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Unformatted text preview: MAT 322: Algebra with Galois Theory Professor Claus Sorensen Problem Set 6 Rik Sengupta November 13, 2009 1. Let E be a finite field extension of F , and let R be a subring of E containing F . Show that R is necessarily a subfield of E . Solution. By the Isomorphism Theorems, we get ( E : F ) = ( E : R )( R : F ), and we know from the hypothesis that ( E : F ) is finite, and so it follows that ( R : F ) must be finite as well. So consider R to be a finite dimensional vector space over F . So choose some α ∈ R such that α 6 = 0. Define ϕ : R → R by ϕ ( x ) = αx . This map is injective since R is an integral domain (so, ϕ ( x ) = ϕ ( y ) = ⇒ αx = αy = ⇒ x = y since α is nonzero, and cancellation holds in an integral domain), and hence a bijection (because the dimensions match). So, it follows that ϕ is also surjective because R is finite dimensional over F . Therefore, from surjectivity, there exists some β ∈ R such that ϕ ( β ) = αβ = 1, and so α has an inverse in R . In particular, then, R is a field. 2. Compute the residue of 117 117 modulo 17 . Solution. We know, from Fermat’s little theorem, since gcd(17 , 117) = 1, and 17 is a prime, we must have 117 16 ≡ 1(mod 17) = ⇒ 117 112 ≡ 1(mod 17) . So now it is enough to consider 117 5 modulo 17. But that is easy once we notice that 117 ≡  2(mod 17), and so 117 5 ≡ ( 2) 5 (mod 17) ≡  32(mod 17) ≡ 2(mod 17) . 1 So, the required residue of 117 117 is 2 modulo 17. 3. Let I be the ideal of Q [ x ] generated by the three polynomials: x 3 + x 2 4 x 4 , x 3 x 2 4 x + 4 , x 3 2 x 2 x + 2 . Does the ideal I contain the polynomial x 2 4 ? Solution. Yes. Call the three polynomials f 1 , f 2 and f 3 . Note, first of all, that the unique monic generator of the ideal is simply the gcd of the polynomials. This can easily be seen from the definition of the gcd. It divides all the three polynomials, and any other divisor of all three polynomials also divides the gcd. Now, to see why the gcd g is the monic generator of the ideal, note that g  f i = ⇒ g  ∑ c i f i , and so every element in the ideal is divisible by g . Furthermore, we may use the division algorithm to show that g itself is in the ideal. And thus, it follows that g is in fact the monic generator of the ideal. We impose the monic condition to force uniqueness. We can easily factorize the given polynomials as follows: x 3 + x 2 4 x 4 = ( x + 1)( x + 2)( x 2) x 3 x 2 4 x + 4 = ( x 1)( x + 2)( x 2) x 3 2 x 2 x + 2 = ( x + 1)( x 1)( x 2) . So it is clear that the gcd of the three polynomials is, in fact, ( x 2). And since this is monic, it is the unique generator of the ideal I . But x 2 4 = ( x + 2)( x 2), and so the monic generator divides x 2 4. So it follows that it must be contained in the ideal....
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This note was uploaded on 05/07/2010 for the course MAT 322 at Princeton.
 '09
 CLAUSSORENSEN
 Algebra

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