Proof that there are No Modulo Invertible Polynomials with f(1) = 0

Asked Jun 07 '16 at 13:45

Active Jun 07 '16 at 15:13

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I was reading an article from the NTRU Cryptosystem (probably the first one): NTRU: A Ring-Based Public Key Cryptosystem

And I don't know how to prove the assertion he makes in parenthesis in paragraph 2.2: that a polynomial satisfying f(1) = 0 can never be invertible.

Does anyone have a proof or a link to one?

Thanks.

edited Jun 07 '16 at 15:13

asked Jun 07 '16 at 13:45

psygo

That should be wrong, consider $f(x)=1-x$. – lattice Jun 07 '16 at 13:50
What does invertible mean in this context? Of course, there is no $g(x)$ for which $f(x)g(x)=1$. – Ben Grossmann Jun 07 '16 at 14:03
Well, in the context of the article... It's modular invertibility basically. – psygo Jun 07 '16 at 15:15
The operation here is the product $(\sum F_i X^i)\circledast (\sum G_i X^i) = \sum H_i X^i$ with $H_k = \sum_{i + j\equiv k\pmod{N}} F_i G_j$. – anomaly Jun 07 '16 at 15:23

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