Tensor product isomorphism 
Suppose that $ E/L $ is an (algebraic) field extension and $L/K$ is a finite field extension. I want to prove
  $$(K[X]/\mathfrak{a})\otimes_K E\cong E[X]/\mathfrak{a}E[X].$$

I alrealdy have that $ K[X]\otimes_K E\cong E[X]$. The other part I began like that. Like
$\mathfrak{a}\xrightarrow{f} K[X]\xrightarrow{g} K[X]/\mathfrak{a}\rightarrow 0$
is an exact sequence,the sequence
$\mathfrak{a}\otimes_K E\xrightarrow{f\otimes Id}K[X]\otimes_K E\xrightarrow{g\otimes Id}K[X]/\mathfrak{a}\otimes_K E\rightarrow 0 $
has to me exact for every $K$-module $E$. This induces the following isomorfism:
$(K[X]/\mathfrak{a})\otimes_K E\cong K[X]\otimes_K E/(Im(f\otimes_K Id)). $
But then, why is  $(Im(f\otimes_K Id))=\mathfrak{a}E[X]$? I probably have to use the other isomorfism I already have proved, but I dont know how to use it (I am not so familiar with tensor products).
Thankful for any help.   
 A: Here is the statement I prove:

Let $ B $ be an $ A $-algebra and $ \mathfrak{X} $ a system of variables and $ \mathfrak{a}\subset A[\mathfrak{X}] $ an ideal. Then there exists an isomormophism $$ (A[\mathfrak{X}]/\mathfrak{a})\otimes_A B\cong B[\mathfrak{X}]/\mathfrak{a} B[\mathfrak{X}].  $$ 

Here is the proof: 
Defining the $ A $-bilinear aplication $ \varphi :A[\mathfrak{X}]\times B\longrightarrow B[\mathfrak{X}] $ by $ \varphi(f,b)=bf $, we infer from the universal property of tensor products that there exists a unique $ A $-linear map $ \phi:A[\mathfrak{X}]\otimes_A  B\longrightarrow B[\mathfrak{X}]$ define by $ \phi(f\otimes b)=bf $. Now we can see that $ \psi: B[\mathfrak{X}]\longrightarrow A[\mathfrak{X}]\otimes_A B $ defined by 
\begin{align*}
\psi\Big(\sum_{\mu} b_\mu \prod_{x\in \mathfrak{X}} x^{\mu_x}\Big)=\sum_{\mu} (\prod_{x\in \mathfrak{X}} x^{\mu_x} \otimes b_\mu)
\end{align*}
satisfies $ \phi\circ \psi=\mbox{id}_{B[\mathfrak{X}]} $ and $ \psi\circ\phi=\mbox{id}_{A[\mathfrak{X}]\otimes_A B} $. Consider the exact sequence
\begin{align*}
0\longrightarrow \mathfrak{a} \xrightarrow{\delta} A[\mathfrak{X}]  \longrightarrow A[\mathfrak{X}]/\mathfrak{a} \longrightarrow 0
\end{align*}
Tensoring on the right side by $ \otimes_A B $, we obtain
\begin{align*}
\mathfrak{a}\otimes_A B \xrightarrow{\delta\otimes\mbox{id}} A[\mathfrak{X}]\otimes_A B  \longrightarrow A[\mathfrak{X}]/\mathfrak{a}\otimes_A B \longrightarrow 0 
\end{align*}
which establishes the isomorphism
\begin{align*}
A[\mathfrak{X}]\otimes_A B/(\delta(\mathfrak{a})\otimes_A B)\cong A[\mathfrak{X}]/\mathfrak{a}\otimes_A B. 
\end{align*}
Thus it remains to prove that $ \delta(\mathfrak{a})\otimes_A B\cong \mathfrak{a}B[\mathfrak{X}] $, since it follows from the above that $ A[\mathfrak{X}]\otimes_A B \cong B[\mathfrak{X}] $. Observe first that we can define the $ A $-bilinear map $ \varphi:\mathfrak{a}\times B\longrightarrow \mathfrak{a}B[\mathfrak{X}] $ defined by $ \varphi(f,b)=fb $, so that there exists a unique $ A $-linear map $ \phi:\mathfrak{a}\otimes_A B\longrightarrow \mathfrak{a}B[\mathfrak{X}] $ such that $ \phi(f\otimes b)=fb $. Now, we can see that $ \ker (\delta\otimes\mbox{id})\subset\ker\ \phi  $ since 
\begin{align*}
\ker (\delta\otimes\mbox{id})=\Big\lbrace \Big(\sum_{\mu} a_\mu \prod_{x\in \mathfrak{X}} x^\mu\Big)\otimes b\ \Big| \ a_\mu b=0, \ \forall a_{\mu}\Big\rbrace. 
\end{align*}
Hence, there exists a unique morphism $ \phi_{\delta}:\delta(\mathfrak{a})\otimes_A B\longrightarrow \mathfrak{a}B[\mathfrak{X}] $ such that $ \phi_{\delta}(f\otimes b)=fb $. Defining $ \psi:\mathfrak{a}B[\mathfrak{X}]\longrightarrow \delta(\mathfrak{a})\otimes_A B $ by 
\begin{align*}
\psi\Big(\Big(\sum_{\mu}a_{\mu}\prod_{x\in\mathfrak{X}}x^{\mu_x}\Big)(\sum_{\gamma}b_{\gamma}\prod_{x\in\mathfrak{X}}x^{\gamma_x}\Big)\Big)=\sum_{\mu,\gamma} \Big(a_\mu  \prod_{x\in\mathfrak{X}} x^{\mu_x+\gamma_x}\Big)\otimes b_\gamma. 
\end{align*}
we can see that this map satisfies $ \phi_{\delta}\circ\psi=\mbox{id}_{\mathfrak{a}B[\mathfrak{X}]} $ and $ \psi\circ\phi_{\delta}=\mbox{id}_{\mathfrak{a}\otimes_A} B $.
I suppose that this proof is right. But if there is something wrong please let me know. 
