find the maximal ideal of the ring ?.

Question

find the maximal ideal of the ring $$ \frac{\mathbb{R}[x]}{ (x^2)} $$

My attempt :here the only proper ideal containing $(x^2)$ are $(x)$ and $(x^2)$, so , we have two maximal ideal that is $(x)$ and $(x^2)$

Is its correct ?

any hints/solution will be appreciated

Answer 1

Let $\varphi:\mathbb{R}[x]\rightarrow R:=\mathbb{R}[x]/(x^2)$ be the canonical homomorphism. There is a bijective correspondence between the ideals of $\mathbb{R}[x]$ that contain the ideal $(x^2)$ and the ideals of R, given by $I \mapsto \varphi (I)$. The only ideals of $\mathbb{R}[x]$ that contains $(x^2)$ are $\mathbb {R}[x]$, $(x)$ and $(x^2)$, so $R$ has only three ideals: $R$, $(\bar x)$ and $(\bar x^2)$. Clearly $(0)=(\bar x^2)\subsetneq (\bar x)\subsetneq R$, which makes it clear that $(\bar x)$ is the only maximal ideal of $R$.

Answer 2

There is a bijective correspondence between the ideals of $x^2$ in $\mathbb{R}[x]$ and in $\dfrac{\mathbb{R}}{x^2}$ by using Lattice Isomorphism Theorem. A maximal ideal in $\dfrac{\mathbb{R}}{x^2}$ corresponds to a maximal ideal in $\mathbb{R}[x]$ that contains $(x^2)$. So, find such ideals. $\mathbb{R}$ has three ideals which are $\mathbb{R},(x),(x^2)$. Clearly, $(x)$ is the maximal ideal.

Answer 3

Hint: The maximal ideal in $\frac{\Bbb R[x]}{(x^2)}$ is clearly $(x)$,because it is the only one containing $(x^2)$ as a proper subset.

Towards a proof, note that $(x^2)$ contains all polynomials without an $x$ or a constant term. Next $(x)$ consists in those polynomials in $x$ with no constant term. It's pretty clear this is as far as it goes, because if there's a polynomial with a nonzero constant term, we can subtract the same polynomial minus that constant (which is also in the ideal by a simple argument, namely that it contains $x$) thus getting a nonzero constant (i.e. a unit).

Let me clarify the last part by an example: so, why wouldn't $(x-6)$ be maximal? Well, multiply $x-6$ by $x$. Get $x^2-6x\equiv -6x$. Then we get $x$, then $6$, then $1$.

Answer 4

Let $P$ be a maximal ideal of $ \frac{\mathbb{R}[x]}{ (x^2)} $. Hence it is a prime ideal and so there exists a prime ideal $Q$ of $ \mathbb{R}[x]$ such that $P=Q/(x^2)$. Thus, $x^2\in Q$. Hence, $x\in Q$. This shows that every maximal ideal of $ \frac{\mathbb{R}[x]}{ (x^2)} $ is contained in $(x)/(x^2)$. Therefore, $(x)/(x^2)$ is the only maximal ideal of $ \frac{\mathbb{R}[x]}{ (x^2)} $.