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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

$\phi(\epsilon)\phi(\epsilon)=\phi(\epsilon^{2})=\phi(0)=0$

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

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    $\begingroup$ 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
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    $\begingroup$ 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
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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".

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  • $\begingroup$ 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
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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 $\mathbb{C}$.

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    $\begingroup$ 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

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