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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?

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  • $\begingroup$ It is by definition that $K[x]/(f)$ is a $\deg(f)$ dimensional $K$-vector space $\endgroup$ – reuns Feb 28 '19 at 3:41
  • $\begingroup$ Can you find two linearly independent elements in your quotient ring? (Yes.) If so, then your proposed map can't be an isomorphism of the quotient map because it has dimension 1. I haven't checked, but it's not obvious to me that your proposed map is even well defined in the quotient ring (which requires you to prove that if $(x^2-1)~|~(p(x)-q(x)$, then $f(p)=f(q)$). $\endgroup$ – Robert Shore Feb 28 '19 at 4:02
  • $\begingroup$ Do you mean to consider it as $\mathbb{Q}$ vector space? Because I don't know that there's a good general definition of the dimension of a ring extension. (I could easily be mis-remembering that, but I would like to know what definition of dimension you're using if you're dealing with rings.) $\endgroup$ – JonathanZ supports MonicaC Feb 28 '19 at 5:05
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  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 $\dim (R)=2$.

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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.)

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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.

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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 $e=0,1$.)

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    $\begingroup$ Isn't idempotent $e^2 =e$, not $e^2 = 1$? $\endgroup$ – JonathanZ supports MonicaC Feb 28 '19 at 15:20
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    $\begingroup$ @JonathanZ Yes, you are correct, thanks. $\endgroup$ – Kyle Miller Feb 28 '19 at 16:49

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