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My guess is no, but I cannot find a counter example. I started off by saying let $x_1,x_2 \in \mathbb{R}[x]/(x^2)$ then consider $x_1x_2 = (a+bx)(c+dx) = (ad+bc)x + ac = 0$ but I'm not sure how to get a contadiction from here

algebar87
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  • Note that the accepted answer (and your question) implicitly assumes that every element of the quotient ring has a unique rep of the form $,a+bx.,$ This requires proof, e.g. see here. However, one can prove the claim without using such normal forms, e.g. see the second proof in my answer. – Bill Dubuque Mar 20 '15 at 02:06

4 Answers4

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$x\in\mathbb{R}[x]/(x^2)$ and $x\neq0$, but $x$ has no inverse since $x(a+bx)=ax\neq 1$ for any $a,b$.

Spenser
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  • Could you expand how that means its not a field – algebar87 Mar 20 '15 at 00:49
  • In a field, every non-zero element is a unit (has an inverse). Also, note that any unit element cannot possibly be a zero divisor, so $x$ gives a counterexample. – user160738 Mar 20 '15 at 00:51
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    In addition, you could note that $(x^2)$ is not a maximal ideal, and hence $\mathbb{R}[x]/(x^2)$ is not a field. – user160738 Mar 20 '15 at 00:52
  • I was trying to find a counter-example like this from my working. Could you perhaps if you have the time continue my working to show that there is no inverse? – algebar87 Mar 20 '15 at 00:53
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    @algebar87: in your example, take $a=c=0$. That is, explicitly, $x \cdot x = 0$ in $\mathbb{R}[x]/(x^{2})$, even though $x$ and $0$ are distinct cosets of the ideal $(x^{2})$ in $\mathbb{R}[x]/(x^{2})$. In general, a zero divisor cannot be invertible. – Alex Wertheim Mar 20 '15 at 00:56
  • why cna't $a = 1/x$ in your example – algebar87 Mar 20 '15 at 01:00
  • Bingo, @AlexWertheim you have it dead-on. – ncmathsadist Mar 20 '15 at 01:03
  • post as an answer and I will upvote it. – ncmathsadist Mar 20 '15 at 01:03
  • I don't understand why 1/x = a does not work – algebar87 Mar 20 '15 at 01:07
  • @algebar87 $1/x$ is not a well defined element of $\mathbb{R}[x]/(x^2)$. Indeed, $\mathbb{R}[x]/(x^2)={a+bx:a,b\in\mathbb{R}}$ and there is no such thing as $1/x$ in this set. – Spenser Mar 20 '15 at 01:10
  • It seems that you are implicitly assuming that every element of the quotient ring has a unique rep of the form $,a+bx.,$ This should be explicitly stated (and it should be proved by the OP if they use it). – Bill Dubuque Mar 20 '15 at 01:27
  • @spenser why not simplify by noting that if $xy=1$ for some $y$, then $0=xxy=x\neq 0$ is a contradiction. It has nothing to do with the structure of the ring beyond that you are working with units and zero divisors. This is far cleaner than the other suggestion that you labor in a proof of uniqueness of representation. – rschwieb Mar 20 '15 at 03:12
  • @rschwieb That also uses uniqueness (or some other way to prove that $,x\neq 0,$ in the quotient ring). There was no suggestion that using uniqueness is the "cleanest" approach (whatever that means) but, rather, only a guess that it was being implicitly assumed in the argument in the answer. – Bill Dubuque Mar 20 '15 at 13:49
  • Dear @BillDubuque : That is true, and proving $x\neq 0$ (if the instructor would even require it) is still going to be easier than uniqueness of representation of elements. "Cleaner" here means that it uses key ideas of units and zero divisors rather than getting sidetracked about what elements in this particular ring look like. Hope this explains things. Regards – rschwieb Mar 20 '15 at 14:14
  • @rschwieb Yes, key ideas that I mention frequently (but not when - as above - discussing how to complete a specific proof). – Bill Dubuque Mar 20 '15 at 14:40
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Hint $\ $ In a field, $\ \, xf = 0\,\Rightarrow\,x=0\,$ or $\,f = 0,\ $ so $\, x^2 = 0\,\Rightarrow\, x = 0\, \ \Rightarrow\!\Leftarrow$

Or $\,\ xf = 1\,$ in $\,\Bbb R[x]/\color{#c00}{xg}\,\Rightarrow\, xf = 1 + \color{#c00}{xg}\, h\ $ in $\,\Bbb R[x]\,\overset{\large x\,=\,0}\Rightarrow\, 0 = 1\,$ in $\,\Bbb R\ \Rightarrow\!\Leftarrow$

Remark $\ $ This is known as the algebra of dual numbers over the ring $\Bbb R.\,$ Such rings and higher order analogs $\,\Bbb R[x]/x^n \;$ prove quite useful when studying (higher) derivations algebraically since such rings provide very convenient algebraic models of tangent / jet spaces. For example, they permit easy transfer of properties of homomorphisms to derivations -- see for example section $8.15$ in N. Jacobson, Basic Algebra II. See this post for further discussion and links.

Community
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Bill Dubuque
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We have a few facts.

  1. Let $\mathfrak a$ be an ideal in a ring $A$. Then $A/\mathfrak a$ is a field if and only if $\mathfrak a$ is a maximal ideal.
  2. Let $\mathfrak a$ be an ideal in a PID $A$. Then $\mathfrak a$ is a maximal ideal if and only if $\mathfrak a$ is a prime ideal.
  3. If $k$ is a field, then $k[x]$ is a PID.
  4. If $k$ is a field with $f\in k[x]$, then $\langle f\rangle$ is a prime ideal in $k[x]$ if and only if $f$ is irreducible.

Can you see how these facts combine to prove that $\Bbb R[x]/\langle x^2\rangle$ is not a field?

Brian Fitzpatrick
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No need of complicated proofs. Let $I=(x^2)$ and consider $x+I$; then $$ (x+I)^2=x^2+I=I $$ Thus the ring is not a domain. We just need to ensure that $x\notin I$, but this is clear because nonzero polynomials in $I$ have degree at least $2$.

egreg
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  • +1 This is the ideal proof if the user realizes fields are domains, although it does seem a bit like the OP may not realize this. Considering that, this is close to Spenser's answer. Markedly more transparent than the remaining solutions, though. – rschwieb Mar 20 '15 at 14:33
  • This is exactly the same as the first argument in my answer yesterday, except using coset notation rather than element notation. – Bill Dubuque Mar 20 '15 at 14:45
  • @BillDubuque Since you're using $x$ with two different meanings, I'd consider your argument at least flaky. – egreg Mar 20 '15 at 16:02
  • @egreg Eh? Where do you see "two different meanings"? – Bill Dubuque Mar 20 '15 at 16:06
  • @BillDubuque Well, $x$ in the question is an indeterminate over $\mathbb{R}$, so surely $x^2\ne0$. Of course I know what you are meaning; the OP may not. – egreg Mar 20 '15 at 16:08
  • @egreg I find it hard to take your critique seriously. Surely you are well aware of this overloaded notation. Everyone uses it, e.g. using the same notation for integers in quotient rings. Otherwise one ends up with precisely formal proofs (like Principia Mathematica) that, while useful for a mechanical proof-checker, only serve to obscure human comprehension. Once one understands the abuse of notation, there is no need to be overly formal. If the OP does not follow then they can certainly ask for elaboration in comments. – Bill Dubuque Mar 20 '15 at 16:18