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$f(x)= a_0+a_1x+...+a_{n-1}x^{n-1} = \sum_{i=0}^{n}a_ix^i$

Verify $Z_7[x]$ satisfies the ring axiom of multiplicative closure:

if $f(x)= \sum_{i=0}^{n}a_ix^i ∈ Z_7[x]$ and $g(x)= \sum_{j=0}^{m}b_jx^j ∈ Z_7[x]$, then $f.g ∈ Z_7[x]$.

I don't know where to even start in going about this and I would really appreciate any help!

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  • $\begingroup$ How about you just go ahead and multiply them? But first of all, fix your notation and naming: two different polynomials shouldn't be given the same name $f(x)$. And they don't have to be of the same degree, so the summation should go up to $n$ in one of them, but up to some other letter (say, $m$) in the other one. $\endgroup$ – zipirovich Jan 18 '17 at 18:53
  • $\begingroup$ If i multiply them I get $(g.h)(x)= \sum_{i=0}^{n+m}(\sum a_ib_j)x^k$ (hopefully that all came out correctly, but I don't know how to turn that into verification. $\endgroup$ – Mike A Jan 18 '17 at 19:01
  • $\begingroup$ Again, some of indexing is messed up, so you need to fix that. Other than that, this is pretty much it. In $\sum_{?=0}^{n+m}(\sum_{??} a_ib_j)x^k$, each $\sum_{??} a_ib_j$ is a new coefficient, which is an element of $\mathbb{Z}_7$, since addition and multiplication are well-defined in it. And since this has the form of a polynomial, it is a polynomial. $\endgroup$ – zipirovich Jan 18 '17 at 19:20
  • $\begingroup$ Huh I thought there would be more to it. I suppose verification is a lot less stringent than a proof. I'll get the notation sorted, thanks for the help. $\endgroup$ – Mike A Jan 18 '17 at 19:24

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