Ideal of the ring of polynomials with even integer coefficients

**matt.qmar** · February 10th 2011, 11:47 AM

Hello!

If we take $Z[x]$ to be the ring of polynomials with integer coefficients, and $I$ to be the subset $I = \{a_nx^n + ... + a_0 \in Z[x] : a_i \in 2Z\}$ .
So I is the ring of polynomials with even integer coefficients
So far, I have shown that I is an ideal.

I am trying to
(i) find & describe the number of cosets in $Z[x]/I$
(ii) Prove that $I$ is not a principle ideal of $Z[x]$
(iii) Decide if $I$ is a prime ideal.

Well, for (i), I am thinking there is infinitely many cosets? As long as any $x^i$ has coefficient 1, it forms it's own coset... for example
$x + I \neq x^3 + I \neq x^4 + x^3 + I$ etc... in terms of coset equality... That seems to be it to me...

(ii) I am not quite too sure how to go about this. If we let $I = <p(x)>$ for some $p(x) \in Z[x]$ and derive a contradiction (show we can't just take multiples of p?)

(iii) I think this will be easy once (i) is figured out... If $Z[x]/I$ is an integral domain, then $I$ is prime. If $Z[x]/I$ is not integral domain, then $I$ is not prime.

Any help appreciated! Thanks in advance!

**tonio** · February 10th 2011, 07:17 PM

Originally Posted by matt.qmar

Hello!

If we take $Z[x]$ to be the ring of polynomials with integer coefficients, and $I$ to be the subset $I = \{a_nx^n + ... + a_0 \in Z[x] : a_i \in 2Z\}$ .
So I is the ring of polynomials with even integer coefficients
So far, I have shown that I is an ideal.

I am trying to
(i) find & describe the number of cosets in $Z[x]/I$
(ii) Prove that $I$ is not a principle ideal of $Z[x]$
(iii) Decide if $I$ is a prime ideal.

Well, for (i), I am thinking there is infinitely many cosets? As long as any $x^i$ has coefficient 1, it forms it's own coset... for example
$x + I \neq x^3 + I \neq x^4 + x^3 + I$ etc... in terms of coset equality... That seems to be it to me...

$x+I=x^3+I\Longleftrightarrow x-x^3\in I$ , which is true...Note that an element

in the quotient ring is uniquely determined by its free coefficient's parity...

(ii) I am not quite too sure how to go about this. If we let $I = <p(x)>$ for some $p(x) \in Z[x]$ and derive a contradiction (show we can't just take multiples of p?)

First show that $I=\langle 2,x\rangle$ , and then suppose $I=\langle f(x)\rangle$ . Check that this

is impossible by noting that both $2,\,x\in I$ and checking the possible degree of $f(x)$

(iii) I think this will be easy once (i) is figured out... If $Z[x]/I$ is an integral domain, then $I$ is prime. If $Z[x]/I$ is not integral domain, then $I$ is not prime.

Indeed, this follows at once from (i)

Tonio

Any help appreciated! Thanks in advance!

.

**matt.qmar** · February 10th 2011, 09:15 PM

Thanks alot!

I am just wondering why we can say:

$x+I=x^3+I\Longleftrightarrow x-x^3\in I$
if I is the ring of integers with even coefficents, how did $x-x^3$ end up in there?

or specifically, why can we say:

$x+I=x^3+I$

If that is true, then there must be only two cosets: 0 + I and 1 + I. Is this true?

**FernandoRevilla** · February 11th 2011, 12:21 AM

Originally Posted by matt.qmar

I am just wondering why we can say: $x+I=x^3+I\Longleftrightarrow x-x^3\in I$

It is the definition of equality of elements in the quotient ring. So, the above "if and only if" is right which implies both sides are right or false, in this case, false.

if I is the ring of integers with even coefficents, how did $x-x^3$ end up in there?

The elements of $\mathbb{Z}[x]/I$ have the form $p(x)+I$ with $p(x)\in \mathbb{Z}[x]$ , not necessarily $p(x) \in I$

or specifically, why can we say: $x+I=x^3+I$

As we said, $x-x^3\not\in I$ so , $x+I\neq x^3+I$ .

Fernando Revilla

**tonio** · February 11th 2011, 04:16 AM

Originally Posted by matt.qmar

Thanks alot!

I am just wondering why we can say:

$x+I=x^3+I\Longleftrightarrow x-x^3\in I$
if I is the ring of integers with even coefficents, how did $x-x^3$ end up in there?

Because this polynomial's free coefficient (zero) is even...

or specifically, why can we say:

$x+I=x^3+I$

If that is true, then there must be only two cosets: 0 + I and 1 + I. Is this true?

Exactly

Tonio

**topspin1617** · February 11th 2011, 04:46 AM

I must me missing something...

