1

Finding the number of ideals in the quotient ring $\mathbb R[x]/\langle x^2-3x+2 \rangle$.

Attempt: $R[x]/\langle x^2-3x+2 \rangle = \{f(x)+\langle x^2-3x+2 \rangle~~|~~f(x) \in R[x]\}$.

Since $(x^2-3x+2)=(x-1)(x-2)$ is a reducible polynomials, the members of $R[x]/\langle x^2-3x+2 \rangle) = f(1)$ or $f(2) $ are essentially real constants.

out of which only $\{0\}$ is an ideal as $a \cdot 0 = 0 \in \{0\}~~\forall~~a \in \mathbb R$

Is my attempt correct?

Thank you for your help.

MathMan
  • 8,974
  • 7
  • 70
  • 135
  • 1
    In line with Rene's answer: Can you show that $f(x)+I\mapsto (f(1),f(2))$ is an isomorphism of rings from $R[x]/I$ to $R\oplus R$? The latter is a ring with componentwise addition and multiplication as operations, so its multiplicative neutral element is $(1,1)$ et cetera. – Jyrki Lahtonen Aug 10 '14 at 17:31

3 Answers3

5

Since the polynomial factors you have

$$\mathbb{R}[x]/(x^2-3x+2)\cong \mathbb{R}[x]/(x-2)\oplus \mathbb{R}[x]/(x-1) \cong \mathbb{R}\oplus \mathbb{R}$$ So what are the ideals of $\mathbb{R}\oplus \mathbb{R}$ ?

Another way of thinking about it is you have a (reducible) variety defined by $x^2-3x+2=0$, this variety has two points.

Rene Schipperus
  • 39,526
1

I didn't understand the argument, but I'm afraid the answer is wrong. The polynomial $x-1$ does not have any inverse in this quotient, so $(x-1)$ is another ideal, for example.

By the way, another way to approach this problem is by the Chinese remainder theorem. Since $(x^2-3x+2)=(x-1)\cdot(x-2)=(x-1)\cap(x-2)$, and $(x-1),(x-2)$ are co-prime, we have

$$\mathbb{R}[x]/(x^2-3x+2)\simeq\mathbb{R}[x]/(x-1)\times\mathbb{R}[x]/(x-2)\simeq\mathbb{R}\times\mathbb{R}.$$

That should simplify the task of finding all ideals.

Amitai Yuval
  • 19,308
  • I am sorry I am a little confused. Do you mean $(x-1)$ or $\langle (x-1) \rangle$ – MathMan Aug 10 '14 at 17:20
  • 1
    Also, $I/A$ is an ideal of $R/A$ if $I$ is an ideal of $R$. – MathMan Aug 10 '14 at 17:21
  • 1
    Almost. If $I$ is an ideal of $R$ that contains $A$, then $I/A$ is an ideal of $R/A$. And this is the point, I guess. In the ring $\mathbb{R}[x]$ you have $(x^2-3x+2)\subset (x-1)$, meaning that the ideal $(x-1)$ becomes a nontrivial ideal in the quotient. – Amitai Yuval Aug 10 '14 at 17:25
1

I'm not sure what you mean by "$\mathbb R [x]/\langle x^2-3x +2 \rangle = f(1)$ or $f(2)$ are essentially real constants." Note that there are nonzero ideals in the quotient:

Hint. For any ring $R$ and any ideal $I$, there is a bijection between ideals of $R$ containing $I$ and ideals of $R/I$.

vociferous_rutabaga
  • 4,591
  • So, $I/A$ is an ideal of $R/A$ if $I$ is an ideal of $R$... So, All i need to find are ideals of $R$? – MathMan Aug 10 '14 at 17:22
  • 1
    In general, you cannot consider $I/A$ unless $A$ is contained in $I$. However, the correspondence $I \mapsto I/A$ gives a bijection between ideals of $R/A$ and ideals of $R$ which contain $A$. So what you need to do is find all ideals of $R$ which contain $A$. – vociferous_rutabaga Aug 10 '14 at 17:24