# 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 $\displaystyle \alpha*\gamma$ with $\displaystyle \alpha \in B$, $\displaystyle \gamma \in K$ already form a subfield of E.
what is $\displaystyle *$ supposed to mean? is it the usual multiplication? if so, then write $\displaystyle \alpha \gamma$ not $\displaystyle \alpha * \gamma.$

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

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