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Given ring $F[X]/(X^2)$ I'm trying to understand why the ideal (X) is the unique maximal ideal of the ring. I have figured out that an element in the ring is either in the ideal (X) or is a unit, but not sure if this helps.

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Your observation directly implies the claim:

Let $R$ be a ring and let $I$ be a proper ideal of $R$ such that every non-unit is contained in $I$. Then $I$ is a maximal ideal, and it is the unique one.

Proof. If $J$ is some ideal, then either $J=R$, or $J$ consists of non-units. Thus, $J \subseteq I$. Thus, if $I \subseteq J$, we get $I=J$, which shows that $I$ is maximal. And if $J$ is maximal, we get $J=I$, so that $I$ is the unique maximal ideal. $\checkmark$

Alternatively, you can use the fact that $R/I^n$ has a unique prime ideal (and therefore a unique maximal ideal) if $I$ is some maximal ideal of a commutative ring $R$. Apply this to $R=F[X]$ and $I=(X)$.

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You are almost there. The (half) following Lemma completes your proof:

Lemma (a) If $I \neq R$ is an ideal in $R$, which contains all non-invertible elements, then $I$ is the only maximal ideal in $R$.

(b) If $I$ is the only maximal ideal in $R$ then $I$ contains all the non-invertible elements of $R$.

The idea of the proof:

(a) If $J \neq R$ is any ideal, then $J$ can only contain non-invertible elements thus $J \subset I$.

(b) If $a$ is any non invertible element in $R$ then $<a>$ is contained in a maximal ideal, thus in $I$.

The part $(a)$ completes your proof, while part (b) emphasizes that your approach is a standard way to show that an ideal is the only maximal ideal in a ring.

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A proper ideal cannot contain a invertible element. Thus if for some ring one has that $R \setminus R^{\times}$ is an ideal, than this ideal is maximal, and the only maximal ideal, as every proper ideal is certainly contained in it, since it does not contain any units.

Combining this with what you have shown gives the result.

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In any commutative ring $R$, the prime (maximal) ideals of a quotient ring $R/I$ correspond bijectively to the prime (maximal) ideals of $R$ which contain the ideal $I$.

In the present case, the prime ideals of $F[X]/(X^2)$ correspond to the prime ideals of $F[X]$ which contain $X^2$, hence which contain $X$. As $X$ is a maximal ideal of $F[X]$, $(X)/(X^2)$ is the unique maximal/prime ideal of $F[X]/(X^2)$.

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I suppose you mean a field by the ring $F$. Suppose that $I$ is a maximal ideal of $F[X]/(X^2)$. We will denote elements in this quotientring by putting a bar over them. Let $a\bar{X} + b \in I$ (with $a, b \in F$), then also $\bar{X}(a\bar{X}+b) \in I$ (by the definition of an ideal) and since $\bar{X}^2= 0$ we find that $b\bar{X} \in I$ and if $b \neq 0$ we find that $X \in I$ and thus also $aX \in I$ and most certainly the subtraction of these elements $b= aX + b - a\bar{X} \in I$, this of course means (since $F$ is a field and $b \neq 0$) that $1 \in I$. Since we supposed $I$ to be a maximal (and thus proper ideal) we find that there are only elements of the form $aX \in I$. Thus $I \subseteq (\bar{X})$. Now suppose that there is some element $a\bar{X}+b$ such that $(a\bar{X} + b)\bar{X} = 1$, then we find that $$ a \bar{X}^2 + b \bar{X} = 1 $$ Thus $$b\bar{X} = 1$$ This is of course impossible. So $(\bar{X})$ is a proper ideal of the quotientring and thus is the unique maximal ideal.

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