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- Jun 22, 2012

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Let [TEX] \mathbb{Q}_\mathbb{Z}[x][/TEX] denote the set of polynomials with rational coefficients and integer constant terms.

Prove that the only two units in [TEX] \mathbb{Q}_\mathbb{Z}[x][/TEX] are 1 and -1.

Help with this exercise would be appreciated.

My initial thoughts on this exercise are as follows:

1 and -1 are the units of \(\displaystyle \mathbb{Z} \). Further the constant terms of the polynomials are from \(\displaystyle \mathbb{Z} \) and so I suspect the units of [TEX] \mathbb{Q}_\mathbb{Z}[x][/TEX] are thus 1 and -1 - but this is not a rigorous proof - indeed it is extremely vague!

Can someone help with a rigorous formulation of these thoughts into a formal proof.

I suspect that such a proof would start as follows:

Units of [TEX] \mathbb{Q}_\mathbb{Z}[x][/TEX] would be those p(x) and q(x) such that

p(x)q(x) = 1

can you just assert now that the only possible polynomials in [TEX] \mathbb{Q}_\mathbb{Z}[x][/TEX] would be 1 and -1 - what reason would you give - is it obvious???

Hope someone can clarify.

Peter

[This exercise is also posted on MHF]

Prove that the only two units in [TEX] \mathbb{Q}_\mathbb{Z}[x][/TEX] are 1 and -1.

Help with this exercise would be appreciated.

My initial thoughts on this exercise are as follows:

1 and -1 are the units of \(\displaystyle \mathbb{Z} \). Further the constant terms of the polynomials are from \(\displaystyle \mathbb{Z} \) and so I suspect the units of [TEX] \mathbb{Q}_\mathbb{Z}[x][/TEX] are thus 1 and -1 - but this is not a rigorous proof - indeed it is extremely vague!

Can someone help with a rigorous formulation of these thoughts into a formal proof.

I suspect that such a proof would start as follows:

Units of [TEX] \mathbb{Q}_\mathbb{Z}[x][/TEX] would be those p(x) and q(x) such that

p(x)q(x) = 1

can you just assert now that the only possible polynomials in [TEX] \mathbb{Q}_\mathbb{Z}[x][/TEX] would be 1 and -1 - what reason would you give - is it obvious???

Hope someone can clarify.

Peter

[This exercise is also posted on MHF]

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