# Principal Ideal

• November 12th 2006, 07:15 AM
JaysFan31
Principal Ideal
Looking for help on the following problem:
Let R, a ring be Z[x] (polynomials with integer coefficients). Let (2,x) be the ideal generated by 2 and x.
1.) Prove that the ideal (2,x) is not principal and conclude that Z[x] is not a principal ideal ring.
2.) Find a homomorphism f: R to Z2 such that (2,x)=Ker(f).

Thank you.
• November 12th 2006, 07:39 AM
ThePerfectHacker
Quote:

Originally Posted by JaysFan31
Looking for help on the following problem:
Let R, a ring be Z[x] (polynomials with integer coefficients). Let (2,x) be the ideal generated by 2 and x.
1.) Prove that the ideal (2,x) is not principal and conclude that Z[x] is not a principal ideal ring.

(I think we should mention a commutative ring with unity first).

We need to show that,
$S=\{2 n+xm|n,m\in \mathbb{Z}[x]\}$ is not equal to some $, r\in \mathbb{Z}[x]$
Let us assume that it is,
$2\in S$ thus, $2=ar$ since 2 is prime and $a,r$ are polynomials we conclude they must be constant and that $a=1,r=2$, $a=2,r=1$.

Thus, if it is an ideal it must be either,
$S=<1>,S=<2>$
But, $<1>=\mathbb{Z}[x]$
And $S$ contains all polynomials with the constant term being even, thus, $S\not = <1>$

Thus, if it is an ideal it must be,
$S=<2>$
But, $<2>$ contains all the polynomials with even coefficients. While $S$ does contains some that are not, for example, $2(1)+x(x)=1x^2+2$. Thus, $<2>\not = S$
An impossibility.

(What we actually shown is that,
$<2>\subset S\subset <1>$)

Now, what ails me is that you say,
"conclude Z[x] is not a principal ideal ring"
That is not true, any commutative ring with unity always is a principal ideal ring. What you probabaly wanted to say is that, "conclude that (2,x) is not a principal ideal ring in Z[x]".

-----------
Quote:

2.) Find a homomorphism f: R to Z2 such that (2,x)=Ker(f).
That is impossible, the kernel of any ring homomorphism is an ideal!
$(2,x)$ is not an ideal in $R$.
• April 8th 2010, 09:01 PM
hamidr
Quote:

Originally Posted by ThePerfectHacker
(I think we should mention a commutative ring with unity first).

We need to show that,
$S=\{2 n+xm|n,m\in \mathbb{Z}[x]\}$ is not equal to some $, r\in \mathbb{Z}[x]$
Let us assume that it is,
$2\in S$ thus, $2=ar$ since 2 is prime and $a,r$ are polynomials we conclude they must be constant and that $a=1,r=2$, $a=2,r=1$.

Thus, if it is an ideal it must be either,
$S=<1>,S=<2>$
But, $<1>=\mathbb{Z}[x]$
And $S$ contains all polynomials with the constant term being even, thus, $S\not = <1>$

Thus, if it is an ideal it must be,
$S=<2>$
But, $<2>$ contains all the polynomials with even coefficients. While $S$ does contains some that are not, for example, $2(1)+x(x)=1x^2+2$. Thus, $<2>\not = S$
An impossibility.

(What we actually shown is that,
$<2>\subset S\subset <1>$)

Now, what ails me is that you say,
"conclude Z[x] is not a principal ideal ring"
That is not true, any commutative ring with unity always is a principal ideal ring. What you probabaly wanted to say is that, "conclude that (2,x) is not a principal ideal ring in Z[x]".

-----------

That is impossible, the kernel of any ring homomorphism is an ideal!
$(2,x)$ is not an ideal in $R$.

Wait a MIN, I thought a PID is of those ring that every ideal in it is of the form <ar> so if <2,x> is an Ideal in Z[X] shouldnt we conclude that z[X] is not PID?
• April 9th 2010, 01:26 PM
FancyMouse
Quote:

Originally Posted by ThePerfectHacker
That is impossible, the kernel of any ring homomorphism is an ideal!
$(2,x)$ is not an ideal in $R$.

This is totally wrong. (2,x) is DEFINED to be the ideal generated by {2,x}. It's just not a principal ideal.

The homomorphism is defined by f(x)|->c mod 2, where c is the constant term of f(x)
• April 9th 2010, 01:27 PM
FancyMouse
Quote:

Originally Posted by hamidr
Wait a MIN, I thought a PID is of those ring that every ideal in it is of the form <ar> so if <2,x> is an Ideal in Z[X] shouldnt we conclude that z[X] is not PID?

An ideal can be generated by two elements does not mean that it cannot be generated by one element. e.g. in ring Z, can you say the existence of (4,6) means that Z is not a PID? No! because (4,6)=(2).
• April 9th 2010, 01:35 PM
FancyMouse
Quote:

Originally Posted by ThePerfectHacker
(I think we should mention a commutative ring with unity first).

We need to show that,
$S=\{2 n+xm|n,m\in \mathbb{Z}[x]\}$ is not equal to some $, r\in \mathbb{Z}[x]$
Let us assume that it is,
$2\in S$ thus, $2=ar$ since 2 is prime and $a,r$ are polynomials we conclude they must be constant and that $a=1,r=2$, $a=2,r=1$.

Thus, if it is an ideal it must be either,
$S=<1>,S=<2>$
But, $<1>=\mathbb{Z}[x]$
And $S$ contains all polynomials with the constant term being even, thus, $S\not = <1>$

Thus, if it is an ideal it must be,
$S=<2>$
But, $<2>$ contains all the polynomials with even coefficients. While $S$ does contains some that are not, for example, $2(1)+x(x)=1x^2+2$. Thus, $<2>\not = S$
An impossibility.

(What we actually shown is that,
$<2>\subset S\subset <1>$)

Now, what ails me is that you say,
"conclude Z[x] is not a principal ideal ring"
That is not true, any commutative ring with unity always is a principal ideal ring. What you probabaly wanted to say is that, "conclude that (2,x) is not a principal ideal ring in Z[x]".

OK I feel that I have to read this proof carefully. The beginning part works. Assume S is principal, then it could only be (1) or (2). Clearly 1 is not in there, so S is not (1). Clearly x+2 is in S but not in (2), so S is also not (2). This itself gives a contradiction and finishes the proof of "Z[x] is not a PID".

The last comment does not make any sense. One really interesting thing is that, if what you said were true, then Lame would have proven Fermat's last theorem years earlier, since the ring of algebraic integers over Q(\zeta_n) is a commutative ring, and PID implies UFD.