# Dimensions of $K(\alpha,\beta)$ and $K(\alpha)(\beta)$: why are they equal?

Suppose $L\supset K$ is a field extension. For $\alpha_1,...,\alpha_n\in L$ we define by $K[\alpha_1,...,\alpha_n]$ the set all elements of $L$ that has the form $f(\alpha_1,...,\alpha_n)$ for some polynomial $f\in K[X_1,...,X_n]$ (the polynomial ring over $K$ in $n$ dimensions embedded in $L$). Also let $K(\alpha_1,...,\alpha_n)$ denote the field of fractions of $K[\alpha_1,...,\alpha_n]$.

Let $\alpha,\beta\in L\setminus K$ that are algebraic and NOT conjugate over $K$ be given. I have a pretty good idea about what sits inside $K(\alpha)$ and $K(\beta)$. If $f=X^n+\sum a_i X^i$ is the defining polynomial of $\alpha$ in $K[X]$ and $g=Y^m+\sum b_j Y^j$ is the defining polynomial of $\beta$ in $K[Y]$ then $1,\alpha,...,\alpha^{n-1}$ form a basis for $K(\alpha)$ over $K$ and $1,\beta,...,\beta^{m-1}$ form a basis for $K(\beta)$ over $K$. Also I understand why $K(\alpha)=K[\alpha]\cong K[X]/\langle f\rangle$ since the latter is already a field.

Now, I want to understand more fully what is inside of $K(\alpha,\beta)$ compared to what is inside of $K(\alpha)(\beta)$ and more specifically how one should regard the dimensions of these. I know they should be equal and both basically contain polynomial expressions in two variables evaluated on $(\alpha,\beta)$.

But what puzzles me is that $f,g\in K[X,Y]$ both map to zero when evaluated on $(\alpha,\beta)$. So the monic polynomial in $K[X,Y]$ of minimal degree mapping to zero when evaluated on $(\alpha,\beta)$ is either $f$ or $g$. I am sure this is wrong, but somehow this suggests to me that $\dim_K(K(\alpha,\beta))=\min(\deg(f),\deg(g))$. On the other hand we must have the tower of extensions $$K\subset K(\alpha)\cong K[X]/\langle f\rangle\subset K(\alpha)(\beta)\cong K(\alpha)[Y]/\langle g\rangle$$ where a dimension argument shows that $\dim_K\left(K(\alpha)(\beta)\right)=\deg(f)\deg(g)=n\cdot m$. Please help me to clarify why and convince me that also $\dim_K(K(\alpha,\beta))=n\cdot m$ wihtout just stating that $K(\alpha,\beta)=K(\alpha)(\beta)$.

Is it because the kernel of ''evaluation at $(\alpha,\beta)$'' is $\langle f,g\rangle$ and how does this lead to the expected dimension of $K(\alpha,\beta)$ over $K$ that I do not yet fully understand?

Your problem is notational. The field $K(\alpha,\beta)$ isn't obtained by adding the pair $(\alpha,\beta)$ to $K$, but is defined as the smallest field that contains $\alpha$ and $\beta$. So you never have to think of polynomials in two variables!

To see that $K(\alpha,\beta)=K(\alpha)(\beta)$, we first see that $K(\alpha,\beta)$ is the smalles field that contains $\alpha$ and $\beta$, and since $K(\alpha)(\beta)$ contains $\alpha$ and $\beta$, we have that $K(\alpha,\beta)\subseteq K(\alpha)(\beta)$. On the other hand, an element in $K(\alpha)(\beta)$ is the quotient of polynomials in $\beta$ with coefficients in $K(\alpha)$, and so is easily seen to be an element of $K(\alpha,\beta)$.

• I'm trying to show a similar identity: $K(A_1 \bigcup ... \bigcup A_n)$ = $K(A_1)...(A_n)$. Could I use a similar argument? Apr 10 '14 at 3:03
• Yes, a similar argument would work. Apr 11 '14 at 11:17

So the monic polynomial in $K[X,Y]$ of min degree vanishing on $(\alpha,\beta)$ is either $f$ og $g$.

Monic in which variable? The problem is that obtaining $K(\alpha,\beta)$ from $K[X,Y]$ means we need to quotient by the ideal containing all polynomial relations satisfied by $\alpha$ and $\beta$, and $K[X,Y]$ is not a principal ideal domain $-$ so even if $\pi(X,Y)$ had minimal degree and was monic in both variables there is no guarantee that $K[X,Y]/\pi(X,Y)$ is isomorphic to $K(\alpha,\beta)$, it may still be too big! In point of fact, $\dim_K K[X,Y]/(\pi)=\deg\pi$ does not hold in two variables. E.g. $K[X,Y]$ quotiented by $\pi=f(X)$ or $\pi=g(Y)$ yields $K(\alpha)[Y]$ or $K(\beta)[X]$ respectively, both infinite-dimensional.

