About a proof of the minimal polynomial of matrix doesn't change when the field changed In Keith. Conrad's lecture notes, he proved the following theorem
Theorem:
  Let $K/F$ be any field extension.
(1) For any $A\in\operatorname{M}_n(F)$, its minimal polynomial in $F[x]$ is its minimal polynomial in $K[x]$.
(2) Two matrices in $\operatorname{M}_n(F)$ are conjugate in $\operatorname{M}_n(F)$ if and only if they are conjugate in $\operatorname{M}_n(F)$.
The following is his proof of (1):
Let $m(x)$ be the minimal polynomial of $A$ in $F[x]$. Since $m(x)\in K[x]$ and $m(A)=O$, $m(x)$
  is divisible by the minimal polynomial of $A$ in $K[x]$. Next we show that if $f(x)\in K[x]$ and $f(A)=O$ then there exists a polynomial in $F[x]$ of the same degree which kills $A$, so $\deg f\geq m(x)$. Therefore, $m(x)$ is the minimal polynomial of $A$ in $K[x]$.
Suppose
  $$A^r+a_{r-1}A^{r-1}+\cdots+a_1A+a_0I=O,$$
  where $a_i\in K$. Consider $A^{i}$ as an $F^{nn}$ vector, then $\{A^r,A^{r-1},\dots,A,I\}$ are linearly dependent or the equation
  $$[I,A,\dots,A^{r-1}]\begin{bmatrix}
    x_1\\
    x_2\\
    \vdots\\
    x_{r}
  \end{bmatrix}=A^r$$
  has a nonzero solution in $K^r$, hence it has a nonzero solution in $F^r$. Therefore, there exists $b_0,\dots,b_{r-1}\in F$ not all zero, such that
 $$A^r+b_{r-1}A^{r-1}+\cdots+b_1A+b_0I=O.$$
My question is: Is the following proof correct?:
Let $m_F(x)$ be the minimal polynomial of $A$ in $F[x]$ and $m_K(x)$ the minimal polynomial of $A$ in $K[x]$. Then we have
    $F[A]\simeq F[x]/(m_F(x))$ and $K[A]=K[x]/(m_K(x))$. Since
  $$K[A]=K\otimes_F F[A]=K\otimes_F F[x]/(m_F(x))=K[x]/(m_F(x))$$. 
We deduce that 
$$m_F(x)=m_K(x)$$.
  But we know that the minimal polynomial of $\sqrt{2}$ over $\Bbb{Q}$ is $x^2-2$ and 
  over $\Bbb{R}$ is $x-\sqrt{2}$. We see that
  $$\Bbb{Q}[\sqrt{2}]\simeq\Bbb{Q}[x]/(x^2-2)$$
  and
  $$\Bbb{R}[\sqrt{2}]=\Bbb{R}\neq\Bbb{R}\otimes\Bbb{Q}[x]/(x^2-2)=\Bbb{R}[x]/(x^2-2).$$
Could anyone explain the difference for me?
 A: In general, it is not true that $$K[a]=K\otimes_FF[a].$$ For example, take $K$ to be a finite extension of F, and take $a\in K\setminus F$. The LHS is then just $K$. The RHS is the tensor product of two vector spaces over $F$, so it is a again a vector space, whose dimension is equal to the product of dimensions of $K$ and $F[a]$ respectively. Since $a\not\in F$, this product is greater than the dimension of $K$ over $F$.
This explains what happens in the example with $\mathbb{Q}(\sqrt{2})$ and $\mathbb{R}$.
However, when working with matrices, the above equation does hold if one defines everything properly, as the following.
First we define $i:F\hookrightarrow M_n(F)$ by mapping every $x\in F$ to $xI$, and denote by $F'$ the image of $i$. $F'$ is clearly a field, isomorphic to $F$. Then, given a matrix $A\in M_n(F)$ with minimal polynomial $m_F(x)$, it is true that $$F'[A]\cong F'[x]/m'_F(x)\cong F[x]/m_F(x),$$where $m'_F(x)$ is the image of $m_F(x)$ under the obvious extension of $i$. If we identify every $x\in K$ with the matrix $xI$, and call the image $K'$, it is then true that $$K'[A]\cong K'\otimes_{F'}F'[A],$$and all the following statements in the posted proof are also true, thus the proof for equality of minimal polynomials is valid.