Why are you saying that a polynomial is in $I$ if and only if its constant term is even? For example, the claim that $x\in I$ . I just don't see it... the coefficient of $x,1$ , is definitely NOT an even integer.

In fact, the ideal $I$ , as it is written in the original problem, IS principal. That should be clear:

$p(x)\in I\Leftrightarrow p(x)=\sum a_ix^i$ , each $a_i\in 2\mathbb{Z}$
$\Leftrightarrow p(x)=2q(x)$ , some $q(x)\in \mathbb{Z}[x]$
$\Leftrightarrow p(x)\in (2)$ .

Are you sure that you have written the original problem correctly?

**tonio** · February 11th 2011, 05:15 AM

Originally Posted by topspin1617

I must me missing something...

Why are you saying that a polynomial is in $I$ if and only if its constant term is even? For example, the claim that $x\in I$ . I just don't see it... the coefficient of $x,1$ , is definitely NOT an even integer.

In fact, the ideal $I$ , as it is written in the original problem, IS principal. That should be clear:

$p(x)\in I\Leftrightarrow p(x)=\sum a_ix^i$ , each $a_i\in 2\mathbb{Z}$
$\Leftrightarrow p(x)=2q(x)$ , some $q(x)\in \mathbb{Z}[x]$
$\Leftrightarrow p(x)\in (2)$ .

Are you sure that you have written the original problem correctly?

Oops, my bad! I misread the definition of the ideal I: I thought it was written $a_0=0\!\!\pmod 2$ and

not $a_i=0\!\!\pmod 2$ , as it actually is.

Anyway, what we said before applies for each coefficient, so that if

$f(x)=\sum\limits^n_{i=0}a_ix^i\,,\,g(x)=\sum\limit s^m_{j=0}b_jx^j$ , then

$f(x)+I=g(x)+I\Longleftrightarrow a_k=b_k\!\!\pmod 2 \mbox{ for }k=0,1,...,min(n,m)\,,\,\,$

$a_k\,(\,\,or\,\,b_k)\,=0\!\!\pmod 2 \mbox{ for } k=min(n,m),min(n,m)+1,...,max(n,m)$ .

So we have a factor $\mathbb{Z}/2\mathbb{Z}$ for each coefficient, and then

$\mathbb{Z}[x]/I\cong \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}\oplus\ldots$

Tonio

**matt.qmar** · February 11th 2011, 05:38 AM

Originally Posted by topspin1617

Are you sure that you have written the original problem correctly?

Yes, I was referring to polynomials with even integer coefficients, not the polynomials which have f(0) = 2, although I know someone else had asked that question recently on the forum, so that is why I was confused that $x^3 - x \in I$

So $I =\ <2>$ and, I had also worked out that $Z[x]/I$ is the direct sum of infinitely many $Z_2$ , which means also that I is not prime, since a direct sum of rings is never an integral domain, as I recall? ie) (1,0)*(0,1) = (0,0) so we have zero divisors in $Z_2 \oplus Z_2$

Thanks for all the replies everyone.

**topspin1617** · February 11th 2011, 02:41 PM

Originally Posted by tonio

$\mathbb{Z}[x]/I\cong \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}\oplus\ldots$

Tonio

I agree.

Originally Posted by matt.qmar

Yes, I was referring to polynomials with even integer coefficients, not the polynomials which have f(0) = 2, although I know someone else had asked that question recently on the forum, so that is why I was confused that $x^3 - x \in I$

So $I =\ <2>$ and, I had also worked out that $Z[x]/I$ is the direct sum of infinitely many $Z_2$ , which means also that I is not prime, since a direct sum of rings is never an integral domain, as I recall? ie) (1,0)*(0,1) = (0,0) so we have zero divisors in $Z_2 \oplus Z_2$

Thanks for all the replies everyone.

Alright... but I'm still confused, as in your part (b) in the original question, it says to prove that $I$ is NOT a principal ideal.

**tonio** · February 12th 2011, 01:19 AM

Originally Posted by topspin1617

I agree.

Alright... but I'm still confused, as in your part (b) in the original question, it says to prove that $I$ is NOT a principal ideal.

Ah, indeed! This part also contributed, I think, to my confusion, as the ideal of all the polynomials with even free

coefficient is not a principal one, but the one we've been dealing with is...

I wonder whether the question is correctly stated.

Tonio

**topspin1617** · February 12th 2011, 09:50 AM

Originally Posted by tonio

Ah, indeed! This part also contributed, I think, to my confusion, as the ideal of all the polynomials with even free

coefficient is not a principal one, but the one we've been dealing with is...

I wonder whether the question is correctly stated.

Tonio

Yeah, I was pretty sure that was part of the whole confusion; if the problem were stated as you just said, of course everything else you said would have been exactly right.

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