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I want to prove $k[x]/(x^2)$ is local. I know it by rather a direct way: $(a+bx)(a-bx)/a^2=1$. But for general case such as $k[x]/(x^n)$, how can I prove it?

Also for 2 variables, for example $k[x,y]/(x^2,y^2)$ (or more higher orders?), how can I prove they are local rings?

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    Why not just exhibit the unique maximal ideal? You know ideals in the quotient are ideals in the ambient ring containing what you mod out by so proving that your candidate is te only maximal ideal shouldn't be hard.2012-12-02

4 Answers 4

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A ring $R$ is local if and only if the associated reduced ring $R_{red} = R/Rad(0)$ is local. In both of your cases this latter ring is the field $k$, hence both of your rings are local.

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Let $R$ be a commutative ring and $I\subset R$ an ideal. If $\sqrt{I}\in\operatorname{Max}(R)$, then $R/I$ is a local ring of Krull dimension $0$.

Proof. Let $P\in\operatorname{Spec}(R/I)$. Then there is $\mathfrak{p}\in\operatorname{Spec}(R)$, $\mathfrak{p}\supseteq I$, such that $P=\mathfrak{p}/I$. Since $\sqrt{I}\subseteq\mathfrak{p}$ and $\sqrt{I}\in\operatorname{Max}(R)$ it follows that $\mathfrak{p}=\sqrt{I}$. Thus the only maximal ideal of $R/I$ is $\sqrt{I}/I$.

In your examples $\sqrt{(X^2)}=(X)$ is maximal in $K[X]$, and similarly $\sqrt{(X^2,Y^2)}=(X,Y)$ is maximal in $K[X,Y]$.

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There are several other answers here, but you'll get another one.

Basic commutative algebra fact: the set of nilpotents $N$ is an ideal, and is equal to $\cap_{\mathfrak{p} \, \mathrm{prime}} \mathfrak{p}$, the intersection of all prime ideals in your ring.

Well, $x$ is nilpotent in your ring, so every prime ideal contains it(and so every maximal ideal contains it), but $(x)$ is maximal! Thus our ring is local.

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You can use the following

Claim: A commutative unitary ring is local iff the set of non-unit elements is an ideal, and in this case this is the unique maximal ideal.

Now, in $\,k[x]/(x^n):=\{f(x)+(x^n)\;\;;\;\;f(x)\in K[x]\,\,,\,\deg(f) , an element in a non-unit iff $\,f(0)=a_0= 0\,$ , with $\,a_0=$ the free coefficient of $\,f(x)\,$, of course.

Thus, we can characterize the non-units in $\,k[x]/(x^n)\,$ as those represented by polynomials of degree less than $\,n\,$ and with free coefficient zero, i.e. the set of elements $\,M:=\{f(x)+(x^n)\in k[x]/(x^n)\;\;;\;\;f(x)=xg(x)\,\,,\,\,g(x)\in k[x]\,\,,\deg (g)

Well, now check the above set fulfills the claim's conditions.

Note: I'm assuming $\,k\,$ above is a field, but if it is a general commutative unitary ring the corrections to the characterization of unit elements are minor, though important. About the claim being true in this general case: I'm not quite sure right now.

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    No, I used $\,k\,$ is a field, but now that you bring this up it may be there are some weird implications for general rings...I meant to fix the characterization of non-units, which must be then changed to $\,f(0)\,$ is a non-unit in the ring. Some care must be put here. Thanks.2012-12-02