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I have some questions concerning field extensions, which I hope someone can help me with.

1) Understanding $K(\alpha)$. Let $L,K$ be fields, where $K$ is a subfield of $L$. I know that $K(\alpha)$ is the smallest field, that contains $K$ and $\alpha$. However, ist there some explicit form of $K(\alpha)?$ I know that for rings $R[\alpha]$, one can view this as the image of a homorphism, but is there a similar method for the field?

2) Example: $\mathbb{R}/\mathbb{Q} , \mathbb{Q}[\sqrt{2},\sqrt{5}]=?$ I know that $\mathbb{Q}(\sqrt{2})=\{a+b\sqrt{2}:a,b \in \mathbb{Q}\}$. I think this is due to $\sqrt{2}$ being a root of the polynomial $f=X^2-2$. Then $\mathbb{Q}(\sqrt{2})$=$\mathbb{Q}[X]/(f)$. Since $f$ is a polynomial of degree $2$, this should contain only polynomials of degree less or equal than $1$, due to the equivalence relation. How do I exactly figure out the set,though? And how does the first set I gave, where i adjoin two elements look? Is there a general method to determine the elements of $R[X]/(aX+b)$, since I only know that it contains polynomials of degree less than $2$ right now, which is basically because of polynomial division and the equivalence relation.

3) In order to determine the degree of the extension, one can use the minimal polynomial. If one has a polynomial $f$ with root $\alpha$, which is irreducible over $K$, does one automatically know that its of lowest degree? If so, why is that the case?

Thank you in advance for answering these questions, it will help me out a lot!

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    $\begingroup$ If $\alpha$ is algebraic over $K$ then $K(\alpha) = K[\alpha]=\sum_{n=0}^{\deg(f)-1}K \alpha^n \cong K[x]/(\ker(K[x]\to K[\alpha])) = K[x]/(f(x))$ where $f$ is the minimal polynomial, the monic polynomial $\in K[x]$ of smallest degree with $\alpha$ in its roots, the polynomial $\gcd$ implies it divides every polynomial $g\in K[x]$ such that $g(\alpha)=0$, if $\alpha$ is transcendental then $K(\alpha) \cong K(x)$ $\endgroup$ – reuns Jan 11 at 0:12
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Take a look at the definition of a polynomial ring. By $R[x]$ we look at all polynomials with coefficients in our ring $R$. Note that its crucial that we not only look at the linear polynomials when "adjoining" $x$, but also at the quadratic, the cubic, etc. So in reality we not only "adjoin" $x$ as an intermediate variable but also $x^2$, and $x^3$, and $x^4$, and so on. Simply said: we look at all powers of our new element.
So at a first glance you can look at $K[\alpha]$ as the polynomial ring over $K$ in the variable $\alpha$ but now we are in the reign of fields and want to have a field extension, saying that adjoining (now I omit the quotation marks as the process is called so) actually yields a field again. So, we not only need $\alpha,\alpha^2,\dots$ but also $\alpha^{-1},\alpha^{-2}\dots$ to regain our field structure. This new field we call $K(\alpha)$, saying we adjoin $\alpha$ to $K$ and also its inverse such that we have a field again. This might shed some light on first question. Also, take a look at this question dealing with the difference between algebraic and transcendental extensions regarding the question of inverses.

Regarding your second question, we are now dealing with a specific field extension. Your notation seems a bit confused so I will just assume that you meant to write $\Bbb Q(\sqrt2,\sqrt5)$. So, first lets start with $\Bbb Q(\sqrt2)$. Let us first think about $\Bbb Q[\sqrt2]$, the polynomial ring over $\Bbb Q$ with $\sqrt2$ as a variable. According to our definition of polynomial rings the elements of $\Bbb Q[\sqrt2]$ are of the form $a_0+a_1\sqrt2+a_2(\sqrt2)^2+a_2(\sqrt2)^3+\cdots+a_n(\sqrt2)^n$ for some positive integer $n$. But now we know in addition $(\sqrt2)^2=2$. So all powers of $(\sqrt2)^k$ can be reduced to either a multiple of $2$ or to a multiple of $2$ times $\sqrt2$. Esssentially all boils down to $\Bbb Q[\sqrt2]=\{a+b\sqrt2~|~a,b\in\Bbb Q\}$, we do not need the rest of the powers to generate all of $\Bbb Q[\sqrt2]$ and this fact is encrypted in the minimal polynomial $f(x)=x^2-2$ over $\Bbb Q$. Thinking for a moment of $\sqrt2$ as just an expression for the solution of $x^2=2$ what we actually do by adjoining $\sqrt2$ is to manually demanding this equation to be solvalbe by adjoining an element, called $\sqrt2$, which is defined to fulfill $f(\sqrt2)=0$. This happens when forming the quotient $\Bbb Q[X]/(x^2-2)$.
It might be helpful to adopt this point of view when thinking about adjoining elements. Regarding $\Bbb Q(\sqrt2)$ we note that $\sqrt2^{-1}=\frac1{\sqrt2}=\frac{\sqrt2}2\in\Bbb Q[\sqrt2]$ and similiar for the powers so we already can conclude $\Bbb Q[\sqrt2]=\Bbb Q(\sqrt2)$. This would not work if we would consider $f(x)=x^2-2$ as a polynomial over $\Bbb Z$ so there we would have to adjoin the inverse of $\sqrt2$ (if needed) separately.

