# Maximal Ideals in polynomial quotient rings

I am having trouble unpacking this problem involving finding maximal ideals.

It would be great to gain some insight into how maximal ideals work in polynomial quotients.

The example I'm trying to understand is $\mathbb{R}[x]/(x^2)$.

The way I went about understanding this problem was to first find out what elements in the quotient looked like:

Elements in $\mathbb{R}[x]$ look like $$a_nx^n + \cdots + a_2 x^2 + a_1 x + a_0$$ for $a_0,...,a_n \in \mathbb{R}$

Elements in $\mathbb{R}[x]/(x^2)$ are cosets of the ideal $I = (x^2)$ which are of the form $$f(x)+I$$ and $$f(x)*I$$ for $f(x) \in\mathbb{R}[x]$.

This is where I get lost and I'm not sure how to proceed, perhaps there are some theorems to do with $(x^2)$ being principal ideal generated by a monic polynomial. Am I approaching this problem in the right way?

• What is your question? Are you trying to find all maximal ideals of $\mathbb{R}(x)/(x^2)?$ Oct 16, 2016 at 22:46
• Thomas, yes, I am trying to find all maximal ideals of $\mathbb{R}[x]/(x^2)$.
– jcm
Oct 16, 2016 at 22:48
• The cosets are of the form $p+I$ for some $p\in\Bbb R[x]$ and $I=(x^2)$ now. The product $p\cdot I$ is a part of $I$ itself (the coset of $0$), and not of $p+I$. Oct 16, 2016 at 23:51

Hints:

1) You can think of the quotient ring $A = \Bbb{R}[x]/(x^2)$ as being like the polynomial ring $\Bbb{R}[x]$ but equipped with a new algebraic law saying that $x^2 = 0$. So every element of $A$ can be written uniquely in the form $a + bx$ for $a, b \in \Bbb{R}$ (and where I'm using $x$ by abuse of notation for the coset $x + (x^2)$). You add in $A$ just as for polynomials and you multiply using the rule $(a + bx)(c + dx) = ac + (ad + bc)x$.

2) An ideal $M$ in a ring $R$ is maximal iff the quotient ring $K = R/M$ is a field. In a field, $t^2 = 0$ implies $t = 0$, so an element like $x \in A$ with $x^2 = 0$ must map to $0$ in $K$, i.e., it must belong to $M$. This leaves you with only one possibility for $M$ in the ring $A$.

• So from that I think I can conclude that the only maximal ideal is (x)? How do we check that there's no other maximal ideals?
– jcm
Oct 17, 2016 at 2:08
• The ideal must contain $(x)$ and as $A$ is a 2-dimensional vector space over $\Bbb{R}$ it can't contain anything else. Oct 17, 2016 at 2:16

There is an inclusion-preserving bijective correspondence between the set of subrings $A$ (containing $I$) of any ring $R$ and the $set$ of subrings $A/I$ in $R/I.$ This is known as the Lattice Isomorphism Theorem.

Since $\mathbb{R}$ is a field, $\mathbb{R}[x]$ is a Principal Ideal Domain. $(0)$ and $\mathbb{R}$ are the only ideals of $\mathbb{R}$ (as $\mathbb{R}$ is a field). The remaining ideals of $\mathbb{R}[x]$ can be generated by the irreducible polynomials of $\mathbb{R}[x].$ Since we are concerned with $\mathbb{R}[x]/(x^2),$ whose structure is isomorphic to $\phi(\mathbb{R}[x])$ with $\phi()$ the function such that $\phi(x^2q(x) + ax + b) = ax + b$ for any polynomial $p(x)=x^2q(x)+ax+b \in \mathbb{R}[x]$ (by the First Isomorphism Theorem), the only ideals to focus upon are, therefore, the ideals $(ax+b).$

But notice in $\mathbb{R}[x]/(x^2),$ for nonzero $b,$ we have $$(ax+b)(ax-b) = a^2x^2-b^2 = -b^2$$ so that the nonzero $-b^2 \in (ax+b)$ in $\mathbb{R}[x]/(x^2).$ Hence $\frac{-1}{b^2}\cdot(-b^2) = 1 \in (ax+b)$ so that $(ax+b)=(1)=\mathbb{R}[x]/(x^2)$ itself. Hence the only ideals of $\mathbb{R}[x]/(x^2)$ are $(0),(x),$ and $\mathbb{R}[x]/(x^2).$ In line with the first paragraph, you can also note that $(x^2) \not \subseteq (x-b)$ for nonzero $b.$

Our only possible choice for a maximal ideal of $\mathbb{R}[x]/(x^2)$ is, therefore, $(x).$ Explain why $\frac{\left(\mathbb{R}[x]/(x^2)\right)}{(x)}$ is maximal (what do you have left over if you treat $x=0$?), and you're all done.

• I follow your logic all the way until the end when you say x=0. you would just have the zero ring if x=0, right?
– jcm
Oct 17, 2016 at 2:39
• $\frac{\mathbb{R}[x]/(x^2)}{(x)}$ would leave you with a field isomorphic to $\mathbb{R}.$ Oct 17, 2016 at 3:10
• Right! since if you adjoin 0, you still have $\mathbb{R}$ since $0 \in \mathb{R}$. thanks
– jcm
Oct 17, 2016 at 4:25