# Thread: Tensor product and Algebras

1. ## Tensor product and Algebras

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2. Originally Posted by dabien
Suppose we have finite field extensions B and K of F of respective degrees m and n, all contained in some larger field of E.

Prove that the set of all finite linear combinations with coefficients in F of elements of E of the form $\alpha*\gamma$ with $\alpha \in B$, $\gamma \in K$ already form a subfield of E.
what is $*$ supposed to mean? is it the usual multiplication? if so, then write $\alpha \gamma$ not $\alpha * \gamma.$

3. i'll show the set of all finite linear combinations of the form $\alpha \gamma, \ \alpha \in B, \ \gamma \in K$ with coefficients in $F$ by $Q.$ clearly $Q$ is a subring of $E$ and $F \subseteq Q.$ also note that every element of $Q$ is algebraic

over $F$ because $B,K$ are algebraic over $F.$ now let $0 \neq q \in Q.$ let $f(x) =\sum_{i=0}^r a_ix^i \in F[x], \ a_r=1,$ be the minimal polynomial of $q.$ so $q^r + \cdots + a_1q + a_0 = 0.$ the assumption $a_0 = 0$ will contradict the minimality of $f(x)$ because $E$ is an integral domain. so $a_0 \neq 0$ and hence $qa_0^{-1}(- q^{r-1} - \cdots - a_1)=1.$ therefore $q$ is invertible in $Q$, because $q^{-1}=a_0^{-1}(-q^{r-1} - \cdots - a_1) \in Q.$