$\Bbb Q[x]/\langle x^3\rangle$ is isomorphic to $R$. Find $R$.

I know about $\Bbb Q[x]/\langle x\rangle$ is isomorphic to $\Bbb Q$ by the isomorphism $T(f(x))=f(x=0)$.

$\Bbb Z[x]/\langle(x-1)(x-2)\rangle$ is isomorphic to $\Bbb Z\times \Bbb Z$ by the isomorphism $T(f(x))=(f(1),f(2))$

Looking these examples $\Bbb Q[x]/\langle x^3\rangle$ must be isomorphic to $\Bbb Q\times \Bbb Q\times \Bbb Q$. But, I am not sure. Help me please.


This is not true.

$\mathbb Q[x]/\langle x^3 \rangle$ has a nilpotent element, namely $x^3=0$. $\mathbb Q \times \mathbb Q \times \mathbb Q$ has no nilpotent elements.

We have $\mathbb Q[x]/\langle (x-a)(x-b)(x-c) \rangle \cong \mathbb Q \times \mathbb Q \times \mathbb Q$ if and only if $a,b,c$ are three distinct numbers.

Approaching the original question, actually in my opinion $\mathbb Q[x]/\langle x^3 \rangle$ is the easiest description for this ring. An other isomorphic description would be $R=\mathbb Q[A] = \{aI+bA+cA^2 | a,b,c \in \mathbb Q\} \subset \operatorname{Mat}(3 \times 3, \mathbb Q)$ with $A=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$, hence all matrices of the form $\begin{pmatrix}a&b&c\\0&a&b\\0&0&a\end{pmatrix}$.

  • $\begingroup$ Just a question: wouldn't the element $(x-a)(x-b)(x-c) \rightarrow 0$ under the second quotient? Does this have to do with multiplicity somehow. $\endgroup$ – frogeyedpeas Apr 20 '16 at 8:14
  • $\begingroup$ Of course, we have $(x-a)(x-b)(x-c)=0$ in the second quotient. $\endgroup$ – MooS Apr 20 '16 at 8:17
  • $\begingroup$ Wait i'm silly I realized that nilpotent isn't just any 0 element. $\endgroup$ – frogeyedpeas Apr 20 '16 at 8:19
  • $\begingroup$ @MooS I came to the same conclusion about this being the simplest description of this ring. I'm really curious what the book/professor is looking for with this question. $\endgroup$ – Alex Mathers Apr 20 '16 at 8:21
  • $\begingroup$ I don't think this necessarily solves the problem so i'll leave it as a comment. The ring that @Moos refers to is tuples $(a,b,c) \in \mathbb{Q}^3$ with addition defined componentwise, and multiplication swapped out with the subtler: $U * V = (u_1 , u_2 ,u_3) * (v_1, v_2, v_3 ) = (u_1 * v_1, u_1*v_2 + v_1 * u_2 + u_3*v_3, u_3*v_1+ v_3*u_1)$ $\endgroup$ – frogeyedpeas Apr 20 '16 at 8:22

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