I need answers with details plz. Problern 4.1. Consider 1-2z E Zlx] C zijz]] and

Question

I need answers with details plz.

Problern 4.1. Consider 1-2z E Zlx] C zijz]] and 1-2r E Z1r], where Z-{0. T, 2, 3) (1) Is 1 - 2x invertible in Zx] Explain why or why not. (2) True or false: 1-2c E U(Z[) with (1-2r)-'-1 + 2 4x2 + 81+ . . . E Zl (3) Is I - 2r invertible in Z4 r? If so, find its inverse explicitly. Hint. (1) As Z is an integral domain, we have deg(fg) = deg(+deg(g) tor all non-zero Hint. (1) As Z is an integral domain, we have deg(fo) degf)+ deglg) for all noa-zero polynomials f, g Zr (3) Note that (2)2 0 Z4 and 2 2EZ

Explanation / Answer

(1) Z[x] is the ring of all polynomials with integral coefficients.

Claim: 1-2x is not invertible in Z[x]

Proof

Suppose 1-2x is invertible with inverse f(x).

Thus (1-2x) f(x) =1. Clearly f(x) cannot be an integer. Let its degree be n

The degree of (1-2x) is 1.

So the LHS product has degree n+1, while the RHS has degree 0.

This contradiction proves that (1-2x) is not a unit (not invertible in Z[x]).

Another method

Consider this as an equality in Q[x] and set x =1/2 on both sides

LHS =0 and RHS is 1. We again get a contradiction.

(2) TRUE

Z[[x]] is the ring of formal power series in x with integral coefficients.

Applying the formal Binomal Theorem (or use Geometric Series with leading term 1 and common ratio 2x)

we obtain the power series for (1-2x) -1 as given in the problem.

(3) In Z4 [x], (1+2x) (1-2x) = 1-2x+2x+4x2 = 1 (as 4 is zero in Z4).

Thus (1-2x) is invertible in Z4[x] and the inverse is (1+2x)

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