Let $K\subseteq L$ be a field extension, and let $V$ be a $K$-vector space. The

**extension of $V$ by scalars in $L$**is the tensor product $E=V\otimes_KL$. I will prove that every $L$-vector spaced is obtained as some extension in this way and that $\dim_L(E)=\dim_K(V)$.**Proposition.**Let $E$ be any $L$-vector space and $K$ a subfield of $L$. Then, $E=V\otimes_KL$ for some $K$-vector space $V$. If $E$ is of finite dimension over $L$, then $\dim_K(V)<\infty$ as well.**Proof.**Let $X\subset E$ be an $L$-basis of $E$. Let $E_K$ be the $K$-span of $X$. Define a map \[ \phi: E \longrightarrow E_K \otimes_K L \] by decreeing that $\phi(x)=x\otimes 1$ for $x\in X$. By $L$-linear extension, it automatically becomes an $L$-linear map. For surjectivity, note that $x\otimes\lambda= \lambda\cdot(x\otimes 1)=\lambda\cdot \phi(x)=\phi(\lambda x)$ for any $x\in X$ and any $\lambda\in L$. These elements generate $E_K\otimes_KL$. To check injectivity, assume that $\phi(\sum_{x\in X} \lambda_x x)=0$. Of course, only finitely many of the $\lambda_x$ are allowed to be nonzero. This means $0 = \sum_{x\in X} x\otimes \lambda_x$ and by the below lemma, we are done. ---**Lemma.**Let $V$ and $W$ be $K$-vector spaces and $w_1,\ldots,w_r\in W$ linearly independent. Inside the tensor product $V\otimes_K W$, the equality $\sum_{i=1}^r v_i\otimes w_i = 0$ implies $\forall i: v_i=0$.**Proof.**Let $f_1,\ldots,f_r\in V^\ast = \mathrm{Hom}(V,K)$ be arbitrary linear functions on $V$. Let $w_j^\ast: W\to K$ be the linear function with $w_j^\ast(w_i)=\delta_{ij}$1 and consider \[ \begin{array}{rcl} \phi: V\times W &\longrightarrow& K \\ (x,y) &\longmapsto& \sum_{j=1}^r f_i(x)\cdot w_j^\ast(y). \end{array} \] Since $F$ is bilinear, there exists a linear function $h:V\otimes_KW\to K$ with the property that \[ \begin{array}{rcl} h(x\otimes y) &=& F(x,y) = \sum_{j=1}^r f_i(x)\cdot w_j^\ast(y), \\ 0 = h\left(\sum_{i=1}^r v_i\otimes w_i\right) &=& \sum_{j=1}^r \sum_{i=1}^r f_j(v_i)w_j^\ast(w_i) = \sum_{i=1}^r f_i(v_i). \end{array} \] Since the $f_i$ were arbitrary, we can conclude that $v_i=0$ for all $i$. --- With the upcoming proposition, we will essentially prove that $\dim_L(E)=\dim_K(E_K)$. Hence, we will have $\dim_L(E)=\dim_K(E)$ if and only if $E_K=E$. In other words, if and only if $E=E_K\otimes_KL=E\otimes_KL$. For $L=E=\mathbb C$ and $K=\mathbb R$, this is not the case: $\mathbb{C}\otimes_{\mathbb R}\mathbb{C}$ is not isomorphic to $\mathbb C$. However,**Proposition.**Let $V$ be a finite-dimensional $K$-vector space and $K\subseteq L$ a field extension. Then, \[ \dim_K(V) = \dim_L(V\otimes_KL) \]**Proof.**We define the $L$-linear homomorphism \[ \begin{array}{rcl} \phi: V &\longrightarrow& V\otimes_K L \\ v &\longmapsto& v \otimes 1. \end{array} \] Let $B\subset V$ be a basis of $V$ as a $K$-vector space. Then, we claim that $\phi(B)$ is an $L$-basis of $V\otimes_KL$. If $v\otimes \lambda\in V\otimes_KL$ is any rank one tensor, then we may write \[ v\otimes\lambda = \left(\sum_{b\in B} b^\ast(v)\cdot b\right)\otimes\lambda = \sum_{b\in B} b^\ast(v)\lambda \cdot (b\otimes 1) \] so $\phi(B)$ generates $V\otimes_KL$. On the other hand, let $X$ be a basis of $L$ as a $K$-vector space. If $\sum_{b\in B} \lambda_b\cdot (b\otimes 1)=0$ for certain $\lambda_b\in L$, then we can write $\lambda_b = \sum_{x\in X} x^\ast(\lambda_b)\cdot x$ with almost all $x^\ast(\lambda_b)=0$. Hence, \[ \begin{array}{rcl} 0 &=& \sum_{b\in B} \lambda_b\cdot (b\otimes 1) = \sum_{x\in X} x\cdot \sum_{b\in B} x^\ast(\lambda_b)\cdot(b\otimes 1) \\ &=& \sum_{x\in X} \left(\sum_{b\in B} x^\ast(\lambda_b)\cdot b\right)\otimes x \end{array} \] Note that this sum is finite. Since the $x\in X$ are linearly independent, we may use the above Lemma to conclude \[ \sum_{b\in B} x^\ast(\lambda_b) \cdot b = 0 \] for all $x\in X$, and as $B$ is a basis of $V$, this means $x^\ast(\lambda_b)=0$ for all $x$ and all $b$, therefore $\lambda_b=0$ for all $b$. This proves that $\phi(B)$ is, in fact, a basis.- The
**Kronecker Delta**, i.e. $\delta_{ii}=1$ and $\delta_{ij}=0$ for $i\ne j$. [↩]