If $a,b\in K$ are algebraic over $k$, then $a\pm b, ab$ and $ab^{-1}$ are also algebraic.

Question

1) Let $K/k$ a field extension. Show that if $a,b\in K$ are algebraic over $k$, then $a\pm b, ab$ and $ab^{-1}$ are also algebraic.

2) Deduce that $\{a\in K\mid a\text{ algebraic over }k\}$ is a field.

3) We denote $k^{alg}$ the algebraic closure of $k$. Show that if $K/k$ is algebraic, then $K^{alg}=k^{alg}$.

My work

1) Since $a$ is algebraic over $k$, then $k(a)/k$ is a finite extension. Since $b$ is algebraic over $k$, it's also algebraic over $k(a)$, and thus $$[k(a,b):k]=[k(a,b):k(a)][k(a):k]<\infty $$ and thus $k(a,b)/k$ is finite and thus algebraic. Since $a\pm b, ab$ and $ab^{-1}$ are in $k(a,b)$, they are algebraic. It looks to easy, is it correct ? and rigorous enough ?

2) It's I think what I proved in 1), isn't it ? But if they ask the question, there is a subtlety that I don't understand.

3) I don't really know how to do. The definition I have is $k^{alg}$ is minimal under the assertion : "For all extension algebraically closed $\Omega /k$, there is a $k-$morphisme $k^{alg}\hookrightarrow \Omega $." I actually don't understand what it mean.

Answer 1

There are some subtle points here.

First, if $a$ is algebraic over $k$ then $k(a)/k$ is finite. It is not obvious. Try to redo the proof and see if you understand why this is the case.

Second, if $c$ is an element of a field $L$ and $L/K$ is finite, then $c$ is algebraic over $L$. To show this, take a basis ( enough a system of generators) of $L/K$, say $e_1$, $\ldots$, $e_m$. Now $c \cdot e_i$ are all in $L$ so we have some equalities of the form $ c e_i = \sum k_{ij} e_j$. From here you conclude that $c$ is an eigenvalue of the matrix $(k_{ij})$ ( more or less), hence algebraic over $k$.

These observations can be used to generate explicit polynomial equations say for $a+b$, $ab$, when knowing equations for $a$ and for $b$.

Answer 2

One of my favourite proofs. $a$ is algebraic over $k$ iff there is a square matrix $A$ over $k$ that has $a$ as an eigenvalue. Suppose $a$ is an eigenvalue of $A$ and $b$ is an eigenvalue of $B$ with eigenvectors $u$, $v$ respectively, where if $b \ne 0$ we may take $B$ to be invertible. Then $a + b$, $a-b$, $ab$ and (if $b \ne 0$) $a/b$ are eigenvalues of $A \otimes I + I \otimes B$, $A \otimes I - I \otimes B$, $A \otimes B$, and (if $b \ne 0$) $A \otimes B^{-1}$, with eigenvector $u \otimes v$.