Dimension of $K[x]/x^{2}$ as a vector space Let $K$ be a field, and $R=K[x]$ the polynomial ring over K. Let $J$ be the ideal generated by $X^{2}$ Show that $R/J$ is a K-space. What is its dimension? 
I know that the torsion submodule of $R/J$ is trivial, but I don't know if it $R/J$ is finitely generated, so I don't know if it has a basis. Hints? 
 A: $K[x]/x^2$ is the vector space comprised by the remainders obtained when dividing a polynomial in $K[x]$ by $x^2$. Clearly you get remainders of the form $ax+b$. Thus, it is a space of dimension 2 with basis $\{1,x\}$
A: A general element of $k[x]$ looks like 
$$a_0 + a_1x+a_2x^2 + a_3x^3 + \cdots + a_nx^n$$
The set $k[x]/(x^2)$ is the set of all elements of $k[x]$, modulo $x^2$. Notice that
\begin{eqnarray*}
a_0 + a_1x+a_2x^2 + a_3x^3 + \cdots + a_nx^n &=& a_0 + a_1x+(a_2 + a_3x + \cdots + a_nx^{n-2})x^2 \\ \\
a_0 + a_1x+a_2x^2 + a_3x^3 + \cdots + a_nx^n &\equiv& a_0 + a_1x \bmod x^2
\end{eqnarray*}
Hence, $k[x]/(x^2)$ has dimension $2$, and has $\{1,x\}$ as a basis.
A: It is a K space of dimension 2. A base is the classes of 1 and X. To see this, divides any polynomial by $X^2$ the polynomial is equivalent to the remainder.
A: Here is further intuition. As others have noted, any coset must have the form $\overline{a + bx} := a + bx + x^2K[x]$ for some $a + bx \in K[x]$. Addition is straightforward, and ring multiplication looks like
$$ \left(\overline{a + bx}\right) \left( \overline{c + dx} \right) = \overline{ac + (ad + bc)x + bdx^2}= \overline{ac + (ad + bc)x}\text.$$
So as a ring, $K[x]/(x^2)$ is just like $2 \times 2$ upper-triangular matrices over $K$ with constant diagonal:
$$\begin{pmatrix} a & b \\ 0 & a \end{pmatrix} \begin{pmatrix} c & d \\ 0 & c \end{pmatrix} = \begin{pmatrix} ac & ad + bc \\ 0 & ac \end{pmatrix} \text.$$
The scalar multiplication in $K[x]/(x^2)$ behaves the same as in this matrix ring. It's clear that a basis is $\begin{pmatrix}0 & 1 \\ 0 & 0 \end{pmatrix} $ and the identity matrix.

In general, if $R$ is a commutative ring and $f(x)$ is a nonconstant monic polynomial in $R[x]$, then every element of $R[x]/(f)$ can be written uniquely as a linear combination of $(\overline{1}, \overline{x}, \overline{x}^2, \dots, \overline{x}^{\deg f - 1})$, with coefficients coming from the embedding of $R$ in $R[x]/(f)$.
