My first instinct is to go to the 1st isomorphism theorem and say that the two rings are isomorphic if there is a field where $(x^2)=(x^2-1)$, which doesn't seem terribly promising.

But then it seems to me that $F[x]/(x^2)$ is the ring you get when you take $F$ and adjoin an element that satisfies the relation $x^2=0$, which would be $0$ and every field's got one. Same with $F[x]/(x^2-1)$ which is $F$ adjoined by an element satisfying $x^2=1$. $1$ does the trick and every field's got one of those too.

Any hints (or answers!) would be appreciated.


  • $\begingroup$ I'm thinking you mean $F[x]/(x^2)$ and $F[x]/(x^2-1)$. $\endgroup$ Feb 13, 2012 at 1:44
  • 1
    $\begingroup$ This is an exact duplicate of this question. $\endgroup$
    – Math Gems
    Feb 13, 2012 at 2:06
  • $\begingroup$ @MathGems: Thanks. I did a quick search but I missed that one. $\endgroup$
    – jobrien929
    Feb 13, 2012 at 2:14
  • $\begingroup$ @MathGems: IMHO the questions are not exact duplicates. They are about the same problem but the OPs reflect distinctly different misunderstandings. $\endgroup$ Dec 19, 2012 at 0:54
  • $\begingroup$ Does this answer your question? $F[x]/(x^2)\cong F[x]/(x^2 - 1)$ if and only if F has characteristic 2 $\endgroup$ Dec 2, 2020 at 9:40

2 Answers 2


Assuming you do mean $F[x]/(x^2)$ and $F[x]/(x^2-1)$, they are isomorphic if the field has characteristic $2$ but not otherwise. When the characteristic is $2$, $x^2-1=(x-1)^2$, and the homomorphism $F[x]\rightarrow F[x]$ given by $x\mapsto x-1$ induces an isomorphism (by passage to the quotient) $F[x]/(x^2)\cong F[x]/(x^2-1)$.

If the characteristic is not $2$, then $F[x]/(x^2-1)$ is isomorphic to $F\times F$ as a ring by the Chinese remainder theorem, while $F[x]/(x^2)$ cannot be because it has a non-zero nilpotent (the image of $x$).

  • $\begingroup$ Is $x \rightarrow x-1$ a homomorphism? $\endgroup$ Nov 24, 2020 at 12:58
  • 2
    $\begingroup$ There is a unique $F$-algebra homomorphism from $F[x]$ to itself sending $x$ to $x-1$. This is an instance of the universal mapping property of the polynomial ring $F[x]$. Of course the map can be written down explicitly: a polynomial $a_dx^d+a_{d-1}x^{d-1}+\cdots+a_1x+a_0$ is sent to $a_d(x-1)^d+a_{d-1}(x-1)^{d-1}+\cdots+a_1(x-1)+a_0$. $\endgroup$ Nov 24, 2020 at 14:04

To add to Keenan's answer and hopefully address the thinking in the OP a little more explicitly:

$F[x]/(x^2)$ is, as you say, the ring you get when you adjoin to a field an element $x$ satisfying $x^2=0$. And, as you say, $0$ already satisfies this relation in $F$. I interpret the second paragraph of the question to be asking, "Doesn't this mean $F[x]/(x^2)$ is just $F$ adjoin $0$?" Another way to ask this could have been: "Is $F[x]/(x^2)$ the same thing as $F$?"

The answer is no, because $x\neq 0$ in $F[x]/(x^2)$. Thus this ring has another element besides zero whose square is zero. Here are two ways to look at why this is true:

(1) By the definition of $F[x]/(x^2)$, it is the ring of additive cosets of the ideal $(x^2)$. The zero coset is the ideal $(x^2)$ itself. Now $x$ is not in this ideal; that means the coset $x+(x^2)$ is different from zero in this quotient ring.

This "nuts and bolts" argument has the advantage that it connects directly to the definition, but the disadvantage that it doesn't address the way of thinking about $F[x]/(x^2)$ that led you to wonder if $x$ was zero. Hopefully that is addressed by -

(2) As you said, we can think of $F[x]/(x^2)$ as $F$ adjoin an $x$ such that $x^2=0$. But this ring is not just some ring containing $F$ and an element $x$ whose square is zero. It's the (unique) "biggest"/"freest" ring that has this. The only relations that it has between $x$ and other elements of the ring are those forced by $x^2=0$ (such as $x^5=0$, $x^2+1=1$, etc.). In a ring (unlike a field), it is possible for an element to square to zero without equaling zero, so for "maximum bigness/freedom", the element $x$ that is adjoined to $F$ in $F[x]/(x^2)$ is different from zero. $x^2=0$ does not force $x=0$ so $F[x]/(x^2)$ does not have this latter relation.

This second argument is an informal statement of the fact that the quotient $F[x]/(x^2)$ is characterized by a "universal property": for any ring $R$ containing $F$ and also an element (say $\alpha$) whose square is zero, $F[x]/(x^2)$ has a unique homomorphism to that ring that is the identity on $F$ and sends $x$ to $\alpha$. Since it has to be able to do this for any ring $R$ with these properties, it can't have any "extra" relations.


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.