Maximal ideals of $k[x_1,\cdots,x_n]$ and degrees of field extensions For the past couple of days i had been working on an interesting homework problem, my interpretation of which is as follows:
Let $m$ be a maximal ideal of $k[x_1,\cdots,x_n]$. Let $m_i = m \cap k[x_i], i=1,\cdots,n$.
Then show that $m$ is the unique pre-image of $m_1,\cdots,m_n$, under the map
$\operatorname{Specm} k[x_1,\cdots,x_n] \rightarrow \prod_i \operatorname{Specm} k[x_i]$ given by $m \mapsto (m_i)_i$, if and only if 
$\left[k[x_1,\cdots,x_n]/m:k\right] = \prod_i \left[k[x_i]/m_i :k\right]$.
Related to this problem are the two questions of mine:
Intersecting maximal ideals of $k[x_1,\dots,x_n]$ with $k[x_i]$ and Conditions for the degree of field extension reaches its upper bound.
PS: My homework is submitted, please feel free to attack to the problem :)

There exists the following counter-example for your assertion.
Suppose $char(k) = p > 0$.
Let $\bar k$ be an algebraic closure.
Let $a \in k$.
Suppose $x^p - a$ is irreducible in $k[x]$.
Let $m_1 = (x_1^p - a)$, $m_2 = (x_2^p - a)$.
$x^p - a$ has a unique root $\alpha$ in $\bar k$.
Hence there exists unique maximal ideal $m$ of $k[x_1, x_2]$ such that $m_i = m \cap k[x_i]$ fo $i = 1, 2$.
Since $[k(\alpha) : k] = p$, $[k[x_1, x_2]/m : k] = p$ and $[k[x_i]/m_i: k] = p$ for $i = 1, 2$.
