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Let $F,K$ be fields. Suppose $a,b\in K$ are algebraic over $F$ with degrees $m,n$, where $(m,n)=1$. I want to show that $F(a,b)$ is of degree $mn$ over $F$, and $F(a)\cap F(b)=F$.

Consider $$[F(a,b):F]=m[F(a,b):F(a)]=[F(a,b):F(a)][F(a):F]=[F(a,b):F(b)][F(b):F]=n[F(a,b):F(b)]$$ so $[F(a,b):F]\geq mn$. On the other hand $\{a^ib^j\}_{0\leq i<m,0\leq j<n}$ is a basis, so $[F(a,b):F]=mn$.

Now, how could we show that $F(a)\cap F(b)=F$?

Kunal
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2 Answers2

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Hint If $c \in F(a) \cap F(b)$ then $[F(c):F]$ must divide $[F(a):F]$ and $[F(b):F]$.

Also, note that in the first part, you can simplify the argument this way:

Since $m,n |[F(a,b):F]$, their least common multiple divides $[F(a,b):F]$. But their least common multiple is $mn$.

N. S.
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There is a classic argument for composite extensions which says that if you have two fields $K,K'$ extensions of the same field $F$, then $$ [KK' : F] = \frac{[K:F][K':F]}{[K \cap K' : F]}. $$ In this case, since you know that $F(a) = K$ has dimension $m$ over $F$ and $F(b) = K'$ has dimension $n$ over $F$, together with $KK' = F(a) F(b) = F(a,b)$ has dimension $mn$ over $F$, you get that $F(a) \cap F(b)$ has dimension $1$ over $F$, i.e. $F(a) \cap F(b) = F$.

You can find the argument in Dummit & Foote's Abstract Algebra, it is Corollary 20 at page 592. In case you wanna look, I really loved using this book to learn Galois theory. Lots of examples and exercises.

While I think about it ; you already have a basis for $F(a,b)$ ; being in $F(a)$ means you're just a polynomial in $a$ with coefficients in $F$, and being in $F(b)$ means you're a polynomial in $b$ with coefficients in $F$. How can you be both? Only if you're in $F$ by linear independence of the powers of $a$ and $b$. (I'm not convinced you know that the powers of $a$ and $b$ form a basis though, unless you already know that they generate the space whose dimension is known by some other argument.)

Hope that helps,

Patrick Da Silva
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  • Thanks, Patrick! If we know that the minimum polynomial of $a$ in $F$ has degree $n$, isn't it always true that $1,a,a^2,\ldots,a^{n-1}$ form a basis of $F(a)$? – Kunal Jan 18 '14 at 01:56
  • @Kunal : Sure, but if you have two elements, it is not always true that ${ a^i b^j }$ (with the right degree bounds on $i$ and $j$) are linearly independent, because there might be relations between $a$ and $b$. In this case you can --show-- that there are no relations, but you need to do so! – Patrick Da Silva Jan 18 '14 at 12:11