# What is an irreducible polynomial in $\mathbb{Z}$ that has root $\sqrt{2}+\sqrt{3}$?

What is an irreducible polynomial in $\mathbb{Z}$ that has root $\sqrt{2}+\sqrt{3}$? Obviously that root is not in $\mathbb{Z}$.

I tried $$(x-(-\sqrt{2}-\sqrt{3})(x-(-\sqrt{2}+\sqrt{3})(x-(\sqrt{2}-\sqrt{3})(x-(\sqrt{2}+\sqrt{3‌}))$$ That doesn't come out with integer terms. If I had to guess the degree of the integer coefficient irreducible polynomial, I'd guess $4$, but I don't know.

• What are your thoughts on the problem? What do you expect the degree of this polynomial to be? – Omnomnomnom Apr 27 '15 at 20:45
• Hint: What is $(\sqrt{2}+\sqrt{3})^4$? What is $(\sqrt{2}+\sqrt{3})^3$? What is $(\sqrt{2}+\sqrt{3})^2$? How can you cancel these therms if you sum/subtract multiples of them? – Bman72 Apr 27 '15 at 20:49
• I tried (x-(-sqrt(2)-sqrt(3))(x-(-sqrt(2)+sqrt(3))(x-(sqrt(2)-sqrt(3))(x-(sqrt(2)+sqrt(3)). That doesn't come out with integer terms... If I had to guess the degree of the integer coefficient irreducible polynomial, I'd guess 4, but I don't know. I'm not sure what else to try. – clay Apr 27 '15 at 20:49
• Multiply $(x-\sqrt{2}-\sqrt{3})(x-\sqrt{2}+\sqrt{3})=(x-\sqrt{2})^2-3=x^2-2\sqrt{2}x-1$ and then $(x^2-1-2\sqrt{2}x)(x^2-1+2\sqrt{2}x)=(x^2-1)^2-8x^2=x^4-10x^2+1$ – Alamos Apr 27 '15 at 20:51

Start with $x=\sqrt 2+\sqrt 3$ and square both sides: $$x^2=2+2\cdot\sqrt 6+3$$ Now isolate the square root and square again: $$(x^2-5)^2=(2\cdot\sqrt 6)^2=24$$ Expand. (Why is the resulting polynomial irreducible?)
• $x^4-10x^2+1=0$. Thanks! – clay Apr 27 '15 at 20:54
• @clay If you want to prove that $x^4-10x^2+1$ is irreducible you can first try all the possible integer roots (What are they?) and see that there are not. Secondly you can look if the polynomial factors as a product of two polynomials of degree $2$. I.e.: $x^4-10x^2+1=(x^2+ax+b)(x^2+cx+d)$ for some $a,b,c,d \in \Bbb{Z}$. Should this factorization be possible? – Bman72 Apr 27 '15 at 20:58
• An alternative way to show irreducibilirty: We know all four roots of the polynomial: $\pm\sqrt 2\pm\sqrt 3$ None of them is rational, hence no linear factor. The sum of $\sqrt2+\sqrt 3$ and another root is rational only for $-\sqrt 2-\sqrt 3$, but then the product is irrational, hence no two of the roots can be the roots of a quedratic factor – Hagen von Eitzen Apr 28 '15 at 5:53
Consider the product: \begin{align*} (x-\sqrt2-\sqrt3)&(x-\sqrt2+\sqrt3)(x+\sqrt2-\sqrt3)(x+\sqrt2+\sqrt3)\\ &=\bigl((x^2-\sqrt2)^2-3\bigr)\bigl((x^2+\sqrt2)^2-3\bigr)\\ &=(x^2-1-2\sqrt2x)(x^2-1+2\sqrt2x)\\ &=(x^2-1)^2-8x^2=x^4-10x^2+1 \end{align*} This polynom is irreducible over $\mathbf Q$, since the above computation showed all possible factorisations over $\mathbf R$ and none of these has rational coefficients.