LEC ALG3 17

Algebraic elements form a subextenion

Let $K/F$ be a field extension, and $\alpha ,\beta \in K$ be algebraic over $F$. Are $\alpha +\beta$, $\alpha \beta$, and $\alpha /\beta$ algebraic over $F$?

Proposition 334.1(Proposition).

Let $K/F$ be a field extension. Then the elements of $K$ algebraic over $F$ form a subextension of $K/F$.

Proof.

Let $\alpha ,\beta \in K$ be algebraic over $F$. Then $F(\alpha ,\beta )/F$ is a finite extension by Prp 322.8. It follows from Rmk 322.3 and Prp 322.2 that $F(\alpha +\beta )/F$ is a finite extension, which implies $\alpha +\beta$ is algebraic by Prp 322.4. Similarly for the difference, product and quotient of $\alpha$ and $\beta$.□

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We can pose the same question for integral extensions.

Remark 334.2(Remark).

Let $S\subseteq R$ be rings, and $\alpha ,\beta \in R$ be integral over $S$. Are $\alpha +\beta$ and $\alpha \beta$ integral over $S$?

We can try to answer this in the fashion of Prp 1: Using Prp 332.2, $S[\alpha ]$ and $S[\beta ]$ are finitely generated $S$-modules. Hence, $S[\alpha ,\beta ]$ is a finitely generated $S$-module (by taking pairwise products of generators). So, $\alpha +\beta$ is algebraic over $S$ iff $S[\alpha +\beta ]$ is a finitely generated $S$-module. Since $S[\alpha +\beta ]\subseteq S[\alpha ,\beta ]$, it would suffice to know the following: a submodule of a finitely generated $S$-module is also finitely generated¹. This statement holds if $S$ is Noetherian, but fails if $S$ otherwise.

However, the Noetherian hypothesis is not required; this statement is true for any $S$. Basically, one has to show that $r\in R$ is integral over $S$ iff $S[r]$ is contained in a finitely generated $S$-module; since $S[\alpha ,\beta ]$ is finitely generated, we would be done. See Conrad (n.d.) Proposition 1.

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Definition 334.3(Definition).

Let $L/F$. Let $K_1/F$ and $K_2/F$ be such that $K_1,K_2\subseteq L$. $K_1{K}_2$ denotes the smallest field containing $K_1$ and $K_2$.

Remark 334.4(Remark).

Clearly, $K_1{K}_2$ is the quotient field of the ring $K_1[K_2]$, the ring generated by the elements of $K_2$ over $K_1$. Note that $K_1[K_2]=K_2[K_1]$.

Lemma 334.5(Lemma).

Let $E_1,E_2$ be extensions of a field $F$, contained in some bigger field $E$, and let $\sigma$ be an embedding of $E$ in some field $L$. Then

$$\sigma (E_1{E}_2)=\sigma (E_1)\sigma (E_2).$$

Proof.

Apply $\sigma$ to an element of $E_1{E}_2=Q(E_1[E_2])$:

$$\begin{array}{r}\sigma \left(\frac{a_1{b}_1+\cdots +a_n{b}_n}{a_1^{\prime }{b}_1^{\prime }+\cdots +a_m^{\prime }{b}_m^{\prime }}\right)=\frac{a_1^{\sigma }{b}_1^{\sigma }+\cdots +a_n^{\sigma }{b}_n^{\sigma }}{a_1^{\prime \sigma }{b}_1^{\prime \sigma }+\cdots +a_m^{\prime \sigma }{b}_m^{\prime \sigma }}\in \sigma (E_1)\sigma (E_2).\end{array}$$

The reverse containment is similarly clear.□

Example 334.6(Example).

Let $p,q$ be prime numbers. Let ${\eta }_p$ and ${\eta }_q$ be roots of unity. $K_1=\mathbb{Q}({\eta }_p)$, $K_2=\mathbb{Q}({\eta }_q)$. Let $1=mp+nq$. Any field containing $K_1$ and $K_2$ would contain $e^{2\pi i/pq}=(e^{2\pi i/p}{)}^m(e^{2\pi i/q}{)}^n$,
so $\mathbb{Q}({\eta }_{pq})\subseteq K_1{K}_2$. Since $\mathbb{Q}({\eta }_p)\subseteq \mathbb{Q}({\eta }_{pq})$ and $\mathbb{Q}({\eta }_q)\subseteq \mathbb{Q}({\eta }_{pq})$, we have $K_1{K}_2=\mathbb{Q}({\eta }_{pq})$.

Footnotes

1. This was taken for granted when working with field extensions; subspaces of finite dimensional vector spaces are finite dimensional ↩

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References

Conrad, B. (n.d.). Integral Ring Extensions. Retrieved November 18, 2025, from https://math.stanford.edu/~conrad/210BPage/handouts/math210b-integral-ring-extensions.pdf