# A question regarding finding the minimal polynomial associated with a field extension . [duplicate]

Say we have the field extension $$\Bbb Q(w,\sqrt{5})$$ over $$\Bbb Q$$, where w is the primitive cubed root of unity. I know that the minimum polynomial of $$\sqrt{5}$$ is $$x^3-5$$. I want to figure out the degree of the extension;

I know we can use the tower law $$|\Bbb Q(w, \sqrt{5});\Bbb Q|=|\Bbb Q(w, \sqrt{5});\Bbb Q(\sqrt{5})||\Bbb Q(\sqrt{5});\Bbb Q|=x.3$$ as $$x^3-5$$ has degree 3

But how does one find the minimum polynomial of the extension of degree x

## marked as duplicate by Dietrich Burde abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Apr 13 at 19:49

• See also this question with $n=3$. So $x=2$. – Dietrich Burde Apr 13 at 19:48
• @DietrichBurde I edited my question to hone in the part of the question I was particularly interested in , I'm not sure the duplicate link properly addresses that point – excalibirr Apr 13 at 19:52
• The duplicate has an answer exaxctly to this, i.e., why the first extension has degree $x=2$. That's your question, right? – Dietrich Burde Apr 13 at 19:54
• $w$ is a root of $x^3-1=(x-1)(x^2+x+1)$. Since it is a primitive root (hence not equal to $1$) it isn't a root of $x-1$, so it is a root of $x^2+x+1$. This is a polynomial of degree $2$ which has no real roots, hence it is irreducible over $\mathbb{Q(\sqrt{5})}$. So this is the minimal polynomial. – Mark Apr 13 at 19:57
• not quite, as an answer below stated $x^2+x+1$ is the minimum polynomial of the extension but I don't understand how this is found, maybe I'm not understanding something in the link but it didn't seem to show clearly how its assumed that this Is the minimum polynomial – excalibirr Apr 13 at 19:57

Clearly $$w \not \in \Bbb Q(\sqrt{5})$$ so the degree is at least $$2$$. Now notice that $$x^2+x+1$$ is the minimum polynomial.
• Why is $x^2+x+1$ the minimum polynomial , how was it found ? – excalibirr Apr 13 at 19:55
• By noticing that $0=w^3-1=(w-1)(w^2+w+1)$. – Marco Vergamini Apr 13 at 20:04
• An interesting generalisation of this problem could be this one: find the degree of the splitting field of $x^p-s$ over $\Bbb Q$ where $p$ is a prime number and $s>1$ is a squarefree integer. – Marco Vergamini Apr 13 at 20:24