Now, if we have two elements adjoined the situation is a polynomial ring in two variabels $\Bbb Q[\sqrt2,\sqrt5]$ where there elements can have all combinations of powers of $\sqrt2$ and $\sqrt5$. But again, examining the first few we notice that eventually we always get rational multiplies of either $\sqrt2,\sqrt5$ or $\sqrt{10}$. So your ring looks like $\Bbb Q[\sqrt2,\sqrt5]=\{a+b\sqrt2+c\sqrt5+d\sqrt{10}~|~a,b,c,d\in\Bbb Q\}$. Again, inverses follows as in the case of only adjoining $\sqrt2$ so $\Bbb Q[\sqrt2,\sqrt5]=\Bbb Q(\sqrt2,\sqrt5)$. To get a more intuitive understanding you should note that you can view a field extension as a vectors space over the base field of dimension the degree of the extension. $\Bbb Q(\sqrt2,\sqrt5)$ has degree $4$, so the vector space is of dimension $4$ and a basis is given by $\mathfrak B=\{1,\sqrt2,\sqrt5,\sqrt{10}\}$. Similiar $\Bbb Q(\sqrt2)$ is of degree $2$, so the vector space is of dimension $2$ and the extension $\Bbb R/\Bbb Q$ is infinite, so $\Bbb R$ can be viewed an infinite dimensional vector space over $\Bbb Q$.

Regarding your last question, you missed a crucial fact: the minimal polynomial has to be monic, i.e. the leading coefficient has to be $1$. That also answers the part where you were asking about $R[x]/(ax+b)$. First, $ax+b$ is not monic, and second even when looking at $R[x]/(x-a)$ you just get $R$ as $a$ already has to be in $R$ as the polynomial you mod out has to be in $R[x]$. But yes, finding an irreducible, monic polynomial $f(x)\in K[x]$ such that $f(\alpha)=0$ guarantees you that this is the minimal polynomial of $\alpha$ over $K$. One can actually show that the minimal polynomial is unique, and thus finding a polynomial that satisfies all requirements already is the minimal polynomial.

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  • $\begingroup$ First off, thank you very much for taking your time and answering my question in this much detail! I think I understood question 1) and 3) fully now. However I still have remaining questions for 2). In the first abstract you said [...] This happens when forming $\mathbb{Q}/(x^2-2)$ why is that? In the second you said that I can view it as a vector space, but I have only very limited information about the object, why can I view it as a vector space, if I don't know how it looks? Furthermore I wondered, why the degree is 4, and why exactly this then forms a basis. Again, thank you! $\endgroup$ – eu271828 Jan 11 at 9:35
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    $\begingroup$ @eu271828 Regarding $\Bbb Q[x]/(x^2-2)$; when forming the quotient $\Bbb Z/p\Bbb Z$ for instance we actually define all multiplies of $p$ as zero, i.e. $pn=0$ for all $\Bbb Z$. In analogy we define all multiplies of $x^2-2$ as zero, i.e. $f(x)(x^2-2)=0$ for all $f(x)\in\Bbb Q[x]$. In particular we have $p=0$ in the first case and $x^2-2=0$ in the second. But $x^2-2=0$ is nothing else then $x^2=2$, so $x$ is a new element with the property that its square equals $2$. For finding how a field extension looks like, you have to recall polynomial rings (in multiple variables if needed)... $\endgroup$ – mrtaurho Jan 11 at 11:13
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    $\begingroup$ @eu271828 ... So you need to look at all combinations of your generators. For $\sqrt2$ and $\sqrt5$ you also have to look at $\sqrt2\sqrt5=\sqrt{10}$, or $\sqrt2(\sqrt5)^2$, or $(\sqrt2)^2\sqrt5$. But again, a lot of them boil down to easier expressions; in particular all can be expressed in terms of rationals and the elements $\sqrt2,\sqrt5,\sqrt{10}$ multplied by rationals. So those form a basis, as all of $\Bbb Q[\sqrt2,\sqrt5]$ may be rewritten as linear combination of the basis elements $\mathfrak B=\{1,\sqrt2,\sqrt5,\sqrt{10}\}$... $\endgroup$ – mrtaurho Jan 11 at 11:16
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    $\begingroup$ @eu271828 ...To find the degree you need to find the minimal polynomial and then the degree of the minimal polynomial, is the degree of the field extension, is the dimension of the field extension as vector space of the base field. Maybe you could restate how "extension degree" and "minimal polynomial" is defined in your notes/book. $\endgroup$ – mrtaurho Jan 11 at 11:18

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