[3]{2}$ is algebraic over $\Bbb Q$ (it is a root of $p(x) = x^3 - 2 \in \Bbb Q[x]$), and we have $7 + \sqrt[3]{2} \in \Bbb Q[\sqrt[3]{2}]$. Hence if $\dim_{\Bbb Q}(\Bbb Q[\sqrt[3]{2}]) = k$, we have that:

$\{1, 7 + \sqrt[3]{2},(7 + \sqrt[3]{2})^2,\dots,(7 + \sqrt[3]{2})^k\}$ must be a $\Bbb Q$-linearly dependent set, that is, there exist:

$c_0,c_1,\dots,c_k \in \Bbb Q$ not all $0$ such that:

$c_0 + c_1(7 + \sqrt[3]{2}) + c_2(7 + \sqrt[3]{2})^2 + \cdots + c_k(7 + \sqrt[3]{2})^k = 0$.

So, if $q(x) = c_0 + c_1x + c_2x^2 + \cdots + c_kx^k$, evidently $q(7 + \sqrt[3]{2}) = 0$, so $7 + \sqrt[3]{2}$ is algebraic.

(Note that this "miraculously" produces our desired result, without actually finding the polynomial we want).