Determine $\text{Aut}(S)$ where $S = \mathbb{Q}[x]/(x^2)$ Here's the problem: if $S = \mathbb{Q}[x]/(x^2)$, compute the group  of automorphisms $\mbox{Aut}(S)$. 
First off, I have the solution and I know it's isomorphic to the multiplicative group $\mathbb{Q}\backslash\{0\}$, so I'm trying to understand it. It uses what is called the substitution principle and the fact that the restriction of any endomorphism $\phi: \mathbb{Q}[x]\to \mathbb{Q}[x]$ to $\mathbb{Q}$ is the identity. It says that these two facts implies that the endomorphisms of $\mathbb{Q}[x]$ are in 1-1 correspondence with $h \in \mathbb{Q}[x]$, where the endomorphism corresponding to $h$ is $\phi_h : \mathbb{Q}[x] \to \mathbb{Q}[x]$ given by $\phi_h(f) = f(h)$ for $f \in \mathbb{Q}[x]$. I guess I just don't understand how it applied the substitution principle? It's probably obvious, but for some reason I am not getting it.
An explanation would be greatly appreciated! Or if you have a different solution, that'd be cool too. Thanks!!
 A: When I try to understand homomorphisms in general, I try to look at where they send some important elements (I'd call them generators, but then I'd be dangerously close to calling $S$ a vector space over $\Bbb Q$, and I don't want to look at it that way right now). For any (non-trivial) endomorphism $\phi:S\to S$, we must have $\phi(1) = 1$. What then remains is to see what $\phi(x)$ can be.
If we want $\phi$ to be an automorphism, then $\phi(x)$ cannot be a constant term $q$, because then you'd have a non-trivial kernel containing $x-q$. So $\phi(x)$ has to be a non-zero multiple of $x$, say $px$. It turns out that as long as this is fulfilled, $\phi$ is an automorphism (with inverse sending $x$ to $\frac{1}{p}x$). So at least as a set, $\Bbb Q\setminus \{0\}$ is canonically bijective to $\mbox{Aut}(S)$ by sending $p$ to the corresponding $\phi$ as over.
Now, let's use the notation $\phi_p$ for the automorphism of $S$ sending $x$ to $px$. It is not difficult to show that $\phi_p\phi_q = \phi_{pq}$, and we then have that the group structure of $\Bbb Q\setminus \{0\}$ and $\mbox{Aut}(S)$ are isomorphic, with one isomorphism being the bijection above.
A: I'm going to ignore the hint. There are two basic ideas involved here:


*

*There is a one-to-one correspondence between homomorphisms $f : R \to S$ such that $f(I) = \{ 0 \}$ and homomorphisms $g : R/I \to S$. The relationship between corresponding functions is the identity $g(a + I) = f(a)$.

*There is a one-to-one correspondence between homomorphisms $f : R[x] \to S$ and pairs consisting of a homomorphism $g: R \to S$ and an element $\xi \in S$. The relationship between corresponding functions is the identity
$$ f\left(\sum a_i x^i \right) = \sum g(a_i) \xi^i $$


So, any homomorphism $f : \mathbb{Q}[x] / (x^2) \to \mathbb{Q}[x] / (x^2)$ can be described by


*

*A homomorphism $g : \mathbb{Q}[x] \to \mathbb{Q}[x]/(x^2)$ such that $g(x^2) = 0$


and equivalently by


*

*A pair consisting of a homomorphism $\mathbb{Q} \to \mathbb{Q}[x] / (x^2)$ and an element $\xi \in \mathbb{Q}[x]/(x^2)$ such that $\xi^2 = 0$.


There is only one homomorphism $\mathbb{Q} \to \mathbb{Q}[x] / (x^2)$, therefore, the endomorphisms of $\mathbb{Q}[x]/(x^2)$ are completely classified by elements $\xi \in \mathbb{Q}[x]/(x^2)$ such that $\xi^2 = 0$.
But how can we find solutions to the equation $T^2 = 0$ in $\mathbb{Q}[x]/(x^2)$? It's easier to lift the problem to $\mathbb{Q}[x]$, where the corresponding problem is to solve
$$ T^2 \equiv 0 \pmod{x^2} $$
or equivalently, $x^2 \mid T^2$. This happens if and only if $x \mid T$.
Therefore, every endomorphism of $\mathbb{Q}[x] / (x^2)$ is determined by an element $x f(x) + (x^2)$
This can simplify, since:
$$ x \sum_{i=0}^n f_i x^i + (x^2) = x f_0 $$
and so each endomorphism is determined by the value $f_0 \in \mathbb{Q}$.
To recap, we've shown every endomorphism of $\mathbb{Q}[x] / (x^2)$ is of the form
$$ f_c(a + bx + (x^2)) = a + bcx + (x^2) $$
for $c \in \mathbb{Q}$. This would generally be abbreviated by saying it maps $x$ to $cx$.
By inspection, we can see the only endomorphism that is not an automorphism is $f_0$. Explicitly, the inverse of $f_c$ is $f_{1/c}$.
