Let $K$ be a field. Let $K[X]$ be a (nice) algebra over a field corresponding to an affine variety. For simplicity say it is $K[x_1,\dots,x_m]$ (usual polynomial ring). Then "points" of $X$ are homomorphisms $K[x_1,\dots,x_m] \to K$ Given by evaluating $x_i \mapsto a_i \in K$. The kernel of this homomorphism is clearly $(x_1-a_1,\dots,x_m-a_m)$. Call this map $\text{pt}_{(a_i)}$. Now let $\sigma \in \text{Gal}(K/k)$ be an automorphism of $K$ over some ground field $k$. Consider the composition $\sigma \circ \text{pt}_{(a_i)}$. I can't get my brain around the following "paradox" involving the composition, and I really hope someone can clear it up for me:
On one hand, since $\text{pt}$ has kernel $\mathfrak{m}=(x_1-a_1,\dots,x_m-a_m)$, the composition of $\text{pt}$ with $\sigma$ still has kernel $\mathfrak{m}$
On the other hand, $\sigma \circ \text{pt}$, being a homomorphism out of a polynomial ring, is determined by the images of the $x_i$, which are by definition $x_i \mapsto a_i \mapsto \sigma(a_i)$, i.e. $x_i \mapsto \sigma(a_i)$: it is evaluation at $(\sigma a_i,\dots, \sigma a_m )$ in other words it is $\text{pt}_{(\sigma(a_i))}$. But the kernel of this map, thought of in this "other" way is $^{\sigma}\mathfrak{m} = (x_1-\sigma a_1,\dots, x_m-\sigma a_m)$, $^{\sigma}\mathfrak{m} \neq \mathfrak{m}$ generally.
Obviously I am interpreting something wrong or misunderstanding something really basic. It would be very helpful if someone would point it out.
Edit I accepted the answer below, however, I have figured out where I have gone wrong specifically in my own thinking and figured I'd share it as well: I was being sloppy with my thinking about "points" and forgot that a ring morphism (emphasis on the fact that it's not a k-algebra map) $K[x] \to K$ is not just determined by the image of $x$! By the universal property of poly rings, in each formulation the above map is the unique ring homomorphism $K[x] \to K$ extending $\sigma: K \to K$ and sending $x_i \mapsto \sigma a_i$. As Cory Griffith points out, if I first applied $\sigma$ to the coefficients of $K[X]$ and then evaluated $x_i \to a_i$ that would be a different map, the unique map extending $\sigma$ and sending $x_i \to a_i$.
What's happening is that you're mixing up the induced isomorphism $\sigma': K[X] \to K[X]$ and $\sigma$ itself. Let $pt': K[X] \to K[X]$ be the map which takes $K[X]$ onto its subfield $K$ in the same way as $pt$. Then your second formulation of the kernel is correct for $\sigma' \circ pt'$. However, for $\sigma \circ pt$, the $x_i$ are not necessarily fixed for $\sigma$, since by the time you apply $\sigma$ they are actual elements of $K$ and not formal symbols. Looked at this way, the second formulation should be that $\sigma \circ pt$ vanishes on $\vec x$ precisely when $\sigma(x_1) - \sigma(a_1), \dots, \sigma(x_m) - \sigma(a_m)$ are all zero, which occurs when $\vec x$ is in the desired kernel.