Algebra 1 Assignment 5 Solutions

Problem 1 We preform long division:

\begin{matrix}3x^{4} − 2x^{3} + 6x^{2} − x + 2 \\ −(3x^{4} + 3x^{3} + 3x^{2}) \\ −5x^{3} + 3x^{2} − x + 2 \\ −(−5x^{3} − 5x^{2} − 5x) \\ 8x^{2} + 4x + 2 \\ −(8x^{2} + 8x + 8) \\ −4x − 6 \end{matrix} \underline{x^{2} + x + 1} ; 3x^{2} − 5x + 8

And we find that q(x) = x^{2} + x + 1, r(x) = −4x − 6.

Problem 2 Same thing, only recall that in ℤ_{2}[x], a = −a and 2 ∗ a = a + a = 0. We find q(x) = x^{3} + x^{2} + 1, r(x) = 0.

Problem 3 Let f : ℤ[x] → ℤ be the map that sends a ploynomial p(x) = a_{0} + a_{1}x + … a_{m}x^{m} to f(p) = a_{0}. To show: f is a homomorphism. Proof:

• The unit and zero elements in ℤ[x] are respectively 1 and 0. f(1) = 1 and f(0) = 0.

• Let q(x) = b_{0}+b_{1}x+… and p(x) = a_{0}+a_{1}x+. . .. Then p(x)+q(x) = (b_{0}+ a_{0})+(b_{1} +a_{1})x+(b_{2} +p_{2})x^{2} +. . .. We get f(p+q) = a_{0} +b_{0} = f(p)+f(q).

• Let p(x) and q(x) be as above. We have that p(x)q(x) = (a_{0}b_{2}) + (a_{0}b_{1} + b_{0}a_{1})x + . . .. It follows that f(pq) = a_{0}b_{0} = f(p)f(q).

So f is a homomorphism.□

Problem 4 We perform the Euclidian algorithm (notice that at each step I can replace the remainders by their monic representative):

\begin{aligned}x^{4} − x^{3} − x^{2} + 1 & = & (x^{3} − 1)(x − 1) & +(−x^{2} + x) \\ x^{3} − 1 & = & (x^{2} − x)(x + 1) & + x − 1 \\ x^{2} − x & = & (x − 1)x & +0 \end{aligned}

And we find gcd(x^{4} − x^{3} − x^{2} + 1, x^{3} − 1) = x − 1.

Problem 5 Again …

x^{4} + 3x^{2} − 2x + 4 = (x^{2} + 1)(x^{2} + 3x − 1) + \underbrace{−5x − 5}_{=0 in ℤ_{5}[x]}

So in fact x^{2} + 1 divides x^{4} + 3x^{2} − 2x + 4 so the gcd is x^{2} + 1.

Problem 6 Let x, y ∈ ℂ we define

M(x, y) = ( \begin{matrix}x & y \\ −ȳ & x̄ \end{matrix} )

We want to show that the set H = {M(z_{1}, z_{2}) ∈ M_{2}(ℂ)|z_{1}, z_{2} ∈ ℂ} is a subring of M_{2}(ℂ). To save work it is useful to have the following identities for complex conjugation: for all α, β ∈ ℂ:

\begin{matrix}\overline{αβ} = ᾱβ̄ & (1) \\ \overline{α + β} = ᾱ + β̄ & (2) \\ ¯ᾱ = α & (3) \\ αβ̄ = \overline{ᾱβ} & (4) \end{matrix}

Proving (1), (2), (3) is straightforward (write α = a + bi, β = c + di and expand both sides). And (4) is a consequence of (1) and (3). Moreover these identities almost immediately show that the map f : ℂ → ℂ where f(x) = x̄ is a an isomorphism, the fact that f is bijective follows from (3).

Let’s now check that H is a subring of R = M_{2}(ℂ) (this is only a sketch of a proof, but all the steps should be straightforward, using the identities you shouldn’t have to write out stuff like z_{1} = a_{1} + b_{1}i, z_{2} = …):

• M(1, 1) ∈ H is the identity element in R.

• M(0, 0) = 0_{R}

• M(z_{1}, z_{2}) + M(w_{1}, w_{2}) = M(z_{1} + w_{1}, z_{2} + w_{2}) (use identity (2)). So H is closed under addition.

• M(z_{1}, z_{2}) + M(−z_{1}, −z_{2}) = 0_{R} (again, by identity (2)). So every element in H has its additive inverse in H.

•

M(a, b) ∗ M(c, d) = ( \begin{matrix}ac − bd̄ & ad + bc̄ \\ −b̄c − d̄ā & −b̄d + āc̄ \end{matrix} ) = A

Using identities (2), (3) and (4) on the lower entries gives

A = M(ac − bd̄, ad + bc̄) ∈ H

So H is closed under multiplication.

So H is a subring of R. I’ll also point out right now that in general using the result from the last item M(c, d) ∗ M(a, b) = M(ca − db̄, cb − dā) ≠ M(a, b) ∗ M(c, d). Because in general ca−db̄ ≠ ac−bd̄. Therefore H is not a commutative ring.

Problem 7 We must show that every nonzero element in H has a multiplicative inverse. That is, using notation from the previous excercise, that for any matrix M(a, b) ∈ H where a, b ≠ 0, there is a matrix M(x, y) ∈ H such that:

( \begin{matrix}a & b \\ −b̄ & ā \end{matrix} ) ( \begin{matrix}x & y \\ −ȳ & x̄ \end{matrix} ) = ( \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} )

This means that the equations for the matrix coefficients have a solution i.e. we can find x, y such that:

\begin{aligned}e_{1}(x, y) = & ax − bȳ & = & 1 \\ e_{2}(x, y) = & ay + bx̄ & = & 0 \\ e_{3}(x, y) = & −b̄x − āȳ & = & 0 \\ e_{4}(x, y) = & −b̄y + āx̄ & = & 1 \end{aligned}

We now try to solve for x and y. Either a or b is not zero. I’ll only do the case where a ≠ 0 (the other case is the same). a ≠ 0 so ā ≠ 0 so we can divide e_{3}(x, y) by ā. We write:

e′_{3}(x, y) = \frac{b}{ā}e_{3}(x, y) = (−bb̄/ā)x − bȳ = 0

Computing e_{1}(x, y) − e′_{3}(x, y) gives us:

1 = ax − bȳ − \underbrace{((−bb̄/ā)x − bȳ)}_{=0} = \frac{aā − bb̄}{ā}x ⇒ x = \frac{ā}{aā − bb̄}

We can then substitute this value of x back into e_{3}(x, y) and easily solve for ȳ and then for y. So there exist solutions x = x_{0}, y = y_{0} for e_{1}(x, y) = 1 and e_{3}(x, y) = 0. And noting that by using our identities from the previous problem,

e_{2}(x_{0}, y_{0}) = −\overline{e_{3}(x_{0}, y_{0})} = −¯0 = 0

and

e_{4}(x_{0}, y_{0}) = \overline{e_{1}(x_{0}, y_{0})} = ¯1 = 1

We have that the x_{0}, y_{0} we found are solutions to the entire system of equations. So this system of equations has a solution so M(a, b) is invertible. Since M(a, b) ∈ H was arbitrary we may infer that H is a non-commutative ring in which every non-zero element is invertible.