I am trying to make sense of what the following ring really is.

I have that $$\mathbb{Q}(\epsilon)=\mathbb{Q}[x]/x^2$$ where $\epsilon$ denotes the coset of $x$ in the quotient ring.

I am still trying to understand this question. My attempt at showing no isomorphism $\phi : Q[\epsilon] \to Q[\sqrt{2}]$ is that


but $\phi(\epsilon) \neq 0$ and so the image of epsilon is non zero in $Q[\sqrt{2}]$

and so $\phi(\epsilon)\phi(\epsilon)$ would give a non zero image in $Q[\sqrt{2}]$ as it is a field and hence has no non zero , zero divisors.

But again, I am really not sure, this is why I post on this site as Id really like to get advice from people with much more knowledge and experience then myself.

Anything will help

  • 2
    $\begingroup$ Subrings of $\mathbb{C}$ or $\mathbb{R}$ have no zero divisors, so $\epsilon^2=0$ immediately implies that $\mathbb{Q}[\epsilon]$ cannot be isomorphic to $\mathbb{Q}(\sqrt{2})$. $\endgroup$ – Slade Nov 15 '15 at 4:45
  • $\begingroup$ Hm thanks, but I am still wanting to understand if my characterization of it is even correct. and how the elements are written, what the cosets are etc $\endgroup$ – Quality Nov 15 '15 at 4:48
  • 1
    $\begingroup$ It's pretty hard to understand most of your post. If your definition of the ring is $\mathbb{Q}[x]/(x^2)$, then yes, elements are cosets $f+(x^2)$, with a natural class of representatives given by $a+bx$, but I have no idea what you mean by expressions like $(x^2)(g(x))$ or $(f(x))$. $\endgroup$ – Slade Nov 15 '15 at 5:10
  • $\begingroup$ Also, we don't say "coset of a ring," we say "coset of an ideal." $\endgroup$ – Thomas Andrews Nov 15 '15 at 5:21
  • $\begingroup$ It is entirely unclear what you mean by $\mathbb Q(\sqrt{2})\mathbb Q(\sqrt{2})$. It seems like you are confused about the notation in general. $\endgroup$ – Thomas Andrews Nov 15 '15 at 5:25

Yes your characterization is correct. Since you're modding out by a principal ideal $(x^2)$, it's easy to see which elements of your ring are 0 - here it is the polynomials which have a term of degree $\ge 2$. For any polynomial $a_nx^n + a_{n-1}x^{n-1} + \cdots + a_2x^2 + a_1x + a_0\in\mathbb{Q}[x]$, you can write it as $$x^2(a_nx^{n-2} + a_{n-1}x^{n-3} + \cdots + a_2) + a_1x + a_0 \equiv a_1x + a_0\mod x^2$$ Your ring is not a field because it has nilpotents: $x\cdot x = 0$, even though $x\ne 0$. In particular this implies that $x$ is not 0 and yet does not have a multiplicative inverse. Since every nonzero element in a field must have a multiplicative inverse, your ring cannot be a field.

Your characterization makes the ring pretty explicit and easy to compute with. For example, $(x+1)(2x-2) = 2x^2 - 2 \equiv -2\mod x^2$.

In more advanced terms, your ring is a 1st order infinitesimal neighborhood of the point $x = 0$. Here, the $n$th order infinitesimal neighborhood is $\mathbb{Q}[x]/(x^n)$. If you imagine letting $n\rightarrow\infty$, the ring $\mathbb{Q}[x]/(x^n)$ gets closer and closer to the ring of power series with coefficients in $\mathbb{Q}$. If you remember from calculus the theory of Taylor expansions, then you will remember that the $n$th coefficient had to do with the $n$th derivative of the function you were taking the Taylor expansion of. In the case of $\mathbb{Q}[x]/(x^2)$, it's like you're taking a full Taylor series expansion, and then chopping off everything after $a_0 + a_1x$. The $a_1$ is then basically the first derivative of your function, so your ring in some sense carries ``first order differential information''. You can try googling "tangent vectors and dual numbers".

  • $\begingroup$ Thanks, I am a bit confused on the relation to fields. Why is this needed to show that there cannot be an isomorphism? Also, could you elaborate on how we know that $\epsilon^{2}$ =0 ? Is it just because of the reason I gave in the original post? $\endgroup$ – Quality Nov 15 '15 at 5:49
  • $\begingroup$ @Quality Let $R$ be your ring $\mathbb{Q}[x]/(x^2)$, and let $x^2R$ denote the ideal $(x^2)$. Then $\epsilon^2 = 0$ because $\epsilon^2 = (x+x^2R)\cdot (x+x^2R) = x^2 + x^2R = x^2R = 0+x^2R$, where the last thing is the 0-coset. $\endgroup$ – oxeimon Nov 16 '15 at 0:12

Any element of $\mathbb{Q}[x]/x^2$ can be written as $a+bx+(x^2)$ where $a,b\in\mathbb{Q}$, can you see that?

And as Slade says, if there is an isomorphism between $\mathbb{Q}[\epsilon]$ and $\mathbb{Q}(\sqrt{2})$ then the image of $\epsilon$ is a zero divisor in $\mathbb{Q}(\sqrt{2})$, which cannot happen since it is a subfield of $\mathbb{R}$ or $\mathbb{C}$.

  • 2
    $\begingroup$ You are perpetuating the OPs error writing $\mathbb Q(\epsilon)$. Parentheses here imply a field. The correct notation is $\mathbb Q[\epsilon]$. $\endgroup$ – Thomas Andrews Nov 15 '15 at 5:24
  • $\begingroup$ Yes I can see the first part. As x^{2} is of degree 2 so we can write the elements of the coset as polynomial with degree with coefficients in rational, I do see that. Hm, the thing is, we have not talked about fields really. I wonder if that is the only way to do this $\endgroup$ – Quality Nov 15 '15 at 5:48
  • $\begingroup$ @Thomas Andrews, you are right of course! $\endgroup$ – gradstudent Nov 15 '15 at 6:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.