$ a={\sqrt 3 } - {\sqrt 5 } $ I have to find minimal polynomial for this under field Q

after few squarings I got (1) $ f(x)= x^4-16x^2+4 $

And after I have factored out it to

$ f(x)=(x - {\sqrt 5 } + {\sqrt 3 } )( x - {\sqrt 5 } -{\sqrt 3 })( x + {\sqrt 5 } +{\sqrt 3 }) ( x + {\sqrt 5 } -{\sqrt 3 }) $

Lets number these p1,p2,p3,p4.

p4(a)=0 p1*p2,p2*p3,p1*p3 dont satisfy for the field Q

Need I continue doing like p1p2p3,p2p3p1 and so on, or it is enough to say that (1) polynom is minimal?


You are on the right track. You could indeed check that all possible (nonempty) products of the linear factors $p_1$, $p_2$, $p_3$ and $p_4$ do not have coefficients in $\Bbb{Q}$, except of course $f=p_1p_2p_3p_4$. But this is a lot of cumbersome work.

You could also note that if some product of the $p_i$ has coefficients in $\Bbb{Q}$, then so does the product of the remaining $p_i$. For example, if $p_1p_2$ has coefficients in $\Bbb{Q}$, then so does $p_3p_4$. This is because we can do division with remainder with polynomials; there exist unique polynomials $q,r\in\Bbb{Q}[x]$ such that $f=qp_1p_2+r$ and $\deg r<\deg p_1p_2$. We know that $f=p_1p_2p_3p_4$ and hence $q=p_3p_4$ and $r=0$. In particular we see that $p_3p_4=q\in\Bbb{Q}[x]$.

By this argument, we only need to check half of the possible products. We can choose the easier polynomials; clearly the $p_i$ do not have coefficients in $\Bbb{Q}[x]$, so neither do the products $p_1p_2p_3$, $p_1p_2p_4$, $p_1p_3p_4$ and $p_2p_3p_4$. Then it suffices to check that the three products $p_1p_2$, $p_1p_3$ and $p_1p_4$ do not have coefficients in $\Bbb{Q}[x]$; it follows that the products $p_3p_4$, $p_2p_4$ and $p_2p_3$ do not have coefficients in $\Bbb{Q}[x]$ either.

Alternatively, you could note that the numbers $$a^0=1,\qquad a^1=\sqrt{3}-\sqrt{5},\qquad a^2=8-2\sqrt{15},$$ are linearly independent over $\Bbb{Q}$, and hence the minimal polynomial of $a$ must have degree at least $3$. Then the argument above (together with your factorization) show that its degree is $4$, and hence $f$ is the minimal polynomial. You could also continue to compute $$a^3=18\sqrt{3}-14\sqrt{5},\qquad a^4=124-32\sqrt{15},$$ and see that there is no linear dependence between the powers of $a$ until you include $a^4$, and some linear algebra will give you the coefficients of $f$.


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