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

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!

• 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 $(x^{2})$? Nov 12, 2015 at 19:41
• 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. Nov 12, 2015 at 19:42
• 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. Nov 12, 2015 at 19:44
• 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. Nov 12, 2015 at 19:47
• 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. Nov 12, 2015 at 19:49

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.

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.

• 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 $\pm 1$? Nov 12, 2015 at 19:46
• @ThomasAndrews you're right, it is nonsense. :) I will either delete it or try to recall what I was thinking of instead. Nov 12, 2015 at 22:50

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.

Is $\mathbb{Z}[x]/(x^2)$ a field? Is it a domain? Is it a reduced ring?

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

• 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. Nov 12, 2015 at 19:59
• Since every field is an integral domain. Then the ring is not a field either? Nov 12, 2015 at 20:03
• Actually, $(x^2)$ is a principal ideal. Did you mean it's not a radical ideal? Nov 12, 2015 at 20:06
• @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). Nov 12, 2015 at 20:11
• 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? Nov 12, 2015 at 20:28