First Isomorphism Theorem (another) 
Use the First Isomorphism Theorem to show that $\mathbb{Q}[x]/I$ is isomorphic to the ring
        \begin{align*}
      \left\{\begin{pmatrix}
      a & b\\
      0 & a
      \end{pmatrix}\in M_2(\mathbb{Q}):a,b\in\mathbb{Q}\right\}
  \end{align*}
  where $I=\langle x^2\rangle\subset\mathbb{Q}[x]$.

I'm not too sure how to go about defining the map. Is it starting from $\mathbb{Q}[x]/I$ or do I have to start from the very beginning (that is, $\mathbb{Q}[x]$)?
A worked out solution would be appreciated!
 A: Note that
$$
\begin{pmatrix} a & b \\ 0 & a \end{pmatrix}
=
a \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
+
b \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}
= a I + bU
$$
where $U^2=0$. The isomorphism with $\mathbb{Q}[x]/\langle x^2\rangle$ is clear: just send $x \mapsto U$.
A: Hint: let $R$ denote your matrix ring, and consider the map $\mathbb Q[x]\to R$ given by $$f(x)\mapsto \begin{bmatrix} f(0) & f'(0) \\ 0 & f(0)\end{bmatrix}.$$
Edit: apologies for the typo: I put the $f'$ in the wrong matrix entry. I have fixed it now.
A: Let us denote the given ring by $T$:
$T = \left \{ \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \in M_2(\Bbb Q 
): a, b \in \Bbb Q \right \}, \tag 1$
and let $N$ be the matrix
$N = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}; \tag 2$
we observe that any
$\begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \in T \tag 3$
may be written in the form
$T = aI + bN, \tag 4$
and that 
$N^2 = 0. \tag 5$
Now consider the map
$\phi:  \Bbb Q[x] \to T, \tag 6$
given by
$\phi(p(x)) = p(N), \tag 7$
that is, if
$p(x) = \displaystyle \sum_0^n p_j x^j, \tag 8$
then
$\phi(p(x)) = p(N) = \displaystyle \sum_0^n p_j N^j; \tag 9$
it is easy to see that $\phi$ is a homomorphism, since it is simply evaluation of $p(x) \in \Bbb Q[x]$ at $x = N \in T$, so we leave the details of the demonstration to the reader.  Furthermore, $\phi$ is surjective, since for any $a, b \in \Bbb Q$, 
$\phi(a + bx) = aI + bN = \begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \in T, \tag{10}$
and I claim that
$\ker \phi = \langle x^2 \rangle; \tag{11}$
for if
$g(x) \in \langle x^2 \rangle, \tag{12}$
there exists
$h(x) \in \Bbb Q[x] \tag{13}$
with
$g(x) = x^2 h(x); \tag{14}$
then
$\phi(g(x)) = \phi(x^2h(x)) = N^2h(N) = 0, \tag{15}$
whence 
$g(x) \in \ker \phi; \tag{16}$
thus
$\langle x^2 \rangle \subset \ker \phi; \tag{17}$
and if
$g(x) \in \ker \phi, \tag{18}$
writing
$g(x) = \displaystyle \sum_0^n g_j x^j$
$= g_0 + g_1 x + \displaystyle \sum_2^n g_ix^i = g_0 + g_1 x + x^2 \sum_2^n g_i x^{i - 2} \tag{20}$
we have
$\phi(g(x)) = \phi \left ( g_0 + g_1 x + x^2  \displaystyle \sum_2^n g_i x^{i - 2} \right )$
$= \phi(g_0 + g_1x) + \phi(x^2) \phi \left (   \displaystyle \sum_2^n g_i x^{i - 2} \right )$
$= g_0I + g_1N + N^2 \displaystyle \sum_2^ng_iN^{i - 2} = g_0 + g_1N \tag{21}$
by virtue of (5); since (18) is equivalent to
$\phi(g(x)) = 0, \tag{22}$
(21) implies
$g_0 = g_1 = 0, \tag{23}$
and thus
$g(x) = \displaystyle \sum_2^n g_ix^i = x^2 \sum_2^n g_i x^{i - 2} \in \langle x^2 \rangle, \tag{24}$
and we have shown that
$\ker \phi \subset \langle x^2 \rangle; \tag{25}$
combined with (17) this yields (11).
I looked up The First Isomorphism Theorem just to refresh my memory and, sure enough, that is the one which affirms that a surjective homomorphism $\theta:R \to S$ of commutatve unital rings becomes an isomorphism
$\bar \theta:R/\ker \theta \cong S \tag{26}$
when passing to the quotient ring $R/\ker \theta$; applying this theorem to $\phi$ immediately yields a quotient isomorphism
$\bar \phi:\Bbb Q[x]/\langle x^2 \rangle \cong T, \tag{27}$
as per request.
First Isomorphism Theorem (another)
