Showing that $A$ is an Ideal of $\mathbb{Q}[x]$ and Expressing it as $<f(x)>$

[mathmari]

Hey!

I have to show that the set $$A=\{f(x) \in \mathbb{Q}[x], f(i)=0\}$$ is an ideal of the ring $\mathbb{Q}[x]$ and I have to express it in the form $<f(x)>$.

I have done the following:

To show that $A=\{f(x) \in \mathbb{Q}[x], f(i)=0\}$ is an ideal of the ring $\mathbb{Q}[x]$, we have to show that:

1. $\forall f(x) \in A, g(x) \in \mathbb{Q}[x], g(x) \cdot f(x) \in A $
2. $\forall f(x) \in A, g(x) \in \mathbb{Q}[x], f(x) \cdot g(x) \in A $
3. $\forall h(x), f(x) \in A, h(x)-f(x) \in A $1. $f(x) \in A$, so $f(x) \in \mathbb{Q}[x]$ and $f(i)=0 $
$g(x) \in \mathbb{Q}[x] $

Since $f(x) \in \mathbb{Q}[x]$ and $g(x) \in \mathbb{Q}[x]$ and since $\mathbb{Q}$ is a ring, it implies that $g(x) \cdot f(x) \in \mathbb{Q}[x]$.

$g(i) \cdot f(i)=g(i) \cdot 0=0 $

Therefore, $g(x) \cdot f(x) \in A$. 2. $f(x) \in A$, so $f(x) \in \mathbb{Q}[x]$ and $f(i)=0 $
$g(x) \in \mathbb{Q}[x] $

Since $g(x) \in \mathbb{Q}[x]$ and $f(x) \in \mathbb{Q}[x]$ and since $\mathbb{Q}$ is a ring, it implies that $f(x) \cdot g(x) \in \mathbb{Q}[x] $.

$f(i) \cdot g(i)=0 \cdot g(i)=0 $

Therefore, $f(x) \cdot g(x) \in A$. 3. $f(x) \in A$, so $f(x) \in \mathbb{Q}[x]$ and $f(i)=0 $
$h(x) \in A$, so $h(x) \in \mathbb{Q}[x]$ and $h(i)=0 $

Since $f(x) \in \mathbb{Q}[x]$ and $h(x) \in \mathbb{Q}[x]$ and since $\mathbb{Q}$ is a ring, it implies that $h(x) - f(x) \in \mathbb{Q}[x]$.

$h(i) -f(i)=0-0=0 $

Therefore, $h(x) -f(x) \in A$. Therefore, $A$ is an ideal.

Is this correct?? (Wondering)How can I write the ideal $A$ in the form $<f(x)>$?? (Wondering)

 

[Deveno]

What you've done so far is OK.

(2) is actually unnecessary, $\Bbb Q[x]$ is a commutative ring, so (1) is all you need.

Personally, I would use another letter besides "$f$" to indicate the generator of $A$, since you are taking $f(x)$ to be a "general" polynomial, and for the generator of $A$ you want a "specific" one.

For of all, you need to know/show that $\Bbb Q[x]$ is a PRINCIPAL IDEAL DOMAIN. The easiest way to do this is to use the fact that any Euclidean domain is a principal ideal domain.

The Euclidean "$d$" function for $\Bbb Q[x]$ is the degree of a polynomial. This is because, for ANY two non-zero polynomials:

$f(x),g(x) \in \Bbb Q[x]$, we can write:

$f(x) = g(x)q(x) + r(x)$ for UNIQUE polynomials $q(x),r(x)$ with $r(x) = 0$ (the 0-polynomial, not the number) or $\text{deg}(r(x)) < \text{deg}(g(x))$.

It is unclear what you mean by "$i$". If you mean a root of $x^2 + 1$, then it is clear that $\langle x^2 + 1\rangle \subseteq A$.

On the other hand, suppose that:

$p(x) \in A$. Write:

$p(x) = (x^2+1)q(x) + r(x)$.

