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 Abstract Algebra Abstract Algebra Math Forum

 April 25th, 2017, 01:01 PM #1 Newbie   Joined: Apr 2017 From: United Kingdom Posts: 1 Thanks: 0 Homework Problem Hello, I am a bit stuck on this question and can't start it.. Any help is very welcome, thanks in advance. Q: In the ring R=ℚ[x]/(x^5) find the inverse element for (1-2(x^2)+x^3). April 25th, 2017, 03:21 PM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Q[x]/x^5, if I understand your notation, correctly, can be interpreted as the field of all polynomials, with rational coefficients, of degree 5 of less with multiplication defined as regular polynomial multiplication except that we drop all powers of x larger than 4. To find the multiplicative inverse of 1+ 2x^2+ x^3, we look for ax^4+ bx^3+ cx^2+ dx+ e such that this multiplication gives 1. That is, we look at (x^3+ 2x^2+ 0x+1)(ax^4+ bx^3+ cx^2+ dx+ e)= ax^7+ (2a+ b)x^6+ (2b+ a)x^5+ (a+ 2c+ d)x^4+(c+2e)x^2+ dx+ e= (a+ 2c+ d)x^4+ (b+ 2d+ e)x^3+(c+2e)x^2+ dx+ e since we drop the x^7, x^6, and x^5 terms. We want that to be equal to 1 so we must have a+ 2c+ d= 0, b+ 2d+ e= 0, c+ 2e= 0, d= 0, and e= 1. Solve those five equations for a, b, c, d, and e. Thanks from topsquark Tags homework, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post TrimHonduras Trigonometry 1 July 21st, 2013 06:40 PM gasteven Algebra 2 February 14th, 2013 09:56 AM andi7 Algebra 1 October 24th, 2009 04:17 AM andi7 Algebra 6 March 13th, 2009 10:55 AM Jamers328 Number Theory 3 September 27th, 2007 08:00 AM

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