What is the dimension of $\Bbb Q[x]/\langle (x+1)^2 \rangle$ over $\Bbb Q$?

Question

What is the dimension of $\Bbb Q[x]/\langle (x+1)^2 \rangle$ over $\Bbb Q$?

First we identify this ring. I know $$\Bbb Q[x]/\langle (x-a)(x-b) \rangle \simeq \Bbb Q \times \Bbb Q$$ where $a,b$ are distinct. But how to deal with this ring ?

Is $$\Bbb Q[x]/\langle (x+1)(x+1) \rangle \simeq \Bbb Q \times \Bbb Q$$ via the map $$f:\Bbb Q[x] \ni p(x) \longmapsto \Big(p(-1)+\langle(x+1)\rangle,p(-1)+\langle(x+1)\rangle\Big) \in \Bbb Q \times \Bbb Q$$ and by using first isomorphism theorem? Any help?

Answer 1

1. Let $R=\mathbb{Q}[X]/I$ where $I=\langle X^2+2X+1\rangle$. Notice that $$\overline{X}^2=-2\bar{X}-1$$ in $R$. This means that you can write any $\overline{X}^k$ as $a\overline{X}+b$ with $a,b\in\mathbb{Q}$. So $\dim (R)\leq 2$.

2. Now prove that $\{\overline{X}, \overline{1}\}$ is a linearly independent set over $\mathbb{Q}$ (which is equivalent to saying that $I$ does not contain any polynomial of degree less than 2).

3. Combine the facts to conclude $\dim (R)=2$.

Answer 2

For any $p(x) \in \Bbb Q[x]$, define $f(p) = \langle r(x) \rangle$, where $p(x) = q(x)(x+1)^2 + r(x)$ and $\operatorname{deg}(r) \leq 1$. If $(x+1)^2|(p_1-p_2)$, then $\exists h(x) \text{ with } p_1(x)-p_2(x)= (x+1)^2h(x)$, so if $p_1(x) = q_1(x)(x+1)^2 + r(x)$, then $p_2(x) = p_1(x)-(x+1)^2h(x) = (x+1)^2(q_1(x)-h(x))+r(x)$ and $f$ is well defined.

The quotient space has zero divisors ($x+1$ is nilpotent) so it is not a field but it is still a vector space under addition and scalar multiplication by elements of $\Bbb Q$. Prove that $f$ preserves addition and scalar multiplication to see that it is a homomorphism. Prove that $\langle 1 \rangle \text{ and } \langle x \rangle$ are a basis for the quotient space. Prove that $1 \in \Bbb Q[x] \text{ and } x \in \Bbb Q[x]$ map to that basis.

Answer 3

If $f$ is a polynomial of degree $d$ then $\Bbb Q[x]/\left<f(x)\right>$ will have dimension $d$. Despite this, $\Bbb Q[x]/\left<(x+1)^2\right> \not\cong \Bbb Q\times\Bbb Q$. For instance $\Bbb Q[x]/\left<(x+1)^2\right>$ has a nonzero nilpotent element, and $\Bbb Q\times\Bbb Q$ does not. Alternatively $\Bbb Q[x]/\left<(x+1)^2\right>$ has one non-trivial ideal, while $ \Bbb Q\times\Bbb Q$ has two. (Non-trivial here excludes the zero ideal and the whole ring.)

Answer 4

The homomorphism $\mathbb{Q}[x]/((x+1)^2)\to \mathbb{Q}[y]/(y^2)$ defined by $x\mapsto y-1$ is an isomorphism. (One reason: the map $\mathbb{Q}[x]\to \mathbb{Q}[y]/(y^2)$ defined in the same way has $((x+1)^2)$ as its kernel.)

By division, $\mathbb{Q}[y]/(y^2)$ is represented by polynomials of the form $a+by$. If $a+by=0$ in this ring for some $a,b\in\mathbb{Q}$, then $a+by=q(y)\,y^2$ in $\mathbb{Q}[y]$ for some polynomial $q(y)$. Just by thinking about degrees, $a=b=0$, so $1$ and $y$ are independent in $\mathbb{Q}[y]/(y^2)$.

Over $\mathbb{R}$ rather than $\mathbb{Q}$, these are called the dual numbers.

(If this ring split as a direct sum, there would be an idempotent element $e$ with $e^2=e$ and $e\neq 0,1$. If $e=a+by$, $e^2=a^2+2aby+b^2y^2=a^2+2aby$. Then, $a^2=a$ and $2ab=b$. So the only idempotents are $e=0,1$.)