Since $p(i) = 0$ (because it's in $A$), we have:

$0 = 0q(i) + r(i) = r(i)$.

If $r(x) \neq 0$, then $r(x)$ has degree 1, so $r(x) = ax + b$ for some rational numbers $a,b$, with $a \neq 0$.

Since $r(i) = 0$, it must be $i$ is a root of $r(x) = ax + b$.

Thus $ai + b = 0 \implies i = -\dfrac{b}{a}$. But this is absurd, since:

$\left(-\dfrac{b}{a}\right)^2 + 1 = \dfrac{b^2}{a^2} + 1 = \dfrac{b^2 + a^2}{a^2} > 0$ (since $a \neq 0$ -this uses the ORDER properties of $\Bbb Q$).

Our only other option is that $r(x) = 0$, in which case $p(x) = (x^2 + 1)q(x) \in \langle x^2 + 1\rangle$.

This shows $A \subseteq \langle x^2 + 1\rangle$, so the two are equal.

 

[mathmari]

For the first part, do you mean that I don`t have to write (2) at all, or do I have to include it into (1)??

 

[Deveno]

(2) follows from (1) by commutativity.

In a commutative ring: left-ideal = right-ideal = two-sided ideal.

$\Bbb Q[x]$ is a commutative ring, for all polynomials $f(x),g(x) \in \Bbb Q[x]$:

$f(x)g(x) = g(x)f(x)$.

 

[mathmari]

What you've done so far is OK.

(2) is actually unnecessary, $\Bbb Q[x]$ is a commutative ring, so (1) is all you need.

Personally, I would use another letter besides "$f$" to indicate the generator of $A$, since you are taking $f(x)$ to be a "general" polynomial, and for the generator of $A$ you want a "specific" one.

For of all, you need to know/show that $\Bbb Q[x]$ is a PRINCIPAL IDEAL DOMAIN. The easiest way to do this is to use the fact that any Euclidean domain is a principal ideal domain.

The Euclidean "$d$" function for $\Bbb Q[x]$ is the degree of a polynomial. This is because, for ANY two non-zero polynomials:

$f(x),g(x) \in \Bbb Q[x]$, we can write:

$f(x) = g(x)q(x) + r(x)$ for UNIQUE polynomials $q(x),r(x)$ with $r(x) = 0$ (the 0-polynomial, not the number) or $\text{deg}(r(x)) < \text{deg}(g(x))$.

It is unclear what you mean by "$i$". If you mean a root of $x^2 + 1$, then it is clear that $\langle x^2 + 1\rangle \subseteq A$.

On the other hand, suppose that:

$p(x) \in A$. Write:

$p(x) = (x^2+1)q(x) + r(x)$.

Since $p(i) = 0$ (because it's in $A$), we have:

$0 = 0q(i) + r(i) = r(i)$.

If $r(x) \neq 0$, then $r(x)$ has degree 1, so $r(x) = ax + b$ for some rational numbers $a,b$, with $a \neq 0$.

Since $r(i) = 0$, it must be $i$ is a root of $r(x) = ax + b$.

Thus $ai + b = 0 \implies i = -\dfrac{b}{a}$. But this is absurd, since:

$\left(-\dfrac{b}{a}\right)^2 + 1 = \dfrac{b^2}{a^2} + 1 = \dfrac{b^2 + a^2}{a^2} > 0$ (since $a \neq 0$ -this uses the ORDER properties of $\Bbb Q$).

Our only other option is that $r(x) = 0$, in which case $p(x) = (x^2 + 1)q(x) \in \langle x^2 + 1\rangle$.

This shows $A \subseteq \langle x^2 + 1\rangle$, so the two are equal.

(2) follows from (1) by commutativity.

In a commutative ring: left-ideal = right-ideal = two-sided ideal.

$\Bbb Q[x]$ is a commutative ring, for all polynomials $f(x),g(x) \in \Bbb Q[x]$:

$f(x)g(x) = g(x)f(x)$.

I understand! (Happy) Thank you very much! (Smile)