In order to obtain repeated algebraic extensions with multivariable polynomials, reason as so:

$$K(\alpha,\beta)\cong K(\alpha)(\beta)\cong\frac{K(\alpha)[Y]}{m_{\beta,K(\alpha)}(Y)}\cong \frac{\left(\frac{K[X]}{m_{\alpha,K}(X)}\right)[Y]}{m_{\beta,K(\alpha)}(Y)}\cong\frac{K[X,Y]}{(m_{\alpha,K}(X),m_{\beta,K(\alpha)}(Y))}$$

where $m_{\lambda,L}(Z)$ stands for the minimal polynomial of $\lambda$ over $L$ in variable $Z$, or symmetrically

$$K(\alpha,\beta)\cong\frac{K[X,Y]}{(m_{\beta,K}(X),m_{\alpha,K(\beta)}(Y))}.$$

If $m_{\alpha,K(\beta)}=m_{\alpha,K}$ (which is stronger than $\alpha,\beta$ are not conjugate: what if one of them is a slight perturbation of a conjugate of the other, like $\sqrt[3]{2}$ and $\omega\sqrt[3]{2}+1$?) then we see that a basis is given by $\{X^jY^k$ with $0\le j<m_{\alpha,K}$ and $0\le k<m_{\beta,K}\}$, hence the dimension is the product of the degrees of $\alpha$ and $\beta$'s minimal polynomials as desired.

The "universal property" behind adjoining algebraics is that $K(S)$ is the smallest field containing the set $S$ (this is well-defined: just take the intersection of all fields containing $S$ and $K$). Every fielding containing $K$ and $S$ will contain $K(S)$. We can reason with this universal property to see these equivalences, as in Robert's answer: if $K(\alpha,\beta)$ is the smallest field containing $\alpha$ and $\beta$ then since it contains $\alpha$ and $K$ it contains $K(\alpha)$, and since it contains $K(\alpha)$ and $\beta$ it contains $K(\alpha)(\beta)$; conversely $K(\alpha)(\beta)$ contains $\beta$ and $K(\alpha)$ and since it contains $K(\alpha)$ it contains $K$ and $\alpha$, since it contains $K$, $\alpha$ and $\beta$ it contains $K(\alpha,\beta)$. Thus $K(\alpha)(\beta)\supseteq K(\alpha,\beta)$ and $K(\alpha)(\beta)\subseteq K(\alpha,\beta)$, which yields the desired equality $K(\alpha)(\beta)=K(\alpha,\beta)$ and symmetrically $K(\alpha,\beta)=K(\beta)(\alpha)$.

Understanding what sits inside $K(\alpha,\beta)$ takes no leap of imagination. At the very least $K(\alpha,\beta)$ includes all polynomial expressions in both $\alpha$ and $\beta$. Since $\alpha$ and $\beta$ are algebraic over $K$ and hence over $K(\beta)$ and $K(\alpha)$ respectively one can see that the set of all polynomial expressions in $\alpha,\beta$ is in fact a field. Its elements can be factored into polynomials in $\alpha$ whose coefficients are themselves polynomials in $\beta$, or as polynomials in $\beta$ whose coefficients are polynomials in $\alpha$. For instance

$$3\alpha^2\beta^2+5\alpha\beta^2-\alpha^2\beta+\alpha+\beta-2 =\begin{cases}(3\beta^2-\beta)\color{Blue}{\alpha^2}+(5\beta^2+1)\color{Blue}{\alpha}+(\beta-2)\color{Blue}{1} \\[3pt] (3\alpha^2+5\alpha)\color{Red}{\beta^2}+(1-\alpha^2)\color{Red}{\beta}+(\alpha-2)\color{Red}{1} \end{cases}$$

With this perspective it should be intuitively "obvious" that $K(\beta)(\alpha)=K(\alpha,\beta)=K(\alpha)(\beta)$.

• I think he was thinking of $(\alpha,\beta)$ as a pair, not as an ideal. Your analysis is right, however. Sep 26 '13 at 15:09
• @RobertAuffarth I don't think OP was thinking of $(\alpha,\beta)$ as an ideal either. He or she was thinking of an ideal though: s/he mentions "the monic polynomial of minimal degree mapping to zero evaluated on $(\alpha,\beta)$" (which is the part of the question I highlighted), and related this idea to dimension of $K(\alpha,\beta)$, so I surmised OP was operating on the belief that this one-variable minpoly trick $K(\alpha)\cong K[X]/f(X)$ generalized to two variables with principal ideals just the same.
– anon
Sep 26 '13 at 15:14
• you're right, I misread what you wrote at the beginning. Sep 26 '13 at 15:15
• Thanks to both of you! anon's answer helped me gain more insight about what is inside of $K(\alpha,\beta)$ regarded as a quotient. Sep 26 '13 at 19:12