U(Z[x]) ?


by 1800bigk
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1800bigk
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#1
Oct25-05, 01:20 PM
P: 42
Hi, I have to find the units of Z[x]. I am little unclear and my book does not go into detail. Does U(Z[x]) mean I need to find every polynomial with integer cooefficients that has a multiplicative inverse? or do I have to find the multiplicative identity? I was thinking about both cases and the number of polynomials with a multiplicative inverse is pretty limited, isn't it? f(x)=1 or f(x)=-1. As for inverses of polynomials, there would be none because if you multiply a polynomial with x by another polynomial with x then the powers of x get bigger.
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Hurkyl
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#2
Oct25-05, 04:17 PM
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Hi, I have to find the units of Z[x]. I am little unclear and my book does not go into detail. Does U(Z[x]) mean I need to find every polynomial with integer cooefficients that has a multiplicative inverse?
It means to find all elements of Z[x] with a multiplicative inverse. (That's what it means to be a unit)

As for inverses of polynomials, there would be none because if you multiply a polynomial with x by another polynomial with x then the powers of x get bigger.
That sounds plausible... can you work it into a rigorous proof?
1800bigk
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#3
Oct25-05, 04:41 PM
P: 42
am I right about f(x)=1 and f(x)= -1 being the only polynomial in U(Z[x]) or is there more? would contradiction be the best way, how would I start it?

tia

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#4
Oct25-05, 04:46 PM
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U(Z[x]) ?


am I right about f(x)=1 and f(x)= -1 being the only polynomial in U(Z[x])
I generally hate answering this question. Learning how and when to be confident in your own work is important! But yes, you are correct.

would contradiction be the best way, how would I start it?
There are lots of ways. I would suggest starting with a literal translation of what you said:

If p(x) and q(x) are nonconstant, then p(x)*q(x) is nonconstant.

(Actually, you made a stronger statement, but I don't want to spoil figuring out how to translate that!)

And see if you can prove this statement and relate it to inverses.
1800bigk
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#5
Oct25-05, 04:51 PM
P: 42
ok thanks!


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