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week08tutsols

# week08tutsols - THE UNIVERSITY OF SYDNEY Math2968...

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THE UNIVERSITY OF SYDNEY Math2968 Algebra (Advanced) Semester 2 Tutorial Solutions Week 8 2008 1. (for general discussion) Let R [ x ] denote the set of all polynomials in an indeterminate x with real coefficients, which has the usual addition and multiplication of polynomials. (a) Let φ : R [ x ] C be the ‘evaluation map’ at i = 1: φ : p ( x ) mapsto→ p ( i ) for all p ( x ) R [ x ] . Clearly φ preserves addition and multiplication. In particular, φ is a homomorphism between addi- tive abelian groups. Describe the kernel and image of φ . Interpret the Fundamental Homomorphism Theorem in this case. (b) Let ψ : R [ x ] Mat 2 ( R ) be the ‘evaluation map’ at M = bracketleftbigg 0 1 1 0 bracketrightbigg : ψ : p ( x ) mapsto→ p ( M ) for all p ( x ) R [ x ] . Again ψ is a homomorphism between additive abelian groups. Describe the kernel and image of ψ . Interpret the Fundamental Homomorphism Theorem in this case. Glue this and the previous case together to find a representation of complex numbers by 2 × 2 real matrices. Solution (a) If p ( x ) ( x 2 + 1) R [ x ] then p ( x ) = ( x 2 + 1) q ( x ) for some q ( x ) R [ x ], so that p ( i ) = ( i 2 + 1) q ( i ) = 0 . Thus ( x 2 + 1) R [ x ] ker φ . Suppose now p ( x ) ker φ , so that p ( i ) = 0. Hence x i is a factor of p ( x ). But the coefficients of p ( x ) are real, so its roots come in complex conjugate pairs. Thus x + i is also a factor, so that x 2 + 1 = ( x i )( x + i ) is a factor of p ( x ). Hence p ( x ) ( x 2 + 1) R [ x ]. This proves ker φ = ( x 2 + 1) R [ x ]. Clearly, if a,b R then a + bi = p ( i ) where p ( x ) = bx + a , so that φ is surjective. Hence im φ = C . By the Fundamental Homomorphism Theorem applied to these additive groups, C = im φ = R [ x ] / ker φ = R [ x ] / ( x 2 + 1) R [ x ] , which is just addition mod x 2 + 1 in disguise. (b) If p ( x ) ( x 2 + 1) R [ x ] then p ( x ) = ( x 2 + 1) q ( x ) for some q ( x ) R [ x ], so that p ( M ) = ( M 2 + I ) q ( M ) = ( I + I ) q ( M ) = 0 . Thus ( x 2 + 1) R [ x ] ker ψ . Suppose now p ( x ) ker ψ , so that p ( M ) = 0. Express p ( x ) = ( x 2 + 1) q ( x ) + r ( x ) for some quotient q ( x ) and remainder r ( x ) = bx + a . Then 0 = p ( M ) = ( M 2 + I ) q ( M ) + r ( M ) = 0 q ( M ) + bM + aI = bracketleftbigg a b b a bracketrightbigg , so that a = b = 0. Hence p ( x ) ( x 2 + 1) R [ x ]. This proves ker ψ = ( x 2 + 1) R [ x ]. To find the image, again use quotients and remainders to see that im ψ = { p ( M ) | p ( x ) R [ x ] } = { ( M 2 + I ) q ( M ) + r ( M ) | q ( x ) R [ x ] , r ( x ) a linear polynomial } = { ( I + I ) q ( M ) + r ( M ) | q ( x ) R [ x ] , r ( x ) = bx + a, a,b R } = { bM + aI | a,b R } = braceleftBig bracketleftbigg a b b a bracketrightbigg vextendsingle vextendsingle vextendsingle a,b R bracerightBig .

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week08tutsols - THE UNIVERSITY OF SYDNEY Math2968...

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