How to formally prove that we cannot find a polynomial in $\textbf Z[x]$ with degree $2$ with such a root?

Question

I am trying to find the kernel of the map from $\textbf Z[x]$ to $\textbf C$. The map is evaluating at $\sqrt 2 + \sqrt 3$.

A solution says that we cannot find polynomials of degree $2$ or $3$ that has such a root. So it skips the procedure of trying degree $2$ and $3$. And the final solution is the ideal in $\textbf Z[x]$ generated by $$x^4 − 10x^2 + 1 = (x − \sqrt 2 − \sqrt 3)(x − \sqrt 2 + \sqrt 3)(x + \sqrt 2 − \sqrt 3)(x + \sqrt 2 + \sqrt 3)$$

A solution says there are $\sqrt 6$'s in $(\sqrt 2 + \sqrt 3)^2$ and $(\sqrt 2 + \sqrt 3)^3$. So we cannot find polynomials with root $\sqrt 2 + \sqrt 3$ of degree $2$ or $3$. And that is what I am confused about: How does it imply the fact?

Thanks in advance!

Just FYI: it is better to use \sqrt{} than √. Then the overbar can extend as far as needed, e.g., \sqrt{10} makes $\sqrt{10}$. — David R., Oct 04 '16 at 21:03

score 3 · Accepted Answer · edited Apr 13 '17 at 12:20

3

Proceed by contradiction.

Assume that there exist integer $a,b,c$ such that

$$ax^2+bx+c=0$$

with $x=\sqrt{2}+\sqrt{3}$.

$$a(5+2\sqrt{6})+b(\sqrt{2}+\sqrt{3})+c=0$$

Note that $1,\sqrt{2},\sqrt{3},$ and $\sqrt{6}$ are linearly independent, which means that there exist no integers $n_1,n_2,n_3,n_4$, not all $0$, such that

$$n_1+n_2\sqrt{2}+n_3\sqrt{3}+n_4\sqrt{6}$$

(A proof of this fact can be found here.) However, this means that

$$(5a+c)+b\sqrt{2}+c\sqrt{3}+(2a)\sqrt{6} \neq 0$$

which is a contradiction.

edited Apr 13 '17 at 12:20

Community

1

answered Oct 04 '16 at 01:47

Carl Schildkraut

32,931

Note that 1,√2,√3, and √6 are linearly independent That's true, but it's essentially equivalent to what the OP asked to prove. – dxiv Oct 04 '16 at 01:51
@dxiv Isn't there a relatively simple way to show it though? – Carl Schildkraut Oct 04 '16 at 01:54
1

@dxiv An easy way to see they are linearly independent is to see that the quotient of any two of them is irrational. I don't gather that the OP was asking for that; however, if he asks for it then it's not long to prove it. – Alex Oct 04 '16 at 01:54
@Alex Yeah, that's what I was thinking. I've edited my answer accordingly. – Carl Schildkraut Oct 04 '16 at 01:55
@Alex the quotient of any two of them is irrational Again, that's true but it's not a proof. To prove that, you have a problem that's essentially equivalent to OP's question. – dxiv Oct 04 '16 at 01:58
@dxiv No, it's not equivalent. It would be a totally different proof: Assume that $\sqrt{y}$ can be written as $a/b$ for $a,b\in\Bbb Z$. Then $y=a^2/b^2$. Then $3\mid y$, therefore $3\mid a$. etc. – Alex Oct 04 '16 at 02:00
1

$1$, $\sqrt{2}$ and $\sqrt{2}-1$ also have the property that the quotient of any two is irrational, but there is a nontrivial integral combination of them equal to zero. – stewbasic Oct 04 '16 at 02:01
@Alex Please write down what you think is the complete proof. – dxiv Oct 04 '16 at 02:02
2

@dxiv I know this isn't what you're looking for, but I refer you to an answer to a question previously posted on here. – Carl Schildkraut Oct 04 '16 at 02:04
@CarlSchildkraut Thanks! :) The solution I had in mind was the one provided by William: http://math.stackexchange.com/a/96964/48061 – Alex Oct 04 '16 at 02:05
@Alex & Carl I said a few times already that the statement is true, but it is neither obvious nor trivial, so it needed to be proved. All that I objected to was handwaving it as "note that..." which was simply not enough (as the actual proof that Carl linked clearly shows). – dxiv Oct 04 '16 at 02:09
@dxiv I was under the impression that this was a well-known statement. Is what I have now sufficient? – Carl Schildkraut Oct 04 '16 at 02:10
@CarlSchildkraut Yes, your answer is technically OK now, thank you. – dxiv Oct 04 '16 at 02:12
@dxiv I bet the OP would be stunned, when he comes back to see such a lengthy discussion about a technicality he didn't even have in mind... :D – Alex Oct 04 '16 at 02:13
1

@Alex Sorry, I disagree with it being called a technicality. That's in fact the meat of the problem. – dxiv Oct 04 '16 at 02:17
Thank you to let me know so much... That's a suprise! – Y.X. Oct 04 '16 at 02:52

Alex · Answer 2 · 2016-10-04T02:10:46.280

3

EDIT: The poster Carl has a more formal answer to what I was typing, when degree is 2.

