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Describe the ring $\mathbb{Z}[x]/(x^2)$.

Attempt: $\mathbb{Z}[x]/(x^2)$ can be thought as the linear polynomials with integer coefficients.

So $\mathbb{Z}[x]/(x^2) = \{a + bx + (x^2) : a,b \in \mathbb{Z}\}$.

What are other things I could describe? Thank you!

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  • $\begingroup$ What you have is probably pretty good for the purposes of the exercise (can you glean any more information from the source text about what the problem is requesting of you?). Otherwise, for interest, can you prove that this ring is not an integral domain and not a field, using properties of the ideal $(x^{2})$? $\endgroup$ – Sinister Cutlass Nov 12 '15 at 19:41
  • $\begingroup$ the book just says "describe (briefly) the ring structure of the ring" . I was not really sure what other things I could mention. I will try to consider what you mentioned. $\endgroup$ – user2942 Nov 12 '15 at 19:42
  • $\begingroup$ The ring structure means describe addition, multiplication, and the additive and multiplicative identity, if any. It can also mean describe properties of the ring beyond that - is commutative, an integral domain, etc. $\endgroup$ – Thomas Andrews Nov 12 '15 at 19:44
  • $\begingroup$ By the way, always put the exact text of the problem in the question, if possible. "Describe the ring structure of the ring..." might seem only slightly different, but it is actually quite different. $\endgroup$ – Thomas Andrews Nov 12 '15 at 19:47
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    $\begingroup$ Your description "the linear polynomials ..." is wrong, or incomplete, because that set is not closed under multiplication. You have to tell us what happens in multiplying. $\endgroup$ – Thomas Andrews Nov 12 '15 at 19:49
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Is $\mathbb{Z}[x]/(x^2)$ a field? Is it a domain? Is it a reduced ring?

What can you deduce from the ideal $(x^2)$?

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  • $\begingroup$ Could I also mention : It is not integral domain nor a PID since the polynomial ring $\mathbb{Z}[x]$ contains nonprincipal ideals. $(x^2)$ is not a principal ideal. $\endgroup$ – user2942 Nov 12 '15 at 19:59
  • $\begingroup$ Since every field is an integral domain. Then the ring is not a field either? $\endgroup$ – user2942 Nov 12 '15 at 20:03
  • $\begingroup$ Actually, $(x^2)$ is a principal ideal. Did you mean it's not a radical ideal? $\endgroup$ – Ottavio Bartenor Nov 12 '15 at 20:06
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    $\begingroup$ @OttavioBartenor is correct. Since $(x^{2})$ can be written as being generated by a single element, i.e., $x^{2}$, the ideal is principal. It is however, not a radical ideal, because while it contains the second power of $x$, it does not contain $x$ itself. On another note, every quotient of a PID is a PID (this is not too difficult to prove). $\endgroup$ – Sinister Cutlass Nov 12 '15 at 20:11
  • $\begingroup$ But as @user2942 pointed out, $\mathbb{Z}[x]$ is not a PID, since some ideals, e.g. $(2,x)$, cannot be generated by a single element. What about $\mathbb{Z}[x]/(x^2)$, could it be a PID anyway? $\endgroup$ – Ottavio Bartenor Nov 12 '15 at 20:28
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This is actually a very interesting ring:

You can consider 'x' an "infinitesimal" in a sense.

Let $A = a + x a'$ and $B = a + x a'$ then we have

  • $A + B = (a+b) + x(a'+b')$
  • $AB = (ab) + x(a'b + ab')$

The first component is the normal operation, but the second component follows the same formulas as differentiation does!

This has applications in automatically differentiating functions in programming.

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It's a vague question, so the answer might also turn out vague.

As you have already done, describe the elements of the ring. What else? You could describe the addition and the multiplication in the ring, and you could figure out which elements have a multiplicative inverse.

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  • $\begingroup$ That last line has me stumped. The ring above has a multiplicative identity. What surjective homomorphism exists with $f:\mathbb Z\to\mathbb Z$, other than multiplication by $\pm 1$? $\endgroup$ – Thomas Andrews Nov 12 '15 at 19:46
  • $\begingroup$ @ThomasAndrews you're right, it is nonsense. :) I will either delete it or try to recall what I was thinking of instead. $\endgroup$ – Mankind Nov 12 '15 at 22:50
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Abstractly, it's the ring $\mathbb Z[u]$ in which $u^2=0$, and so its elements are of the form $a+bu$, with $a,b\in\mathbb Z$.

In this sense, it's like $\mathbb Z[i]$ in which $i^2=-1$. But they are very different rings, of course.

If you want to understand $\mathbb Z[u]$ a little better, you may want to find its units, its zero divisors, its nilpotent elements, its idempotents, its ideals, etc.

In particular, you'll find that there are no non-trivial idempotents in $\mathbb Z[u]$ and so it cannot be expressed as the product of two rings.

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