For degree 3, note that $(\sqrt{2}+\sqrt{3})^3=11\sqrt{2}+9\sqrt{3}$. Again, trying to solve $$ax^3+bx^2+cx+d=0$$ would show no non-zero value for $a$ and $c$, would make $n_2$ and $n_3$ both equal to zero.

(where, as in his proof, $n_1+n_2\sqrt{2}+n_3\sqrt{3}+n_4\sqrt{6}=0$)

edited Oct 04 '16 at 02:10

answered Oct 04 '16 at 01:50

Alex

2,449

score 2 · Answer 3 · answered Oct 04 '16 at 21:42

Note that once you have that $\sqrt 2 + \sqrt 3$ is a root of $x^4-10x^2+1$, you can prove that the polynomial is irreducible over $\Bbb Z$ and thus $\sqrt 2 + \sqrt 3$ is not a root of a polynomial in $\Bbb Z[x]$ of degree lesser than $4$.

To show irreducibility, an elementary approach is to first note that it cannot be factored into form $(x+\ldots)(x^3+\ldots)$ because the polynomial has no rational roots by rational root test. Thus, if it could be factored, it would have to be of form $(x^2+ax\pm 1)(x^2+bx\pm 1)$. Expanding leads to system \begin{align} ab \pm 2 &= -10\\ a + b &= 0 \end{align} which has no solutions in $\Bbb Z$.

score 1 · Answer 4 · answered Oct 04 '16 at 22:12

Let's work, for the moment, in $\mathbb{Q}[x]$.

The number $b=\sqrt{2}+\sqrt{3}$ is certainly a root of $f(x)=x^4-10x^2+1$, because $(b-\sqrt{2})^2=3$, so $b^2-1=2b\sqrt{2}$ and, squaring again, $b^4-2b^2+1=8b^2$.

Therefore there is a monic polynomial $p(x)$ of minimal degree (with coefficients in $\mathbb{Q}$) which $b$ is a root of; in particular, $\deg p(x)\le 4$. If we do Euclidean division, we get $$ f(x)=p(x)q(x)+r(x) $$ with $r(x)=0$ or $\deg r(x)<\deg p(x)$. Evaluating at $b$ gives $$ f(b)=p(b)q(b)+r(b) $$ and, since $f(b)=p(b)=0$, we obtain $r(b)=0$. By minimality of $\deg p(x)$, we conclude that $r=0$. In particular, $p(x)$ is a factor of $f(x)$ with rational coefficients.

If we prove that $f(x)$ is irreducible in $\mathbb{Q}[x]$, we have that $f(x)=p(x)$ and, in particular, that no nonzero polynomial in $\mathbb{Q}[x]$ having $b$ as root has degree less than $4$.

If $f(x)$ is reducible, it can either be split into a product of two polynomials of degree $2$ or into a product of a degree $1$ polynomial and one of degree $3$.

The second possibility is dismissed, because the degree $1$ factor would provide a rational root of $f(x)$, which has none (the only rational roots can be $1$ and $-1$).

Let's try the other one: $f(x)=(x^2+Ax+B)(x^2+Cx+D)$ (it is not restrictive to assume the factors to be monic). This gives $$ \begin{cases} A+C=0\\ B+D+AC=-10\\ AD+BC=0\\ BD=1 \end{cases} $$ Hence $C=-A$ and, substituting in the third equation, $A(D-B)=0$.

The case $A=0$ leads to $B+D=-10$ and $BD=1$, which has no rational solutions. With $B=D$ we obtain $B^2=1$ and $2B-A^2=-10$. If $B=1$, we get $A^2=12$; with $B=-1$ we get $A^2=8$. Neither case leads to rational solutions.

Since $f(x)$ is irreducible in $\mathbb{Q}[x]$ and is monic, it is also irreducible in $\mathbb{Z}[x]$.

Now, suppose $g(x)\in\mathbb{Z}[x]$ belongs to the kernel of the homomorphism, meaning $g(b)=0$. We can still do Euclidean division by $f(x)$, because it is monic. Therefore $$ g(x)=f(x)q(x)+r(x) $$ with $r(x)=0$ or $\deg r(x)<\deg f(x)$. Evaluating at $b$ gives $r(b)=0$, so, as before, $r(x)=0$. Hence $g(x)$ belongs to the ideal generated by $f(x)$. The converse inclusion is obvious.

How to formally prove that we cannot find a polynomial in $\textbf Z[x]$ with degree $2$ with such a root?

4 Answers4